Topic 5 - Physical Chemistry and Transition Elements

Question 1

Kc =

Answer 1

[products] / [reactants]

Question 2

[ ] =

Answer 2

equilibrium concentrations in moldm^-3

Question 3

If the equation is aA + bB cC + dD, what is the Kc equation?

Answer 3

Kc = ([C]c x [D]d) / ([A]a x [B}b)

Question 4

What is Kc?

Answer 4

Kc is a constant but is temperature dependent

Question 5

How does temp do to Kc on an exothermic forward reaction?

Answer 5

Temp increase so concentration of reactants increase and products decrease so Kc drops
Temp decrease so concentration of reactants decreases and products increases so Kc rises

Question 6

Size of Kc meaning?

Answer 6

Large Kc, equilibrium on the right

Small Kc, equilibrium on the left

Question 7

If an equilibrium mixture is homogeneous, what does this mean? and what happens to Kc?

Answer 7

If an equilibrium mixture is homogeneous (all the reactants and products are in the same physical state) all the species go into the Kc expression

Question 8

If an equilibrium mixture is heterogeneous, what does this mean? and what happens to Kc?

Answer 8

If an equilibrium mixture is heterogeneous (reactants and products are in different physical states) only gaseous and aqueous species go into the Kc expression

Question 9

When writing Kc expressions for heterogeneous equilibria we leave out any species which are solid or liquid as …

Answer 9

… Their concentrations are essentially constant

Question 10

For gaseous equilibria it is more convenient to use ______ ________ of reactants and products instead of equilibrium concentration in moldm-3

Answer 10

partial pressures

Question 11

if aA(g) + bB(g) cC(g) + dD(g), then Kp =

Answer 11

((pC)c x (pD)d) / ((pA)a x (pB)b)

Question 12

If N2(g) + 3H2)g) 2NH3(g), then Kp =

Answer 12

Kp = (pNH3)2 / ((pN2) x (pH2)3)

Question 13

In Kp, what is ‘p’?

Answer 13

P = partial pressure in Pa/kPa

Question 14

If Kp>1 there are …

Answer 14

… more of the products to reactants at equilibrium

Question 15

If Kp<1 there are …

Answer 15

… more of the reactants to products at equilibrium

Question 16

Partial pressure is the …

Answer 16

… individual pressures of the components in the mixture

Question 17

Kp is a constant at …

Answer 17

constant temperature

Question 18

How does temperature effect Kp in an exothermic reaction?

Answer 18

Suppose temperature increases, an increase in temperature favours the endothermic reaction, which in this case is the reverse reaction, so the equilibrium responds by moving to the left hand side, in other words the partial pressure of the products drop, this drops the value of Kp.
Suppose temperature decreases, a decrease in temperature favours the exothermic reaction, which in this case is the reverse reaction, so the equilibrium responds by moving to the right hand side, in other words the partial pressure of the reactants drops, this increases the value of Kp.

Question 19

How do we know partial pressure is proportional to the mole fraction?

Answer 19

A(g) + B(g)  C(g) at a fixed volume and constant temperature
pV = nRT

Question 20

Partial pressure =

Answer 20

mol fraction x total pressure

Question 21

Mole fraction =

Answer 21

mol / total moles

Question 22

What is Kp?

Answer 22

the equilibrium constant in terms of partial pressure

Question 23

Why is equilibria involving gases are usually expressed in terms of Kp?

Answer 23

This is because it is easier to measure the pressure of gas rather than the concentration

Question 24

The mole fraction of a gas is the same as its …

Answer 24

… proportion by volume to the total volume of gases in a gas mixture

Question 25

For gas A in a mixture:
	Mole fraction (A) =

Answer 25

moles (A) / total number of moles in gas mixture

Question 26

Mole fraction of gases in the air?

Answer 26

Air has approx. 78% N2 , 21% O2 and 1% other gases
Mole fraction of N2 x(N2) = 78/100 = 0.78
Mole fraction of O2 x(O2) = 21/100 = 0.21
Mole fraction of other x(other) = 1/100 = 0.01

Question 27

Sum of mole fractions always =

Answer 27

1

Question 28

Calculating Mole Fraction from Gas Volumes?

Answer 28

Volume = moles x 24

This shows that volume is proportional to the moles of gas

Question 29

Sum of partial pressures =

Answer 29

total pressure

Question 30

How are the units for Kp found?

Answer 30

in the same way as Kc

Question 31

Nitrogen, hydrogen and ammonia coexist in a homogeneous equilibrium ;
N2 (g) + 3 H2 (g)  2 NH3 (g)
An equilibrium mixture at 400 oC contains 18 mol N2, 54 mol H2 and 48 mol of NH3. The total equilibrium pressure is 200 atms. Use this information to calculate the value of Kp.

Answer 31

Step 1 : find the mole fraction of N2, H2 and NH3.
Total number of gas moles = 18+54+48 = 120 mol
x(N2) = 18/120 = 0.15 x(H2) = 54/120 = 0.45 x(NH3) = 48/120 = 0.40
Step 2 : find the partial pressures of each gas.
p(N2) = 0.15 x 200 = 30 atms p(H2) = 0.45 x 200 = 90 atms p(NH3) = 0.40 x 200 = 80 atms
Step 3 : Calc Kp Kp = p(NH3)2 = 802 = 2.9 x 10-4 atms-2
p(N2) x p(H2)3 30 x 903
Units = (atm)2 = 1 = atm-2
(atm) x (atm)3 atm2

Question 32

Describe and explain how you find kP in heterogeneous equilbria

Answer 32

Equilibrium contains different phases (solid and gas)
Equilibrium: CaCO3(s) ⇌ CaO(s) + CO2(g)
Kp expression contains only gaseous species
Kp = p(CO2)
Solid species are omitted
(solids have no gas pressure)
Therefore, if a question arose that said:
In the equilibrium mixture, CO2 has a partial pressure of 2.5 x 10-2 atm at 600oC. Calculate Kp.
Kp = p(CO2) = 2.5 x 10-2 atm.

Question 33

K is a constant and only changed by

Answer 33

A change in temperature

Question 34

Equilibrium is achieved within a system when

Answer 34

the concentrations/partial pressures of the species at equilibrium give the value of K for that temperature when placed in the equilibrium constant expression

Question 35

For forward exothermic reactions what happens when temp increases in terms of Kp?

Answer 35

As temperature increases, Kp values decrease, showing that the equilibrium position shifts to the left

Question 36

For forward endothermic reactions what happens when temp increases in terms of Kp?

Answer 36

As temperature increases, Kp values increase, showing that the equilibrium position shifts to the right

Question 37

Why is there no change in K for changes in concentration or pressure?

Answer 37

as the equilibrium moves to keep the equilibrium constant at the same value

Question 38

Explain how a change in pressure effects Kp?

Answer 38

For a change in pressure, if the pressure is increased, when looking at K equation, the part of the fraction that will be most affected by a change in the system will be the part with the highest powers. This increase in pressure will increase the value of Kp and mean that the system is no longer in equilibrium, to restore Kp to its value for that temperature, the partial pressure of the substance with less powers need to increase so the equilibrium shifts in that direction.

Question 39

What is a reaction rate?

Answer 39

Chemists define the rate of reaction as the rate at which a reactant(s) is used up or the rate at which a product formed

Question 40

when we look at concentration and rates of reaction it is important to state which substance we are following, why?

Answer 40

.For example if we consider the reaction A + 2B C
.The concentration of reactants A and B decrease with time and the concentration of product C increases with time
.The reaction tells us that for every A that reacts two Bs are required
.So the concentration of B decreases twice as fast as the concentration of A

Question 41

.We talk about the amount of reactants or products in terms of their what?

Answer 41

Concentration

Question 42

.Concentration is measured in what?

Answer 42

mol dm-3

Question 43

.We use the symbol [ ] to write about what?

Answer 43

Concentration

Question 44

if we were looking at the concentration of A we would write what?

Answer 44

[A]

Question 45

.Chemical equations tell us what?

Answer 45

what is involved in the reaction

Question 46

the stoichiometric values tell us what?

Answer 46

the quantitative relationship between the different reactants and products

Question 47

chemical equations gives us no information about what?

Answer 47

whether the reaction will actually happen or how fast it will occur

Question 48

.The rate of a chemical process can depend on several factors, what are they?

Answer 48

–	temperature of the reaction
–	concentration of the reactants
–	the surface area
•	only relevant with solid reactants
–	the presence of a catalyst

Question 49

.The chemical kinetics of a reaction allow chemists to what?

Answer 49

control reactions by altering the conditions

Question 50

.The reaction rate is what?

Answer 50

the change in concentration of a reactant (or product) per unit time

Question 51

The reaction rate describes what?

Answer 51

– describes how quickly reactants are used up

– or how quickly products are formed

Question 52

.We can look at the __________ for an entire reaction or look at the rate a specific point during the reaction – __________________

Answer 52

Average rate

the instantaneous rate

Question 53

.The average rate is what?

Answer 53

the total change in concentration of reactants (or products) divided by the time taken for the reaction to take place

Question 54

.The instantaneous rate of a reaction is what?

Answer 54

the rate of change occurring at a specific time in the reaction

Question 55

.One method to calculate the rate of change at a given point is to what?

Answer 55

find the gradient of the tangent at the point you are interested in

Question 56

The units for a rate of reaction is

Answer 56

mol dm-3s-1

Question 57

Rate of reaction =

Answer 57

change in concentration / time

Question 58

.In the reaction: A → B, we can calculate either …

Answer 58

… the average rate of A being used or the average rate of B being produced

Question 59

.Rate is always a positive value so a minus sign can be added to [A] as the change in concentration is …

Answer 59

… negative

Question 60

.The instantaneous rate of reaction is …

Answer 60

… the amount of change in concentration of a product or reactant at a specific time during a reaction

Question 61

.A graph can be used and the rate determined by …

Answer 61

… drawing a tangent to the graph at any time and finding the slope of that tangent

Question 62

.Calculate the gradient by

Answer 62

dividing the change in y by the change in x

Question 63

.Changing the concentration often changes the what?

Answer 63

rate of a reaction

Question 64

.The rate of reaction is proportional to the …

Answer 64

… concentration of a particular reactant raised to a power

Question 65

.For example, for reactant [A] and power n, rate is proportional to …

Answer 65

… [A]^n

Question 66

.For each reactant, the power is (in reference to order of reaction)

Answer 66

the order of reaction for that reactant

Question 67

.In a reaction, different reactants can have different orders and each may affect the rate in different ways
.Common orders are:

Answer 67

zero order (0), first order (1), second order (2)

Question 68

Tell me about the zero order

Answer 68

.When the concentration of a reactant has no effect on the rate, the reaction is zero order with respect to the reactant
.With zero order, rate is proportional to [A]0
.In a zero order reaction:
- Any number raised to the power zero is 1
- Concentration does not influence the rate

Question 69

Tell me about the first order

Answer 69

.A reaction is first order with respect to a reactant when the rate depends on its concentration raised to the power of one
.With first order, rate is proportional to [A]1
.In a first order reaction:
- If the concentration of A is doubled, the reaction rate increases by a factor of 21 = 2
- If the concentration of A is tripled, the reaction rate increases by a factor of 31 = 3

Question 70

Tell me about the second order

Answer 70

.A reaction is second order with respect to a reactant when the rate depends on its concentration raised to the power of two
.With second order, rate is proportional to [A]2
.In a second order reaction:
- If the concentration of A is doubled, the reaction rate increases by a factor of 22 = 4
- If the concentration of A is tripled, the reaction rate increases by a factor of 32 = 9

Question 71

.The rate equation gives the what?

Answer 71

mathematical relationship between the concentrations of the reactants and the reaction rate

Question 72

.For two reactants, A and B, the rate equation is:

Answer 72

• Rate = k [A]m [B]n

Question 73

.The rate constant k is the what?

Answer 73

proportionality constant, it is the number that mathematically converts between the rate of reaction and concentration and orders

Question 74

Outline the principles in Le Chatelier’s principle

Answer 74

• Conc increase of a species, equilibrium moves in direction to reduce that concentration
• Pressure increase favors side with the least number of molecules
• Temperature increase favors the endothermic direction

Question 75

• At constant temperature, K does not change if you change …

Answer 75

… pressure, concentration or use a catalyst

Question 76

• K will only change if …

Answer 76

… the temperature is changed

Question 77

For an exothermic reaction, Kp decreases with increasing temperature and the equilibrium position shifts to the left, why?

Answer 77

an increase in temperature favors the endothermic backward reaction

Question 78

Essentially, why does temperature cause the equilibrium to shift?

Answer 78

A change in the K value, which is caused by a change in temperature, has to be met by a change in the concentrations or partial pressures of the reactants and products - which means the equilibrium is shifting

Question 79

Effect of Increasing Temperature on Equilibrium

Exothermic

Answer 79

K decreases

Equilibrium shifts to the left

Question 80

Effect of Increasing Temperature on Equilibrium

Exothermic Endothermic

Answer 80

K increases

Equilibrium shifts to the right

Question 81

Explain Why K Doesn’t Change with Concentration using :
.N2O4 (g)  2 NO2 (g)
.At constant temperature, [NO2] = 0.4 moldm-3 [N2O4] = 0.01 moldm-3
as an example

Answer 81

Kc = [NO2 (g)]2 = 0.4002 = 16.0 mol dm-3
[N2O4 (g)] 0.010
If [N2O4] increased to 0.02 then Kc = 0.4002 = 8.0 mol dm-3
0.02
The system is now no longer in equilibrium.
.To return the value of Kc ratio back to 16.0 mol dm-3 then [NO2] must increase & [N2O4] must decrease (top bigger bottom smaller)
.N2O4 (g)  2 NO2 (g) the equilibrium shifts to the right
.Using Le Chatelier’s principle, we would expect the equilibrium to shift to the right so that the equilibrium can decrease the [N2O4] if extra is added.
.Le Chatelier’s principle only works because Kc controls the relative concentration of reactants and products present at equilibrium to maintain a constant value.

Question 82

Explain Why K Doesn’t Change with Pressure by using :
.N2O4 (g)  2 NO2 (g)
.Le Chatelier : increase in pressure favours shift to the left due to lower number of molecules
.At constant temp; p(NO2) = 9.6 atms p(N2O4) = 0.24 atms
as an example

Answer 82

.Kp = p(NO2)2 = 9.62 = 384 atms
p(N2O4) 0.24
.If total pressure doubled p(NO2) doubled to 19.2 atms and p(N2O4) = 0.48 then Kp = 19.22 / 0.48 = 768 atms
.To reduce Kp back to 384, p(NO2) must decrease and p(N2O4) must increase (top smaller / bottom bigger), so equilibrium shifts from right to left
.The shift to the left (fewer gaseous molecules) is directed by the value of Kp being restored

Question 83

Increasing Pressure on Ammonia Production, explain what happens with reference to K values?

Answer 83

.N2 + 3 H2  2 NH3
.Increase pressure favors shift from left to right
. Kc = [NH3]2 If pressure increased at constant temperature, then Kc must remain constant
[N2] x [H2]3
.As pressure increased then the concentration of N2, H2 and NH3 will increase (same moles in smaller volume)
.As denominator (bottom) power x4 / numerator (top) x2 then value of Kc will decrease as [ ]’s increase
.To restore Kc back to original value i.e., maintain constant Kc then equilibrium will shift to increase top [NH3] and reduce bottom [N2][H2] – shift from left to right

Question 84

Effect of catalysts on K values

Answer 84

.Equilibrium constants are unaffected by catalysts

.They affect the rate of a reaction, not equilibrium position (speed up both forward and back reaction by same factor)

Question 85

If a species is zero order:

Answer 85

• It means that changing the concentration of a substance has no effect on the rate
• If the concentration doubles, the rate stays the same
• If the concentration triples, the rate stays the same

Question 86

If a species is first order:

Answer 86

• It means that changing the concentration of a substance changes the rate to the power 1
• If the concentration doubles, the rate doubles (21 = 2)
• If the concentration triples, the rate triples (31 = 3)

Question 87

If a species is second order:

Answer 87

• It means that changing the concentration of a substance changes the rate to the power 2
• If the concentration doubles (x2), the rate quadruples (x4) (22 = 4)
• If the concentration triples (x3), the rate increases by 9 (32 = 9)

Question 88

A + 2B + C  D + E
[A] is zero order, [B] is 1st order, [C] is 2nd order
Rate =
Overall rate =

Answer 88

K [B] [C]2

1 + 2 = 3

Question 89

What does ICE stand for?

Answer 89

Initial
Change
Equilibrium

Question 90

2NO2 2NO + O2
0.56 moles of NO2 are left to decompose in a vessel with a volume of 0.5dm3. 0.48 moles of NO are formed. What is the value of Kc?

Answer 90

17.28M

Question 91

How to work out K and the units for K

Answer 91

Rearrange the rate equation and input values from the table to work out the value of K
Write the units of each part of the equation for K and work out the units by cancelling etc.

Question 92

What is Kc?

Answer 92

a mathematical representation of the ration of products:reactants

Question 93

What do homogeneous and heterogenous mean?

Answer 93

Homogeneous – same state

Heterogeneous – different states

Question 94

The concentrations of pure solid and pure liquids are excluded from equilibrium expressions, why?

Answer 94

as their concentrations are constant

Question 95

What is the unit of rate constant for an overall order of reaction of 0

Answer 95

Mol dm-3 s-1

Question 96

What is the unit of rate constant for an overall order of reaction of 1

Answer 96

s-1

Question 97

What is the unit of rate constant for an overall order of reaction of 2

Answer 97

Mol-1 dm3 s-1

Question 98

What is the unit of rate constant for an overall order of reaction of 3

Answer 98

Mol-2 dm6 s-1

Question 99

What is the unit of rate constant for an overall order of reaction of 4

Answer 99

Mol-3 dm9 s-1

Question 100

.For gaseous equilibria, it is more convenient to use ________ of reactants and products instead of equilibrium concentrations in mol dm-3

Answer 100

partial pressures

Question 101

Generic Kp formula –

aA + bB cC + dD

Answer 101

Kp = ((pC)c x (pD)d) / ((pA)a x (pB)b)

Question 102

Haber process equation and Kp formula

Answer 102

N2 + 3H2 2NH3

Kp = (pNH3)2 / ((pN2) x (pH2)3)

Question 103

What is p in Kp?

Answer 103

p = partial pressure in Pa/kPa

Question 104

If Kp > 1 there are

Answer 104

more products than reactants

Question 105

If Kp < 1 there are

Answer 105

more reactants than products

Question 106

What is partial pressure

Answer 106

The partial pressure is the pressure of an independent component of the equilibria

Question 107

Like Kc, Kp is constant at what?

Answer 107

a constant temperature

Question 108

In the haber process, Suppose the pN2 is doubled, how does the equilibrium respond?

Answer 108

The denominator will be bigger, so Kp will get smaller, so to get bigger again it increases the numerator, so pNH3 increases, so the equilibrium moves to the right

Question 109

Effect of temperature –
Delta H = -92 kJmol-1
what would happen in a temperature increase?

Answer 109

Suppose a temperature increase, it would favour the endothermic reaction, so the equilibrium would shift to the left, so Kp would get smaller

Question 110

Effect of temperature –
Delta H = -92 kJmol-1
what would happen in a temperature decrease?

Answer 110

Support a temperature decrease, it would favour the exothermic reaction, so the equilibrium would shift to the right, so Kp would get larger

Question 111

A + B C
Fixed volume at a constant temperature
Why is partial pressure proportional to the number of moles?

Answer 111

pV = nRT

Question 112

Partial pressure is proportional to the what?

Answer 112

mole fraction

Question 113

The sum of the mole fractions equals what?

Answer 113

1

Question 114

Partial pressure =

Answer 114

mol fraction x total pressure

Question 115

Kp is what?

Answer 115

The equilibrium constant in terms of partial pressures

Question 116

Equilibria involving gases are usually expressed in terms of what?

Answer 116

Kp

Question 117

.It is easier to measure ____ of gas rather than concentration

Answer 117

pressure

Question 118

.Concentration & pressure are __________ to each other

Answer 118

directly proportional

Question 119

.Under the same conditions, the same volume of different gases will contain the same number of __________

Answer 119

moles of gas molecules

Question 120

.The mole fraction of a gas is the same as its proportion by volume what?

Answer 120

to the total volume of gases in a gas mixture

Question 121

mole fraction x(A) =

Answer 121

number of moles of A / total number of moles in gas mixture (N)

Question 122

Partial pressure, p, is what?

Answer 122

the contribution of a gas towards the total pressure

Question 123

Why do mol fractions have no units?

Answer 123

1. Mol fractions are calculated with moles on the top and bottom of the equation, so the units cancel out

Question 124

Partial pressure, p, is the what?

Answer 124

contribution of a gas towards the total pressure

Question 125

Partial pressure (p) =

Answer 125

mole fraction x Total pressure (P)

Question 126

To work out the order of the reaction for the concentration, you have to use experimental data, to get this you have two options, what are they?

Answer 126

1. Continuously monitor the change in concentration of A against time
2. Use initial rates method to find out how the initial rate changes as you vary the concentration of A

Question 127

Concentration-time graphs can be plotted by ______________________________________________________________________, it is called continuous monitoring

Answer 127

continuously monitoring the concertation of a reactant through the course of the reaction

Question 128

How can you gain the concentration of a reactant through the course of the reaction?

Answer 128

Can use loss in mass, gas syringe, or measuring cylinder
Also can use pH changes or colorimetry
Can’t use every method for every experiment

Question 129

What can the shape of concentration time graphs tell us?

Answer 129

The shape can be used to tell you the order for that particular reactant

Question 130

For concentration time graphs, a zero order reaction produces ….

Answer 130

a straight line with a negative gradient, the reaction rate does not change at all through the course of the reaction

Question 131

For concentration time graphs, a first order reaction produces …

Answer 131

a downward curve with decreasing gradient over time, as the gradient decreases with time, the reaction gradually slows down

Question 132

For concentration time graphs, a second order reaction has a …

Answer 132

downward curve but steeper at the start and tailing off more slowly

Question 133

For concentration-time graphs, you can find the rate using: gradient =

Answer 133

delta (y) / delta (x)

Question 134

How to find gradient of concetration-time graphs

Answer 134

.For zero orders, draw a triangle to the start and end

.For first order, draw a triangle touching the curve at one point (a tangent)

Question 135

Apart from rate, what else can you find from conc-time graphs?

Answer 135

K

Question 136

How can you tell between first and second order conc-time graphs?

Answer 136

.We can tell if we have a first order curve by measuring half life
Half life (t1/2) is the time taken for half of the reactant to be used up
First order reactions have a constant half-life independent of the initial concentration
This pattern is called exponential decay

Question 137

what does ln(2) / t1/2 equal?

Answer 137

k

Question 138

Equilibria involving gases are usually expressed in terms of what?

Answer 138

Kp – constant in terms of partial pressure

Question 139

Easier to measure pressure of ___ rather than concentration

Answer 139

gas

Question 140

_________ and pressure are directly proportional to each other

Answer 140

concentration

Question 141

Mole fraction x(A) =

Answer 141

number of moles of A / total number of moles in gas mixture (N)

Question 142

Volume = mole x __

Answer 142

24

Question 143

Partial pressure is what?

Answer 143

the contribution of a gas towards the total pressure

Question 144

Partial pressure (p) =

Answer 144

mole fraction x total pressure (P)

Question 145

Kp is similar to Kc but …

Answer 145

… partial pressures are used instead of concentration

Question 146

Briefly explain le Chatelier

Answer 146

.Conc increase of a species equilibrium moves in direction to reduce that conc
.Pressure increase favours side with the least number of molecules
.Temperature increase favours the endothermic reaction

Question 147

.Le Chatelier works because of what?

Answer 147

equilibrium constants

Question 148

k=1, k<1, k>1 meaning

Answer 148

.k=1 equilibrium half way
.k=100 equilibrium lies will over to products (to the right)
.K=0.01 equilibrium lies well over to the reactants (to the left)

Question 149

.At constant temperature, K does no change if

Answer 149

you change pressure, conc, or use a catalyst

Question 150

.K will only change if

Answer 150

the temperature is changed

Question 151

How changes in conc and pressure affect equilibrium constants

Answer 151

.The value of the equilibrium constant K is unaffected by changes on concentration and pressure.
.This may seem strange as you know from Le Chatelier’s principle that the position of the equilibrium can be shifted by changing concentration or pressure.
.The equilibrium shift actually takes place from the very fact that the equilibrium constant does not change

Question 152

Increasing pressure on ammonia production exam answer

Answer 152

.Increase pressure favors shift from left to right. If pressure increased at constant temp, then Kc must remain constant. As pressure increased then the concentration of N2, H2 and NH3 will increase (same moles in smaller volume). As denominator (bottom) power x4 / numerator (top) x2 then value of Kc will decrease as [ ]’s increase. To restore Kc back to original value i.e., maintain constant Kc then equilibrium will shift to increase top [NH3] and reduce bottom [N2][H2] – shift from left to right.

Question 153

.Equilibrium constants are _________ by catalysts. …

Answer 153

unaffected
…. they affect the rate of a reaction, not equilibrium position (speed up both forward and back reaction by same factor).

Question 154

Define acid

Answer 154

dissociates in water and releases H+ ions

Question 155

Define alkali

Answer 155

dissociates in water to release OH- ions

Question 156

Neutralisation equation

Answer 156

H+ + OH-1  H2O

Question 157

Bronsted Lowry Acid:

Answer 157

Proton donor

Question 158

Bronsted Lowry Base:

Answer 158

Proton acceptor

Question 159

pH = 1 (x 10 the concentration of H+ ions) pH =

Answer 159

2

Question 160

What makes an acid strong?

Answer 160

It fully dissociates

Question 161

As a Strong acid fully dissociates, pH of a strong acid can be calculated from

Answer 161

concentration of acid

Question 162

pH =

Answer 162

-log10[H+]

Question 163

Sulphuric acid is dibasic, what does this mean?

Answer 163

H2SO4  2H+ + SO42-

Question 164

[H+] =

Answer 164

10-pH

Question 165

Weak acids form an acid dissociation constant called

Answer 165

Ka

Question 166

Ka =

Answer 166

[products] / [reactants]

Question 167

[H+] =

for weak acids

Answer 167

root([HA] Ka)

Question 168

[HA] is what?

Answer 168

concentration of weak acid

Question 169

A+B C

What happens when you Increase pressure?

Answer 169

equilibria moves to the side with fewer molecule/moles, 2 moles on RHS and 4 on LHS, equilibria moves to the right

Question 170

A+B C

What happens when you Lowering Temperature?

Answer 170

Forward reaction is exothermic (negative energy change), lowering temperature moves equilibria to exothermic reaction (RHS)

Question 171

A+B C

What happens when you Increase A and B?

Answer 171

Equilibria removes addition of A and B, equilibria moves to RHS to remove this extra concentration of A and B

Question 172

A+B C

What happens when you Decrease C?

Answer 172

Equilibria moves to make more of what is removed, equilibria moves to RHS to make more C

Question 173

Meaning of Kc>1?

Answer 173

means more products, equilibria is towards the RHS

Question 174

Meaning of Kc<1?

Answer 174

means more reactants, equilibria towards the LHS

Question 175

Meaning of Kc=1?

Answer 175

means LHS=RHS

Question 176

.When a reversible reaction establishes an equilibria at a certain temperature, K will not change if, ________, _______,
____________ are changed/added

Answer 176

concentration, pressure, or a catalyst

Question 177

.K is only changed by ________

Answer 177

temperature

Question 178

In the equation 2SO2 + O2 2SO3
delta H = -197
As temperature increases Kp decreases, why does this happen?

Answer 178

Kp = p(SO3)2 / ( p(SO2)2 x p(O2) )
Partial pressure of products decrease and reactants increase
Equilibria moves to the left

Question 179

Using N2O4 2NO2 as an example, explain why Kc and Kp do not change due to concentration, pressure or addition of a catalyst?

Answer 179

.If N2O4 concentration was increased at same temperature, Kc value would be lowered (denominator increased), equilibria moves to the RHS and Kc returns to normal (numerator increase, denominator decrease)
.If we increased partial pressure then the numerator would be increased more than the denominator due to the difference in powers, this would cause Kp value to increase (system not at equilibria), equilibria moves left to increase value of the denominator (N2O4) and restore the value of Kp, system would be back in equilibrium
.Catalyst does not change equilibria (Kc/Kp = same) as both forward and reverse reaction rates are changed the same, this means position of equilibria does not move

Question 180

What does the Arrhenius theory define bases and acids as?

Answer 180

The Arrhenius theory of acids and bases defines acids as a substance that releases protons in solution, and bases as a substance that releases hydroxide ions.

Question 181

How can we test for acids?

Answer 181

We can test for acids using universal indicator paper which turns red or orange depending on the strength of the acid
Blue litmus paper also turns red in the presence of an acid

Question 182

HCl + H2O –>

Answer 182

H3O+ + Cl-

Question 183

What is H3O+

Answer 183

A hydronium ion, also known as an oxonium ion

Question 184

What is water acting as in the equation:

HCl + H2O  H3O+ + Cl-

Answer 184

a base

Question 185

What is a bronsted-lowry base?

Answer 185

A Bronsted Lowry base is a substance which accepts protons in solution

Question 186

What is a bronsted-lowry acid?

Answer 186

A Bronsted Lowry acid is a substance which releases or donates protons in solution

Question 187

What does a bronsted-lowry acid-base reaction involve?

Answer 187

the transfer of a proton from one base to another

Question 188

What does ammonia do in water? include equation

Answer 188

Ammonia, in water, accepts a proton

NH3 + H2O  NH4+ + OH-

Question 189

In the equation,

NH3 + H2O  NH4+ + OH-, what is water acting as?

Answer 189

An acid

Question 190

For a bronsted-lowry base, what is the pH of an acid?

Answer 190

So for the Bronsted Lowry theory, a base does not need to have a pH>7

Question 191

What are substances that can act as either a base or an acid called?

Answer 191

Substances that can act as either an acid or a base are amphoteric

Question 192

What are some old defnitions of acids?

Answer 192

.The definition of acids has developed since the time of the ancient Greeks
.Simpler ideas involved substances that had a sour taste, contained hydrogen, hydrogen ions or had a pH lower than 7

Question 193

.When a chemical reacts with an acid it is the __________ from the acid which is driving the reaction forward

Answer 193

hydrogen ion, H+,

Question 194

HCl (aq) + NaOH (aq) 

Answer 194

NaCl (aq) + H2O (l)

Question 195

How did the Bronsted-Lowry name come about?

Answer 195

Johannes Nicolaus Bronsted and Martin Thomas Lowry did not work together but both chemists formulated the idea that acids are proton donors and bases are protons acceptors

Question 196

The Brønsted–Lowry acid–base theory is a development of these earlier ideas and states that:

Answer 196

.An acid is a proton donor

.A base is a proton acceptor

Question 197

.How many protons, neutrons and electrons does a Hydrogen ion, H+ have?

Answer 197

.A hydrogen ion is just a proton

Question 198

We know that HCl forms a covalent bond between the hydrogen and chlorine so what happens when it dissolves in water to become hydrochloric acid?

Answer 198

.HCl(g) + aq  H+(aq) + Cl-(aq)

Question 199

HCl(g) + aq  H+(aq) + Cl-(aq)

.On closer inspection is?

Answer 199

HCl(g) + H2O(l)  H3O+(aq) + Cl-(aq)

Question 200

.H3O+(aq) is known as the hydronium ion, also known as _______ ion

Answer 200

hydroxonium

Question 201

.In this equation:
HCl(g) + H2O(l)  H3O+(aq) + Cl-(aq)

what is HCl and H2O acting as?

Answer 201

The HCl is a proton donor (a Brønsted–Lowry acid)

.The H2O is a proton acceptor (a B-L base)

Question 202

In the reverse equation of HCl(g) + H2O(l)  H3O+(aq) + Cl-(aq) what happens to the roles?

Answer 202

.In the reverse equation, the roles reverse:

H3O+(aq) + Cl-(aq)  HCl(g) + H2O(l)

Question 203

once an acid has ‘donated’ a proton it would become able to ‘accept’ a proton back and hence act as a base, what is this called?

Answer 203

We call these pairs of chemicals conjugate acid-base pairs

Question 204

Give an example of a conjugate acid-base pair?

Answer 204

HCl and Cl- = Acid and conjugate base

Question 205

What are the acid base pairs in HCl + H2O H3O+ + Cl-

Answer 205

HCl = acid 1
Cl- = base 1

H2O = base 2
H3O+ = acid 2

Question 206

What are the acid base pairs in CH3COOH + H2O CH3COO- + H3O+?

Answer 206

CH3COOH = acid 1
CH3COO- = base 1

H2O = base 2
H3O+ = acid 2

Question 207

What are the acid base pairs in NH3 + H2O NH4+ + OH-?

Answer 207

Base 2 Acid 1 Acid2 Base 1

Question 208

What are the acid base pairs in HCO3- + HCl H2CO3 + Cl-?

Answer 208

Base 2 Acid 1 Acid 2 Base 1

Question 209

Complete the equation for the conjugate acid-base pair: CH3CH(OH)COOH + CH3CH2CH2COOH

Answer 209

CH3CH(OH)COO- + CH3CH2CH2COOH2+

Question 210

List the roles of H+ in reactions

Answer 210

spectator ions:
Acid + metal  salt + hydrogen

Solid carbonates and soluble carbonates:
Acid + carbonate  Water + carbon dioxide

base (metal oxides): Acid + base  salt + water

Alkali: H+ + OH-  H2O

Question 211

What did Soren Sorenson do?

Answer 211

introduced simple numbers to represent the colours of indicators using an electrochemical cell to measure the hydrogen content

Question 212

What did Soren Sorenson find?

Answer 212

.He found the ion concentration had a very large range of values of powers of 10 (10-1 to 10-14)

Question 213

What is a strong acid? With general equation

Answer 213

.A strong acid is one which completely dissociates into ions in a solution
HA  H+ + A-

Question 214

So for a strong acid, [H+] =

Answer 214

concentration of the acid ([HA])

Question 215

pH =

Answer 215

-log10[H+]

Question 216

What does the equation pH = -log10[H+] tell us?

Answer 216

.It tells us the relative hydrogen ion concentration of a given solution

Question 217

What must we remember with the equation pH = -log10[H+]?

Answer 217

.THIS ONY WORKS FOR STRONG ACIDS

Question 218

.An increase in H+ x10, it reduces the pH by how much?

Answer 218

1

Question 219

What is pH of 1M HCl?

Answer 219

0

Question 220

What is Sorenson’s pH scale?

Answer 220

.The logarithmic scale means that a shift of one pH unit means a 10x change in the acidity and alkalinity of the solution
.Theoretically there is no limit to the pH scale

Question 221

What type of acid is HCl, what does this mean?

Answer 221

monobasic = [H+]

Question 222

What type of acid is H2SO4, what does this mean?

Answer 222

dibasic = 2[H+]

Question 223

What type of acid is H3PO4, what does this mean?

Answer 223

tribasic = 3[H+]

Question 224

[H+] =

Answer 224

10-pH

Question 225

What is the [HCl] of a solution with pH 1.8?

Answer 225

0.016 moldm-3

Question 226

What is the [H2SO4] of a solution with pH 1.8?

Answer 226

0.008 moldm-3 as dibasic acid

Question 227

50cm^3 of 0.1M HCl is diluted to 100cm^3 with water, what is the change in pH?

Answer 227

Before dilution, pH = 1
0n dilution, HCl conc is halved to 0.05M
After dilution, pH = 1.30
Change = 0.3

Question 228

rate-concentration graphs can be plotted using two different methods, what are they?

Answer 228

1. Continuously monitor the change in concentration of a substance over time and work out the gradient at set point on the graph
2. Use initial rates method to find out how the initial rate changes as you vary the concentration of A

Question 229

We can produce a rate-concentration graph by producing a concentration time graph for the experience at different concentrations and then what?

Answer 229

using a tangent to work out the initial rate (the rate at t=0)

Question 230

We can also use an initial rate reaction called a what?

Answer 230

clock reaction

Question 231

The shape of rate-concentration graph can tell us what?

Answer 231

the order of the reaction

Question 232

Describe a zero order rate-concentration graph

Answer 232

.Produces a horizontal line with zero gradient
.This shows the rate doesn’t change
.Rate = k
.To work out the rate constant, it is the intercept on the y axis

Question 233

Describe a first order rate-concentration graph

Answer 233

.Produces a straight line through the origin
.This shows the rate is directly proportional to the concentration
.Rate = k[A]
.To work out the rate constant you should work out the gradient of the straight line

Question 234

Describe a second order rate-concentration graph

Answer 234

.Produces an upward curve with increasing gradient
.Rate = k[A]2
.This curve doesn’t actually confirm its second order, just that it’s not 0 or 1
.You have to plot another graph from this one to show its 2nd order
.This means you can’t work out the rate constant directly from this curve

Question 235

How to work our rate constant from second-order rate-concentration graph?

Answer 235

.To work out the rate constant you would need to plot a second graph of rate against concentration squared
.If I is truly second order it should produce a straight line
.To work out the rate constant you would work out the gradient of the line (like first order)

Question 236

What is a clock reaction?

Answer 236

.A clock reaction is a more convenient way of obtaining initial rate
.We use a reaction with a visual change – usually a colour change
.We time from the start of the experiment to the visual change
.We repeat the experiment each time with different concentration meaning the time for the colour change each time

Question 237

For a clock reaction, what is assumed?

Answer 237

.We assume that the average rate of the reaction is the same as the initial rate

Question 238

For a clock reaction, the initial rate is proportional to what?

Answer 238

.The initial rate is proportional to 1/t

Question 239

What is an iodine clock?

Answer 239

.This is a clock reaction which relies on the formation of iodine
.Aqueous iodine is orange brown so we can measure the colour change
.We usually use an indicator (starch) as that is a intense black-blue colour with iodine so it is very easy to see the colour change

Question 240

How can you work out the order of hydrogen perioxide in a reaction using an iodine clock?

Answer 240

.Hydrogen Peroxide (H2O2) reacts with iodide ions in an acid solution to form iodine
H2O2 + 2I- + 2H+  I2 + 2H2O
.Lets say we start with 0.05M of H2O2
.We would add some acid (to provide the H+ ions) and some KI to provide the I ions
.We would also add some starch (for the indicator) and some sodium thiosulfate, Na2S2O3, to provide S2O32- ions
Reaction 1:
H2O2 + 2I- + 2H+  I2 + 2H2O
Reaction 2:
2S2O3 2- + I2  S4O6 2- + 2I-
.We time how long it takes to see a colour change (the blue-black to start to appear), this occurs when all of the sodium thiosulfate has been used up
.We would then repeat this for the next concentration (e.g. 0.04 mol dm-3 and so on)
You would end up with a table of results with concentration of H2O2 and time
.You can then use the principle that initial rate = 1/t
.So we work this out and add to the table
You can then plot a graph of [H2O2] against 1/t, and from that work out its order

Question 241

How accurate is the iodine clock?

Answer 241

.In the clock reaction we are measuring the average rate over a period of time, the shorted the amount of time, the less the rate will change over that time
.The clock reaction is classed as accurate as long as less than 15% of the reaction has taken place (in terms of time)

Question 242

What is a reaction mechanism?

Answer 242

A reaction mechanism describes the one or more steps involved in a reaction in a way which makes it clear how certain bonds are broken and made

Question 243

What is the rate determining step?

Answer 243

The slowest step is called the rate determining step

Question 244

What is The overall rate of reaction (the one you could measure if you did some experiments) controlled by?

Answer 244

the rate of the slowest step

Question 245

What can the rate equation tell you in terms of rate determining steps?

Answer 245

.If the species is in the rate equation it is taking part in the rate determining step
.If the species is not in the rate equation it is not taking part in the rate determining step
.The power on the substance in the rate equation is the number of that substance in the rate equation

Question 246

The steps involved in the reaction must add up to what?

Answer 246

the overall equation in the reaction

Question 247

Rate = k[NO2]2
Overall = NO2 + CO  NO + CO2

what are the two steps involved?

Answer 247

R.D.S = NO2 + NO2  NO3 + NO

Other step = NO3 + CO  NO2 + CO2

Question 248

In any chemical reaction, some bonds are broken and new ones are made
Often these changes are too complicated to happen in one single stage, instead, the reaction may involve what?

Answer 248

a series of smaller steps one after the other

Question 249

BrO3- + 6H+ + 5Br-  3Br2 + 3H2O

why must this reaction take multiple steps?

Answer 249

it is unlikely that all the ions will collide at the same time

Question 250

This reaction happens in two steps, the steps also happen at different rates, one step is slow and one step is _____

Answer 250

fast

Question 251

Rate = k [Cl.][O3]
Overall equation = O3 + O  2O2

what are the two steps involved?

Answer 251

RDS = Cl. + O3  ClO. + O2

Other Step = ClO. + O  O2 + Cl.

Question 252

What is a strong acid?

Answer 252

A strong acid is one which completely dissociates into ions in a solution

Question 253

WHat does pH = -log10[H+] only work for?

Answer 253

ONLY WORKS FOR STRONG ACIDS

Question 254

[H^+] equation for changing concentraation

Answer 254

[H+] = [H+]old x (old volume / new volume)

Question 255

How are weak acids different to strong ones?

Answer 255

.Weak acids do not dissociate fully like strong acids

Question 256

What happens when a weak acid dissociates?

Answer 256

An equilibrium is established between the ions and the acid

Question 257

As an equibrlium is established when weak acids dissociate, what can we calculate?

Answer 257

we can calculate an equilibrium constant Ka (Acids dissociation cons tat)

Question 258

What does Ka equal for, HA H+ + A-?

Answer 258

Ka = ( [H+] [A-] ) / [HA]

Question 259

For the Ka equation what two assumptions need to be made?

Answer 259

.A pure acids will split evenly into equal quantities of [h+] and [A-]
.The concentration of the acid doesn’t change as the equilibrium lies so far to the left it can be considered equal

Question 260

does [H+] = [H+ aq]?

Answer 260

yes

Question 261

.A sample of ethanoic acid of concentration 0.04 moldm-3

.What is the value of Ka?

Answer 261

2.5x10-9 moldm-3

Question 262

Steps to calculating pH of weak acid

Answer 262

.Step 1: Calculate [H+] from ka and [HA]

.Step 2: Calculate pH

Question 263

Are the approximations for weak acids justified?

Answer 263

.The first approximation assumes that the dissociation from water is negligible
.[H+(aq)]eqm = [A-(aq)]eqm - if pH>6 then water dissociates and [H+] from water will be more significant than from dissociation of acid
.This approximation breaks down for very weak acids or very dilute solutions
.Second approximation assumes that the concentration of the acid [HA] is much greater than the [H+] concentration at equilibrium
.[HA]start&raquo_space; [H+]eqm [HA]eqm = [HA]start – [H+]eqm so [HA]eqm = [HA]start
.Not valid as acid gets stronger as [H+] becomes more significant and real difference between [HA]eqm = [HA]start – [H+]eqm
.Not justified for stronger weak acids / very dilute solutions.

Question 264

.The value of Ka for weak acids is almost always very small and difficult to compare numbers with negative indices, how do we get around this?

Answer 264

create a scale of more useable numbers we often use the value of pKa, when talking about the aciditing of weak acids

Question 265

What is pKa often used for?

Answer 265

to compare acids in biological systems

Question 266

pKa equation

Answer 266

pKa = -log10Ka

Question 267

What do the values of pKa mean?

Answer 267

The higher the value of pKa, the weaker the acid

The lower the value of pKa, the stronger the acid

Question 268

How does wine show behaviour of typical dibasic and tribasic acids?

Answer 268

Wines often contain traces of sulphurous acid, H2SO3, added as a preservative
.Sulphurous acid is dibasic and its dissociation is shown as:

H2SO3 H+ + HSO3 - (pKa = 1.92)

HSO3 - H+ + SO3 2- (pKa = 7.18)

.For the first dissociation, H2SO3, acts as a weak acid
.From the pKa values, HSO3, is a far weaker acid than H2SO3
.The behaviour is typical of dibasic and tribasic acids

Question 269

What can increase the rate of a reaction?

Answer 269

.Increase temperature
.Increase concentration/pressure
.Increase surface area
.Catalyst
.Increase temperature
.Increase concentration/pressure
.Increase surface area
.Catalyst

Question 270

What needs to happen for a successful reaction?

Answer 270

.The particles must collide
.Have sufficient energy
.Be in correct orientation

Question 271

What must you remember when drawing a maxwell-boltzman distribution?

Answer 271

SHOULD NEVER TOUCH X AXIS

Question 272

Why should the maxwell-boltzman distribution go through the origin?

Answer 272

no molecules with no energy

Question 273

Is the rate constant actually a constant?

Answer 273

.The rate constant is only actually constant if we change the concentration of reacantn
.If we change other factors then it changes

Question 274

How can we show that the rate constant can change?

Answer 274

This can be shown mathematically with an equation, called Arrhenius:

K = A(e^(-Ea/RT))

Question 275

What are the different parts of the Arrhenius equation?

Answer 275

.K = rate constant
.A = pre-exponential frequency factor
.e = natural log constant
.Ea = activation energy
.R = gas constant (8.31)
.T = temperature (K)

Question 276

what is e^(-Ea/RT) called?

Answer 276

the exponential factor

Question 277

what does e^(-Ea/RT) represent?

Answer 277

This represents the proportion of particles that exceed Ea and have sufficient energy to take place

Question 278

In the Arrhenius equation, what does A do?

Answer 278

‘A’ takes into account the frequency of collisions with the correct orientation, it does increase slightly with temperature but is essentially constant over a small temperature change

Question 279

How can we show the effect of temperature by using the Arrhenius equation?

Answer 279

.You can use the arrhenius equation to describe the effect of temperature on the rate equation
.Make up some numbers plug them into the equation to get a value for the exponential factor
.Increase the temperature value by 10
.State the effect that this has had on K, and consequently the rate
.This links to the Boltzmann distribution, as the increase in temperature shifts to distribution to the right, the number of particles above Ea increases, the particles move faster and they collide more frequently at the correct orientation.

Question 280

Increasing temperature effects reactions by increasing the number of particles above Ea, but also increasing the number of collisions - which has the larger effect?

Answer 280

.There is more effect of temperature caused by more particles exceeding Ea than from increased collisions

Question 281

How can we show the effect of a catalyst by using the Arrhenius equation?

Answer 281

.You can use the arrhenius equation to describe the effect of a catalyst on the rate equation
.Make up some numbers plug them into the equation to get a value for the exponential factor
.Decrease the activation energy value (as catalysts offer an alternate pathway, and so lower the Ea)
.State the effect that this has had on K, and consequently the rate

Question 282

Summarise how using the Arrhenius equation can show the effect of changing temperature and catalyst presence

Answer 282

.Increasing the temperature increases the rate constant and therefore increases the rate
.Decreasing the activation energy (by using a catalyst) increases the rate constant and therefore increases the rate

Question 283

What is the logarithmic form of Arrhenius?

Answer 283

ln(k) = -Ea/RT + ln(A)

Question 284

What is a major use of the Arrhenius equation?

Answer 284

We can use this equation to determine Ea and A graphically

If we plot a graph of ln(k) against 1.T we will get a straight line of the type y=mx+c

Question 285

How does the Arrhenius equation link to y=mx+c

Answer 285

ln(k) = -Ea/RT + ln(A)
ln(k) = y
-Ea/R = m
1/T = x
Ln(A) = c
m = gradient so working it out gives -Ea/R
c = intercept on the y axis

Question 286

Steps to plotting an Arrhenius graph

Answer 286

.Work out ln(k) and 1/T
.Plot a graph of ln(k) against 1/T
.Workout the gradient, and consequently the activation energy
.Calculate A from the intercept ( ln(A) )

Question 287

How to work out stuff from Arrhenius graph if its a broken scale?

Answer 287

Use the equation

Question 288

Where strong acids completely dissociate to release all H+ ions into solution, weak acids only partially dissociate, explain this in terms of equations

Answer 288

Strong: HA → H+ + A-
Weak: HA ⇌ H+ + A-

Question 289

As weak acids form an equilibrium their dissociation can be represented by the acid dissociation constant, Ka, whats the equation for this?

Answer 289

Ka = “[H+(aq)][A-(aq)]” /”[HA(aq)]”

Question 290

Explain why a larger value for Ka means a lower pH

Answer 290

.The larger the value for Ka the more the equilibria lies to the right and so the lower the pH i.e. higher [H+]

Question 291

What alters Ka?

Answer 291

Temperature

Question 292

.As the values for Ka are very small it is easier to give their negative logarithm, pKa, instead, what is the equation for this?

Answer 292

pKa = - logKa

Question 293

pKa = - logKa, what is the inverse equation for this?

Answer 293

The inverse for this equation is: Ka = 10-pKa

Question 294

This means that the weaker an acid the:

._____ Ka
._____ pKa

Answer 294

.Smaller Ka

.Larger pKa

Question 295

.The [H+] of a weak acid depends on the value of what?

Answer 295

Ka and [HA]

Question 296

.The equilibrium concentrations, [ ]eqm, can be used to determine a value for Ka, what is the equation for this?

Answer 296

Ka = “[H+]eqm [A-]eqm” /”[HA]eqm”

Question 297

There are two approximations that need to be made when calculating Ka, what are they?

Answer 297

.HA dissociation forms equal [H+] and [A-]

.The change in [HA] is negligible so [HA]eqm = [HA]start

Question 298

Using the approximations, Ka can be determined by what equation?

Answer 298

Ka = “[H+]eqm 2” /”[HA]start”

Question 299

How can pH be calculated for a weak acid?

Answer 299

[H+] = √(“Ka x [HA]” ) → pH = -log[H+]

Question 300

.A value for Ka can be determined experimentally, how?

Answer 300

by using a pH meter to get the pH of a standard solution

Question 301

.The larger the value for Ka, the greater the _________

Answer 301

dissociation

Question 302

When calculating values for Ka there are issues with the approximations made, what are these issues?

Answer 302

• At pH values >6 water dissociation is significant
Therefore doesn’t work for very weak acids or very dilute solutions
• If [H+] concentration is significant there will be a difference between [HA]eqm and [HA]start
Therefore doesn’t work stronger weak acids with Ka > 10-2 mol dm-3 or very dilute solutions

Question 303

All aqueous solutions contain which ions?

Answer 303

H+ and OH- ions

Question 304

H2O

Answer 304

H+ + OH-

Question 305

In ____ [H+] > [OH-]
In ____ [OH-] > [H+]
In ____ [H+] = [OH-]

Answer 305

In acids [H+] > [OH-]
In alkalis [OH-] > [H+]
Neutral [H+] = [OH-]

Question 306

For every 500,000,000 H2O molecules, only _ dissociates

Answer 306

1

Question 307

For every 500,000,000 H2O molecules, only 1 dissociates, what does this mean for the equilbrium?

Answer 307

the equilibrium is on the left hand side

Question 308

Do the Kc equation for water equilibrium

Answer 308

Kc = ([H+][OH-]) / [H2O]

Question 309

[H2O] x Kc =

Answer 309

[H+] x [OH-]

Question 310

[H2O] is such a large excess it can be classed as a ____

Answer 310

constant

Question 311

What is Kw?

Answer 311

ionic product of water

Question 312

Kw = [H2O] x Kc, so replace [H2O] x Kc with Kw in the correct equation

Answer 312

Kw = [H+] x [OH-]

Question 313

At 298K, Kw =

Answer 313

1x10-14 mol2dm-6

Question 314

Why is the pH of pure water at 298K, 7?

Answer 314

Kw = [H+] x [OH-]
1x10-14 = [H+] x [OH-]
The concentrations are the same and so it can be written as:
1x10-14 = [H+]2
1x10-7 = [H+]
pH = 7

Question 315

Effect of temperature on Kw

Answer 315

The dissociation of water is endothermic
Temperature increase will move the equilibrium to the right, and so Kw will increase
Temperature decrease will move the equilibrium to the left, and so Kw will decrease
Water will always remain neutral at all temperatures, pH may change with temperature, but [H+] = [OH-] all the time and so it is neutral

Question 316

.Water ionises very slightly, acting as both an acid and as a base – setting up an ________

Answer 316

equilibrium

Question 317

.Water dissociates a very, very small amount according to the equation – it must do, otherwise it would not _____________

Answer 317

conduct electricity

Question 318

.1 dm3 (1000g) of water is mainly ________ H2O

Answer 318

undissociated

Question 319

. [H2O(l)] = 1000/18 = 55.6 mol dm-3 (a constant), why?

Answer 319

.1 dm3 (1000g) of water is mainly undissociated H2O

Question 320

.If we know the concentration of hydroxide ions we can rearrange Kw to give us the value of what?

Answer 320

[H+]

Question 321

What is the pH of 0.4 mol dm-3 NaOH?

Answer 321

Kw = [H+][OH-] = 1.00 × 10–14 mol2dm–6 
	(1.00 × 10_14 )/([0.4])=[H+] = 2.5 x 10-14 mol dm-3	    
 			       pH 	= -log10[H+]
				= -log10[2.5 x 10-14 ]
				= 13.60

Question 322

What is the pH of a solution with [OH-] = 2 x 10-2 mol dm-3 at 25oC ?

Answer 322

Step 1 : calculate [H+] from Kw and [OH-]
Kw = [H+][OH-] = 1.00 x 10-14
[H+] = Kw = 1.00 x 10-14 = 5.00 x 10-13 mol dm-3
[OH-] 2.00 x 10-2
Step 2 : Use calculator to find pH
pH = - log [H+] = -log (5.00 x 10-13) = 12.30

Question 323

What are the concentrations of H+ (aq) and OH- (aq) in a solution of pH 3.25 at 25oC ?

Answer 323

Step 1 : Use calculator to find [H+(aq)]
[H+] = 10-pH = 10-3.25 = 5.62 x 10-4 mol dm-3
Step 2 : Calculate [OH-] from Kw and [H+]
Kw = [H+][OH-] = 1.00 x 10-14
[OH-] = Kw = 1.00 x 10-14 = 1.78 x 10-11 mol dm-3
[H+] 5.62 x 10-4

Question 324

For pH values that are whole numbers, it is easy to work out the [H+] and [OH-] concentrations as the indices add up to what?

Answer 324

-14

Question 325

.The pH of weak bases can be calculated via a similar method to that used for ______

Answer 325

weak bases

Question 326

.In an aqueous solution, there will always be both H+ (aq) and OH- (aq) ions present such that ________

Answer 326

[H+ (aq)][OH- (aq)] = Kw

Question 327

.A solution is ______ when [H+ (aq)] > [OH- (aq)]
.A solution is ______ when [H+ (aq)] = [OH- (aq)]
.A solution is ______ when [OH- (aq)] > [H+ (aq)]

Answer 327

.A solution is acidic when [H+ (aq)] > [OH- (aq)]
.A solution is neutral when [H+ (aq)] = [OH- (aq)]
.A solution is alkaline when [OH- (aq)] > [H+ (aq)]

Question 328

.So a solution that is acidic will still contain __ ions, it is just that there are more __ions (and vice versa in an ______ solution)

Answer 328

OH-
H+
Alkaline

Question 329

.The value of Kw controls the ______ of each ion

Answer 329

Concentrations

Question 330

• Kw can also tell us the pH of ______

Answer 330

pure water

Question 331

• As the water splits into equal concentrations of OH- and H+ ions we can make what assumption?

Answer 331

Kw = [H+][OH-] = 1.00 × 10–14 mol2dm–6 
	     = [H+]2 = 1.00 × 10–14
	     = [H+] = 1.00 × 10–7
 			     	  pH 	= -log10[H+]
					= -log10[1.00 x 10-7 ]
					= 7

Question 332

When is neutral not neutral?

Answer 332

.A neutral solution is defined by a equal number of moles of H+ ions and OH- NOT by a pH of 7 as you may have been taught
.Whilst this value is 7 at approximately 298K, the value of Kw increases with temperature

Question 333

.The pOH scale measures what?

Answer 333

the concentration of hydroxide ions

Question 334

.Therefore you can solve fpr pH, pOH and [OH-] given just ___

Answer 334

[H+]

Question 335

pH + pOH =

Answer 335

14

Question 336

.Kw = [H+][OH-] =

Answer 336

1x10-14

Question 337

Define Enthalpy of reaction ΔrH

Answer 337

enthalpy change that accompanies a reaction in the molar quantities shown in a chemical equation under standard conditions, with all reactants and products in their standard states

Question 338

Define Enthalpy of formation ΔfH

Answer 338

enthalpy change when one mole of compound is formed from its constituent elements under standard conditions

Question 339

Define Enthalpy of combustion ΔcH

Answer 339

enthalpy change when one mole of substance is burnt completely in excess oxygen under standard conditions

Question 340

Water slightly ionises, what equation will show this?

Answer 340

H2O(l) H+(aq) + -OH(aq)

Question 341

Ka of :
H2O(l) H+(aq) + -OH(aq)
=

Answer 341

[H+][-OH] / [H2O]`

Question 342

Ka x [H2O] =

Answer 342

Kw

Question 343

Kw =

Answer 343

[H+][-OH]

Question 344

What is Kw?

Answer 344

Kw is the ionic product of water, at 25’C it equals 1x10-14 mol2dm-6

Question 345

How does ionic bonding work?

Answer 345

1. Loss of an electron(s) by an element
2. Gain electrons by a second element
3. Attraction between positive and negative ions

Question 346

Na + Cl, ionisation electron addinity and latice equations

Answer 346

Ionisation Energy – 
Na  e- + Na+ 	+496kJmol-1
Electron Affinity – 
e- + Cl  Cl- 	-349kJmol-1
Lattice Enthalpy – 
Cl- + Na+  NaCl	-766kJmol-1

Question 347

What is electron affinity?

Answer 347

.The enthalpy change when one mole of electron is added to one mole of atoms in the gaseous phase to form one mole of -1 ions

Question 348

Is repulsion between two electrons exothermic or endothermic, why?

Answer 348

.Repulsion between two negatively charged things requires energy so exothermic

Question 349

What is lattice enthalpy?

Answer 349

.ΔHlatt is the enthalpy change when 1 mole of ionic substance is formed from its gaseous ions under standard conditions

Question 350

.The strength of the ionic bond is related to the lattice enthalpy, how?

Answer 350

more exothermic the greater the ionic bonding

Question 351

Lattice enthalpy can’t be measured, why?

Answer 351

Cannot be measured directly as cannot form one mole of ionic lattice from gaseous ions

Question 352

What are the factors affecting lattice enthalpy? How?

Answer 352

Charge –
.The greater the charge on the ions, the stronger the attraction – therefore, more exothermic lattice enthalpy
Size –
.Smaller ions can pack together more tightly, therefore there is greater attraction and more exothermic lattice enthalpy

Question 353

Why, when moving down group two, do the temperatures required to break the lattice increase?

Answer 353

.They have decreasing charge densities
.Mg2+ is a smaller ion than Ba2+ , so the +2 charge occupies a smaller volume – this means Mg2+ has a higher charge density than Ba2+
.Mg2+ can distort the electron clouds within the CO32- ion (called polarisation), this weakens the covalent bonding in the ion and reduces its decomposition temperature

Question 354

What is a buffer?

Answer 354

a solution that minimises pH change when a small amount of acid or alkali is added

Question 355

What two types of buffers are they?

Answer 355

Weak Acid and Salt of Weak Acid

Excess Weak Acid and Strong Base

Question 356

Suggest and explain a Weak Acid and Salt of Weak Acid buffer

Answer 356

Weak Acid – Ethanoic Acid –
CH3COOH(aq) CH3COO-(aq) + H+(aq)
Salt of Weak Acid – Sodium Ethanoate –
CH3COONa(s) + (aq) CH3COO-(aq) + Na+(aq)
¬Buffer Contains –
CH3COOH(aq) CH3COO-(aq) + H+(aq)
When add H+, H+ reacts with conjugate base, equilibrium will move to the left to reduce the amount of H+, pH is constant
Adding –OH (alkali) –
-OH + H+  H2O
Conc of H+ decreases, equilibrium moves to the right to increase the conc of H+, pH is constant

Question 357

Suggest and explain an excess Weak Acid and strong base buffer

Answer 357

CH3COOH(aq) + NaOH(aq) –> CH3COONa(aq) + H2O(l)
CH3COONa(aq) CH3COO-(aq) + Na+(aq)
CH3COOH(aq) CH3COO-(aq) + H+(aq)

When add H+, H+ reacts with conjugate base, equilibrium will move to the left to reduce the amount of H+, pH is constant
Adding –OH (alkali) –
-OH + H+  H2O
Conc of H+ decreases, equilibrium moves to the right to increase the conc of H+, pH is constant

Question 358

Excess methanoic acid is reacted with potassium hydroxide, explain how a buffer solution is produced and how pH is controlled when 5cm3 of HCl is added.

Answer 358

The solution now contains HCOOK(aq) HCOO- (aq) + K+(aq) and HCOOH(aq) HCOO-(aq) + H+ equilibriums in solution – the buffer. When 5cm3 of HCl is added, it dissociates into H+ and Cl-, the H+ ions increase the concentration of the H+ already in the solution, so the HCOOH(aq) HCOO-(aq) + H+ point of equilibrium moves to the right – decreasing the conc of H+ in solution, and so keeping the pH constant.

Question 359

Ka x ( [HA]/[A-] ) = ?

what does each part mean?

Answer 359

[H+] = Ka x ( [HA]/[A-] )
[HA] = concentration of weak acid
[A-] = concentration of conjugate base

Question 360

50cm3 of 1.2M NaOH reacts with 250cm3 of ethanoic acid (1M), Ka = 1.74x10-5, what is the pH?

Answer 360

CH3COOH  CH3OO- + H+
CH3COOH + NaOH  CH3COO- + Na+ + H2O
1.	Moles weak acid = 250x1 / 1000 = 0.25mol
2.	Moles NaOH = 50x1.2 / 1000 = 0.06mol
3.	Moles A- = 0.06mol
4.	Moles HA = 0.19mol
5.	[H+] = Ka ([HA] / [A-]) = 1.74x10-5 x (0.19/0.06) = 5.51x10-5
6.	pH = 4.26

Question 361

Equation to work out lattice enthalpy from born haber cycle

Answer 361

L.E = FORMATION – sum(ATOM + I.E + E.A)

or

L.E = FORMATION – sum(REST)

Question 362

What are Born Haber Cycles used for?

Answer 362

Born Haber cycles can be used to calculate a measure of ionic bond strength based on experimental data

Question 363

Draw the born haber cycle for NaCl

Answer 363

check notes or google

Question 364

Draw the born haber cycle for MgCl2

Answer 364

check notes or google

Question 365

Draw the born haber cycle for CuO

Answer 365

check notes or google

Question 366

Blood must contain a pH of what?

Answer 366

Blood must contain a pH of 7.40 +- 0.05

Question 367

What is the most important buffer in blood?

Answer 367

the carbonic acid – hydrogencarbonate

Question 368

Carbonic acid dissociation equation

Answer 368

H2CO3 H+ + HCO3-

Question 369

If blood becomes too acidic (acidosis), what presents?

Answer 369

fatigue, shortness of breath, shock, death

Question 370

If blood becomes too alkaline (alkalosis), what presents?

Answer 370

spasms, light-headed, nausea

Question 371

What happens if the body produces more acidic products?

Answer 371

.The H+ concentration in the blood will increase, the equilibrium of the hydrogencarbonate dissociation will shift to the left, the H+ ions will be used up to form more hydrogencarbonate, the H+ concentration will return to normal, the pH will have had minimal change
.If the acid level continued to rise, fatigue, shortness of breath, shock, or death would eventually set in
.This would occur when we have ran out of HCO3- ions

Question 372

Carbon dioxide dissolves in water to form what? What is the equation for this?

Answer 372

carbonic acid:

CO2 + H2O H2CO3

Question 373

Inhalation of high levels of CO2 means what?

Answer 373

Inhalation of high levels of CO2 mean that the equilibrium in the blood must shift to the right to form more H2CO3, this means that the level of H2CO3 in the blood increase, so in the equilibrium H2CO3 H+ + HCO3- the equilibrium shifts to the right to counter this, this forms more H+ ions in the blood, making it more acidic, and resulting in acidosis which would lead to fatigue, shortness of breath, shock, and eventual death. Increased [H+] means enzymes would also be denatured.

Question 374

Define enthalpy of solution, with an example

Answer 374

DeltasolutionH is the enthalpy change when one mole of ionic compound is completely dissolved in water under standard conditions.

e.g. NaCl(s)  Na+(aq) + Cl-(aq)

Question 375

What is enthalpy of hydration with example

Answer 375

• Delta Hhydration¬ is the enthalpy which takes place when one mole of gaseous ions is dissolved in water forming one mole of aqueous ions under standard conditions
• Na+(g) + aq  Na+(aq)

Question 376

Why is enthalpy of hydration always exothermic?

Answer 376

These are exothermic as bonds are formed between the ions and water molecules

Question 377

Factors Affecting Enthalpy of Hydration

Answer 377

Charge – the higher charge on the ion, the greater the attraction for the H2O molecules, therefore a more exothermic hydration
Size – smaller ions have a greater charge density compared to the larger ions, this creates a greater attraction for H2O molecules, therefore a more exothermic hydration

Question 378

Draw the general cycle of enthalpy of solution, hydration and lattice enthalpy

Answer 378

check notes

Question 379

Equation that links lattice enthalpy, enthalpy of solution, and enthalpy of hydration

Answer 379

lattice enthalpy + enthalpy of solution = sum of the enthalpy of hydrations

Question 380

Draw the titration curve for when HCl is added to NaOH

Answer 380

check notes

Question 381

Draw the titration curve for when NaOH is added to HCl

Answer 381

check notes

Question 382

Draw strong acid strong base titration curve

Answer 382

check notes

Question 383

Draw strong acid weak base titration curve

Answer 383

check notes

Question 384

Draw weak acid strong base titration curve

Answer 384

check notes

Question 385

Draw weak acid weak base titration curve

Answer 385

check notes

Question 386

The shape of the titration curve depends on what?

Answer 386

• Substance being titrated i.e. acid or base

- Type of acid or base i.e. strong or weak

Question 387

Describe and explain parts of titration curve

Answer 387

1. Excess of base: pH decreases as acid added
2. Vertical section: acid base concentration similar, pH alters rapidly
3. Excess of acid: pH decreases slightly as acid is added
4. The equivalence point: halfway up the vertical section

Question 388

What is the equivalence point?

Answer 388

The point halfway between the two horizontal lines on the titration curve

The equivalence point is the volume required to have completely reacted the acid and base toetehr, stoichiometric volumes
The pH at the equivalence point depend son the type of acid and base

Question 389

Where is the equivalence point on each titration curve?

Answer 389

Equivalence point = 7 on SA-SB
Equivalence point < 7 on SA-WB
Equivalence point > 7 on WA-SB
No equivalence point on WA-WB

Question 390

Why is the equivalence point 0 on SASB titration curves?

Answer 390

• H+ ions completely neutralised by OH- ions, so only water and salt present, neutral

Question 391

Why is the equivalence point < 7 on SAWB titration curves?

Answer 391

• A weak base, like NH3, will have a strong conjugate acid, NH4+ , which will react with water to produce H3O+ so the pH is less than 7

Question 392

Why is the equivalence point > 7 on WASB titration curves?

Answer 392

• A weak acid, HA, will have a strong conjugate base, A-, which can react with water to produce OH- ions, so the pH is more than 7

Question 393

What are acid-base indicators?

Answer 393

Acid-Base indicators are weak acids that have different coloured conjugate bases
In acidic conditions the indicator equilibrium is shifted towards the weak acid (HIn)
As the system becomes more basic the equilibrium shifts towards the conjugate base (In-), altering the colour

Question 394

What is the end point?

Answer 394

The end point is when equal [HIn] and [In-] are present, the colour will therefore lie between both extremes
Each indicator will have a different pH value for the end point as they each have different ka values

Question 395

HIn

Answer 395

H+ + In-

Question 396

How is an acid-base indicator chosen?

Answer 396

The indicator required for a titration must have a colour change that lies within the vertical section of a titration curve

Question 397

A chemical reaction will proceed when?

Answer 397

A chemical reaction will proceed if the products are energetically more stable than the reactants

Question 398

What is entropy?

Answer 398

Entropy is a measure of the dispersal of energy in a system, the more disordered a system the greater the dispersal of energy = higher entropy

we define entropy as a measure of disorder

Question 399

What happens to entropy over time?

Answer 399

Entropy must increase over time

Question 400

When is a system in a state of high entorpy?

Answer 400

When its degree of disorder is high

Question 401

As order within a system increases, its entropy decreases, why?

Answer 401

This can be explained in terms of probability: disordered states are simply more likely to exist (or emerge) than ordered states.
The spontaneous direction of change is from a less probable to a more probable state

Question 402

What does entropy always do?

Answer 402

The total entropy always increases, and the process is irreversible

Question 403

What is the unit for entropy

Answer 403

S

in JK^-1mol^-1

Question 404

Why is S (entropy) alwasy potisive?

Answer 404

All substances process some degree of disorder because particles are always in constant motion

Question 405

How does the entropy of each state differ?

Answer 405

Solid has lowest entropy, and gas has the highest

Question 406

Draw a graph of waters change in enthalpy as temperaurer increases

Answer 406

check notes

Question 407

Systems that are more chaotic have a _____ entropy value

Answer 407

higher

Question 408

(s)  (s) + (g) , what is delta S

Answer 408

+

Question 409

(g)  2(g) , what is delta S

Answer 409

+

Question 410

What is delta S?

C2H5OH(l)  C2H5OH(g)

Answer 410

+

Question 411

What is delta S?

C2H2(g) + 2H¬2(g)  C2H6(g)

Answer 411

-

Question 412

What is delta S?

NH4Cl(s) + aq  NH4Cl(aq)

Answer 412

+

Question 413

What is delta S?

4Na(s) + O2(g)  2Na2O(s) deltaS

Answer 413

-

Question 414

Define the standard entropy change

Answer 414

The standard entropy change is the entropy change that accompanies a reaction in the molar quantities expressed in the equation, under standard conditions

Question 415

Sum for delta S^theta

Answer 415

delta S^theta = sum(S^theta products) – sum(S^theta reactants)

Question 416

Why are exothermic reactions more preferable in nature?

Answer 416

as the products are more stable than reactants

the key is not the decrease in energy but the associated increase in entropy of the surroundings.

Question 417

deltaSsurroundings is proportional to what?

Answer 417

-deltaHsystem

Question 418

deltaSsurroundings =

Answer 418

(-deltaHsystem)/T

Question 419

deltaStotal =

Answer 419

deltaSsystem + deltaSsurroundings

Question 420

deltaStotal >

Answer 420

0

Question 421

What can we learn about the entropy of the sun by the photosynthesis equation?

Answer 421

6CO2(g) + 6H2O(l)  UV light  C6H12O6(s) + 6O2(g)
Negative entropy change
Entropy change in sun must be so positive it outweighs every plant on earth

Question 422

What is gibbs free energy equation?

Answer 422

deltaG = deltaH – T deltaS < 0

Question 423

How do you get to gibbs free energy equation?

Answer 423

deltaStotal¬ = deltaSsystem + deltaSsurroundings > 0
deltaStotal¬ = deltaSsystem – (deltaHsystem)/T > 0
T deltaStotal = T deltaSsystem – deltaHsystem > 0
-T deltaStotal = -T deltaSsytem + deltaHsystem < 0
deltaG = deltaH – T deltaS < 0

Question 424

Using deltaG explain when a reaction is and is not feasible

Answer 424

deltaG must be negative (<0) for a reaction to be feasible i.e. proceed
If deltaG is positive (>0) then a reaction is not feasible

Question 425

What is deltaG at the points of feasibility?

Answer 425

At the point of feasibility, we can say deltaG = 0 (assume that deltaH and deltaS don’t vary with temperature)

Question 426

At low temperatures what does deltaGsystem equal? why?

Answer 426

• At low temperatures, deltaGsystem = deltaH (-T deltaS becomes negligible) so for a reaction to occur it needs to be exothermic

Question 427

At high tempertaures deltaGsystem ewuals what? why?

Answer 427

• At high temperatures, deltaGsystem = -T deltaS (deltaH becomes negligible) so for a reaction to occur it needs to have a positive deltaS as – T deltaS needs to be less than 0

Question 428

Limitations to deltaG equation

Answer 428

just because the value is negative and so feasible it doesn’t mean it occurs, the reaction rate might be incredibly slow or the activation energy too high

Question 429

What is an oxidation number?

Answer 429

An oxidation number shows the charge of an atom if all of its bonds were considered totally ionic

Question 430

What are the oxidation number rules?

Answer 430

1. Elements in their natural state = 0
2. The total oxidation states in a molecule = 0
3. The total oxidation states in an ion = the charge

Groups 1, 2, 3 -
State +1, +2, +3

Fluorine -
State -1

Hydrogen -
State +1
-1 in metal hydrides

Oxygen -
State -2
-1 in peroxides

Chlorine -
State -1
+ (varies) in chlorates

Question 431

What is an oxidising agent?

Answer 431

takes electrons from what is oxidised

Question 432

What is a reducing agent?

Answer 432

gives electrons to what is reduced

Question 433

Write the half equations and full equation for

4Na + O2  2Na2O

Answer 433

4Na  4Na+ + 4e-
O2 + 4e-  2O2-
4Na + O2  4Na+ + 2O2-

Question 434

Write the half equations for

Cu2O + H2SO4  Cu + CuSO4 + H2O

Answer 434

Cu+  Cu2+ + e-

2Cu+  Cu + Cu2+

Question 435

Steps of a redox titration using iron sulphate and sulphuric acid and potassium manganate (VII)

Answer 435

1. Make up a 150cm3 standard solution using 7g of iron sulphate by weighing by difference in a volumetric flask, record your mass measurements in a table
2. a. Pipette 25cm3 of the standard solution into a conical flask
   b. Add 10cm3 of 1M H2SO4 to the conical flask
   c. Fill the burette with a standard solution of 0.02M potassium manganate (VII)
   d. Carry out the titrations until the ed point is reached, the first permanent pink colour, record your results in a table
   e. Repeat until you have two concordant results
   f. Calculate the mean titre

Question 436

Reduction: MnO4- + 8H+ + 5e-  Mn2+ + 4H2O
Oxidation: Fe2+  Fe3+ + e- or 5Fe2+  5Fe3+ + 5e-
MNO4- + 8H+ + 5Fe2+  Mn2+ + 4H2O + 5Fe3+

What is the oxidising and reducing agent?

Answer 436

Fe2+ is the reducing agent

MnO4- is the oxidizing agent

Question 437

Mass of impure hydrated iron II sulphate = 6.97g
25cm3 of hydrated iron (II) sulphate used taken from a 250cm3 volumetric flask full of the dissolved iron tablets
Mean titre of 0.02M potassium manganate (VII) = 23.4cm3

Calculate the percentage purity of the impure sample of FeSO4.7H2O

Answer 437

93.3%

Question 438

What are the redox equations for the iodine/thiosulphate titrations?

Answer 438

I2 + 2e-  2I-
2S2O32-  S4O62- + 2e-
I2 + 2S2O32-  2I- + S4O62-

Question 439

What colour is iodine?

Answer 439

Iodine is a yellow brown solution

Question 440

What colour is iodide?

Answer 440

Iodide is a straw colour

Question 441

How do you know when an iodine/thiosulphate titration is complete?

Answer 441

Iodine is a yellow brown solution
Iodide is a straw colour
The above colour change is the indication for the redox reaction to be complete

Question 442

Is an anode positive or negative?

Answer 442

Positive

Question 443

Is a cathode positive or negative?

Answer 443

Negative

Question 444

Draw a general cell

Answer 444

check notes

Question 445

What is a cell? What is it made of? What does it do?

Answer 445

• A cell has two half cells
• The two half cells must be connected with a salt bridge
• Simple half cells will consist of a metal (acts as an electrode) and a solution of a compound containing that metal
  o For example, Cu and CuSO4
• These two half cells will produce a small voltage if connected into a circuit (become a battery or cell)

Question 446

Draw the standard hydrogen half-cell

Answer 446

Check notes

Question 447

What is the standard hydrogen half-cell?

Answer 447

• H+/H2 half-cell is chosen to produce standard potentials
• H+ + e-  ½H2
• 1/2H2  H+ + e-
• Conditions
  o 298K
  o 1atm
  o 1M H+
• Both equations have a voltage of 0.0 volts

Question 448

Why is platinum used in the standard hydrogen half-cell?

Answer 448

Platinum is inert and so does not take part in the reaction

Question 449

How can you find the value of any electron potential?

Answer 449

If the standard hydrogen half-cell is connected to another cell,
you can find the value of any electron potential (as the standard is
0.0 volts).

Question 450

Why a salt bridge? Using potassium chloride and copper as an example

Answer 450

• The salt bridge is used to connect up the circuit
• The free moving ions conduct the charge
• A salt bridge is usually made from a piece of filter paper (or material) soaked in a salt solution, usually potassium nitrate
• The salt should be unreactive with the electrodes and electrode solutions
  o For example, potassium chloride would not be suitable for copper systems as chloride ions can form complexes with copper ions, a wire is not used because the metal wire would set up its own electrode system with the solutions

Question 451

What is standard electrode potential?

Answer 451

The emf of a half cell compared with a standard hydrogen electrode at standard conditions of 298K, concentration 1.0M and 1 atm of pressure.

Question 452

What is emf?

Answer 452

The emf is the difference in the positive and negative sides of the cell.

Question 453

Describe a two ion cell system?

Answer 453

Both of the ions in the equation will be at a concentration of 1M, platinum electrode is used as it is INERT.
2 aqueous ions in a half equation, means they both have to be 1.0M solutions in the same half cell

Question 454

Draw a manganese/lead system

Answer 454

check notes

Question 455

General properties of d-block elements

Answer 455

The d-block elements have high melting and boiling points.

The d-block elements are good conductors of both electricity and heat.

Question 456

What elements have been used in coinage for many years?

Answer 456

copper, silver, nickel, and zinc

Question 457

What element is used extensively in construction and production of tools?

Answer 457

Iron

Question 458

What element is used for electrical cables and water pipes?

Answer 458

Copper

Question 459

Titanium has great strength, what are the elements applications?

Answer 459

It has many aerospace and medical applications (for example joint replacement).

Question 460

Sc orbital configuration

Answer 460

[Ar] 4s2 3d1

Question 461

Fe orbital configuration

Answer 461

Fe – [Ar] 4s2 3d6

Question 462

Ni orbital configuration

Answer 462

Ni – [Ar] 4s2 3d8

Question 463

Zn orbital configuration

Answer 463

Zn – [Ar] 4s2 3d10

Question 464

Cr orbital configuration

Answer 464

Cr – [Ar] 4s1 3d5

Question 465

Cu orbital configuration

Answer 465

Cu – [Ar] 4s1 3d10

Question 466

How do copper and chromium minimize repulsions?

Answer 466

Copper and chromium minimize repulsions by being half full or full, chromium has 4s and 3d orbitals half full, copper has 3d full

Question 467

Fe2+ orbital configuration

Answer 467

Fe2+ - [Ar] 4s0 3d6

Question 468

Fe3+ orbital configuration

Answer 468

Fe3+ - [Ar] 4s0 3d5

Question 469

Cu2+ orbital configuration

Answer 469

Cu2+ - [Ar] 4s0 3d9

Question 470

Cr3+ orbital configuration

Answer 470

Cr3+ - [Ar] 4s0 3d3

Question 471

Mn2+ orbital configuration

Answer 471

Mn2+ - [Ar] 4s0 3d5

Question 472

Mn4+ orbital configuration

Answer 472

Mn4+ - [Ar] 4s0 3d3

Question 473

Sc3+ orbital configuration

Answer 473

Sc3+ - [Ar] 4s0 3d0

Question 474

Zn2+ orbital configuration

Answer 474

Zn2+ - [Ar] 4s0 3d10

Question 475

Define a transition element

Answer 475

A transition element is a d-block element that forms at least one ion with an incomplete d sub-shell.

Question 476

Which ‘d’ block elements do not fit the transition element definition?

Answer 476

• Scandium and zinc

Question 477

What are some characteristic properties of transition metals and there compounds, with examples. (not conduction, melting, boiling etc.)

Answer 477

• They form compounds in which the transition element has different oxidative states
  o Fe2+ = +2
  o Fe3+ = +3
• They form colored compounds
• The elements and their compounds can act as catalysts
  o Fe in Haber process
  o Ni in hydrogenation of alkenes

Question 478

A species containing a transition element in its highest oxidation state is often a what?

Answer 478

strong oxidizing agent

Question 479

What does the observed colour of a solution depend on?

Answer 479

The observed colour of a solution depends on the wavelengths absorbed

Question 480

Why does copper sulphate solution appear blue?

Answer 480

Copper sulphate solution appears blue because the energy absorbed corresponds to red and yellow wavelengths, wavelengths corresponding to blue light aren’t absorbed.

Question 481

Draw the exam colour chart

Answer 481

check notes

Question 482

If the colour we observe is red, what colour has been absorbed?

Answer 482

Cyan

Question 483

What colour is cu2+

Answer 483

white

Question 484

What colour is [Cu(H2O)6]2+

Answer 484

blue

Question 485

Transition metals form complex ions or ?

Answer 485

coordination compounds

Question 486

WHat do transition metals form when they make complex ions?

Answer 486

ligand forms bonds with the central transition metal ion

Question 487

Examples of complex ions

Answer 487

[Cr(H2O)6]3+ [CuCl4]2-

Question 488

What is a ligand?

Answer 488

a molecule or ion that can donate a pair of electrons with the transition metal ion to form a coordinate bond

Question 489

What does monodentate mean?

Answer 489

Monodentate ‘one tooth’ means each ligand donates just one pair of electrons.

Question 490

Examples of ligands and how many loan pairs they have

Answer 490

H2O (oxygen has two loan pairs) NH3 (nitrogen has a loan pair) Cl- (chloride has a loan pair)
CN- (carbon has a loan pair) OH- (oxygen ha a loan pair)

Question 491

Draw [Cu(H2O)6]2+

Answer 491

check notes

Question 492

Describe the structure and bonding in [Cu(H2O)6]2+

Answer 492

• The central ion is Cu2+
• The ligands are water molecules
  o Each molecule donates a pair of electrons from the O atom to the Cu2+ to form a co-ordinate bond
• The co-ordination number is 6
  o This indicates the number of coordinate bonds to the central metal ion

Question 493

What is the oxidation of Co in [Co(H2O)5Cl]+

Answer 493

+2

Question 494

How to name complex ions?

Answer 494

• The name gives the metal ions and its oxidation state last, and the name/number of ligands before
• Pre-fixes di, tri, tetra, penta, hexa used
• Ligands are listed alphabetically, with prefixes not allowed to alter this order

Question 495

[Cr(H2O)4Cl2]+ name

Answer 495

¬¬Tetraaquadichlorochromium (III) ion

Question 496

[Co(H2O)5Cl]+ name

Answer 496

Pentaaquamonochlorocobalt (II) ion

Question 497

If the complex ion is an anion, what do you do?

Answer 497

the suffix ‘-ate’ follows the metal

Question 498

Name of [Fe(CN)6]4-

Answer 498

Hexacyanoferrate (II) ion

Question 499

What does cobalt become in an anion complex ion?

Answer 499

cobaltate

Question 500

What does aluminium become in an anion complex ion?

Answer 500

aluminate

Question 501

What does chromium become in an anion complex ion?

Answer 501

chromate

Question 502

What does vanadium become in an anion complex ion?

Answer 502

vanadate

Question 503

What does copper become in an anion complex ion?

Answer 503

cuprate

Question 504

What does iron become in an anion complex ion?

Answer 504

ferrate

Question 505

What does nickel become in an anion complex ion?

Answer 505

nickelate

Question 506

What does the value of pKa show?

Answer 506

The higher the value of pKa, the weaker the acid

The lower the value of pKa, the stronger the acid

Question 507

Equation for finding hydrogen ion concentration in buffer

Answer 507

[H+] = ka x ([HA]/[A-])

Question 508

In an electrochemical cell, where does oxidation happen?

Answer 508

the half-cell with the most negative standard electrode potential

Question 509

What is the relationship between electrode potential and metal reactivity?

Answer 509

More negative electrode potential = more reactive the metal

Question 510

Which equation links E (cell), E(positive electrode), and E(negative electrode)

Answer 510

E (cell) = Eo (positive electrode) – Eo (negative electrode)

Question 511

The more reactive a metal is, the more likely it is to what?

Answer 511

to lose electrons and form a positive ion

Question 512

More reactive metals have more negative electrode potentials and vice versa for what?

Answer 512

Non-metals

Question 513

in general if the electrode potential for the reaction is positive it is regarded as being what?

Answer 513

feasible

Question 514

What is the problem with predicting the feasibility of electrode reactions?

Answer 514

• The value of E says something about the feasibility of the reaction under standard conditions only
• The value of E says something about the feasibility of the reaction, but does not say anything about the rate of the reaction

Question 515

What is the anticlockwise rule?

Answer 515

Arrange the redox half equations so that:

• The electrons are on the left
• The largest negative electrode potentials are at the top
• The reaction will then take place anticlockwise around the half equations
• E (cell) = E (bottom) – E(top)

Question 516

When does predicting feasibility of electrode reactions go wrong?

Answer 516

When the conditions are not standard, this includes:
- A change in concentration
- If the half equations are in equilibrium
If the kinetics are not favorable:
- The rate of reaction may be slow so reaction does not appear to happen
- If the reaction has a high activation energy

Question 517

Define catalyst

Answer 517

a substance that increases the rate of a chemical reaction by providing an alternative reaction pathway of lower activation energy (Ea)

Question 518

What happens when Al3+, Ca2+, Mg2+

Add NaOH or KOH or NH4OH

Answer 518

white precipitate

Question 519

Description of solution, observation with NaOH, and equation for Copper Cu2+

Answer 519

Transparent blue Pale blue precipitate

Insoluble in excess Cu2+(aq) + 2OH-(aq)  Cu(OH)2(s)

Question 520

Description of solution, observation with NaOH, and equation for Iron(II) Fe2+

Answer 520

Pale green Dark green precipitate
Turns brown on contact with air
Insoluble in excess Fe2+¬(aq) + 2OH-(aq)  Fe(OH)2(s)

Question 521

Description of solution, observation with NaOH, and equation for Iron(III) Fe3+

Answer 521

Orange/brown Orange/brown precipitate

Insoluble in excess Fe3+(aq) + 3OH-(aq)  Fe(OH)3(s)

Question 522

Description of solution, observation with NaOH, and equation for Chromium(III) Cr3+

Answer 522

Violet Grey-green precipitate
Soluble in excess giving dark green solution Cr3+ + 3OH-(aq)  Cr(OH)3(s)
Cr(OH)3(s) + 3OH-(aq)  [Cr(OH6]3-(aq)

Question 523

Description of solution, observation with NaOH, and equation for Manganese(II) Mn2+

Answer 523

Pale pink Off white precipitate
Rapidly turning brown on contact with air
Insoluble in excess Mn2+(aq) + 2OH-(aq)  Mn(OH)2(s)

Question 524

• Cu2+(aq) + 2e- Cu(s)

If we increase the concentration of Cu2+ ions then:

Answer 524

• Equilibrium moves to oppose the charge
• Electrons are removed from the system
• The electrode potential becomes more positive

Question 525

What shape are four coordinate complex ions usualy? with examples

Answer 525

• Tetrahedral is the most common shape

- E.g. [CuCl4]2- and [CoCl4]2-

Question 526

When not tertrahedral, what shape are 4 coordinate complexes, with examples

Answer 526

• Some 4 co-ordinate complex ions are square planar in shape, with the ligands arranged at the corners of a square
• E.g. [Ni(NH3)2Cl2] (cis and trans)

Question 527

When does a square planar shape take place in complex ions? example

Answer 527

• These occur in complexes with 8-d electrons in the d subshell.
• E.g. Pt(II), Pd(II), Au(III)

Question 528

What is cis platin used in?

Answer 528

• Cancer treatment in testicular cancer and useful for ovarian, head and neck, and lung cancer
• Extremely toxic

Question 529

Why did they change from cisplatin to carboplatin?

Answer 529

• Improved chemical stability relative to cisplatin due to chelation by cyclobutane dicarboxylic acid
• Essentially equivalent antitumour activity to cisplatin

Question 530

WHat is oaplatin used in?

Answer 530

treatment of colorectal cancer

Question 531

Types of cancer therapy

Answer 531

• Surgery
• Radiotherapy
• Chemotherapy
  o Cytotoxic
  o Targeted
   Anti-endocrine
   Novel targeted agents
• Immuno-therapy
• Gene therapy

Question 532

What is mustard Gas? What came about because of mustard gas?

Answer 532

• Potent vesicant agent that burns eyes, skin and respiratory tract
  Mustard Gas = war gas
  Nitrogen Mustard = anticancer drug

Question 533

How does cisplatin treat cancer?

Answer 533

• The cisplatin binds to DNA and causes a critical structural change n the DNA – a bend of 45 degrees
• This stops cell replication and leads to apoptosis (cell death)

Question 534

Key points of lactic acid

Answer 534

• 2-hydroxy propanoic acid
• Also known as lactic acid
• The second carbon is a chiral center
• The mirror image is non-superimposable
  o One is found in sour milk
  o The other is found in anaerobic respiration

Question 535

What is a fuel cell?

Answer 535

a fuel cell is a device that converts chemical energy into electrical energy, water, and heat through electrochemical reactions.

Question 536

How ddoes a fuel cell work? Are there many or one usually and why?

Answer 536

• Fuel and air react when they come into contact through a porous membrane (electrolyte) which separates them
• This reaction results in a transfer of electrons and ions across the electrolyte from the anode to the cathode
• If an external load is attached to this arrangement, a complete circuit is formed and a voltage is generated from the flow of electrical current
The voltage generated by a single cell is typically rather small (< 1 volt), so many cells are connected in series to create a useful voltage.

Question 537

Differences between a fuel cell and a battery

Answer 537

Hydrogen Fuel Cell –
• Open system
• Anode and cathode are gases in contact with a platinum catalyst
• Reactants are externally supplied, no recharging required
Galvanic Cell (Battery) –
• Closed system
• Anode and cathode are metals
• Reactants are internally consumed, need periodic recharging

Question 538

Differences between a fuel cell and an internal combustion engine

Answer 538

Fuel Cell – 
•	Output is electrical work
•	Fuel and oxidant react electrochemically
•	Little to no pollution produced
Internal Combustion Engine – 
•	Output is mechanical work
•	Fuel and oxidant react combustively
•	Use of fossil fuels can produce significant pollution

Question 539

Similarities between a fuel cell and in internal combustion engine

Answer 539

• Both use hydrogen-rich fuel
• Both use compressed air as the oxidant
• Both require cooling

Question 540

Draw an alkaline fuel cell

Answer 540

check notes

Question 541

What are the half equations, volatages, cell potential, and overall equation for an alkaline fuel cell?

Answer 541

Half Equations –
2H20 (l) + 2e-  H2 (g) + 2 OH- (aq) E = -0.83V
½ O2 (g) + 2e-  2 OH- (aq) E = +0.40V
Cell Potential –
= 0.4- -0.83 = 1.23V
Overall Equation –
H2 + ½ O2  H2O

Question 542

Why Methanol not Hydrogen?

Answer 542

Some new fuel cells use methanol rather than hydrogen as the fuel because

• Liquid methanol is easier to store then hydrogen gas
• Methanol can be generated from biomass

Question 543

Define electrochemical reaction

Answer 543

A reaction involving the transfer of electrons from one chemical substance to another

Question 544

Define electrode

Answer 544

An electrical terminal that conducts an electric current into or out of a fuel cell (where the electrochemical reaction occurs).

Question 545

Define electrolyte

Answer 545

A chemical compound that conducts ions from one electrode to the other

Question 546

What is an electrochemical cell consisted of?

Answer 546

An electrochemical cell consists of 2 electrodes + 1 electrolyte

Question 547

What is rhe equation for the number of optical isomers

Answer 547

• The number of isomers follows the equation: 2n, where n is the number of chiral centres

Question 548

Why are optical isomers called such?

Answer 548

The isomers are called optical isomers as they can rotate plane-polarized light (light which only travels in one plane)

Question 549

WHat makes a 6 co-ordinate compound cis or trans?

Answer 549

The bond angle between the two ligands which are different, e.g. 90’ cis, 180’ trans

Question 550

What is a bidentate ligand? Example

Answer 550

Bidentate – ‘two tooth’ ligands

Most common is ethane-1, 2-diamine

Question 551

Draw [Ni(NH2CH2CH2NH2)3]2+

Answer 551

Check notes

Question 552

Describe and explain EDTA4-

Answer 552

A Hexadentate Ligand

• EDTA4- has 6 lone pairs, each of which can form a co-ordinate bond
• 1 EDTA ion reacts with 1 metal ion
• Ethylenediaminetetraacetic acid (EDTA)

Question 553

What is ligand substitution?

Answer 553

The addition of another ligand to a solution containing the aqua transition metal ion results in a substitution reaction

Question 554

What occurs for ligand substitution?

Answer 554

• One or more ligands is exchanged for another
• A change in colour of the solution is observed
• Sometimes the complex ion changes shape/coordinate number

Question 555

Colour of [Cu(H2O)6]2+, and its colour with dropwise and excess, ammonia and hydrochloric acid

Answer 555

Complex Ion
[Cu(H2O)6]2+

Pale blue solution

Addition of Ammonia

Dropwise: Pale blue precipitation of copper(II) hydroxide
Excess: Blue precipitate redissolves, forming a deep blue solution

Addition of Concentrated HCl

Dropwise: Begins to turn green

Excess: Begins to turn yellow

Question 556

Colour of [Cr(H2O)6]3+, and its colour with dropwise and excess, ammonia

Answer 556

Complex ion

[Cr(H2O)6]3+

Violet solution

Ammonia

Dropwise: Grey/green precipitate

Excess: Precipitate redissolves to produce a purple solution

Question 557

Equation for addition of NaOH to [Cu(H2O)6]2+

Answer 557

[Cu(H2O)6]2+ + 2OH-  [Cu(H2O)4(OH)2]2+ + 2H2O

Question 558

Equation for addition of NH3 to [Cu(H2O)6]2+ in dropwise and excess

Answer 558

[Cu(H2O)6]2+ + 2NH3  [Cu(H2O)4(OH)2]2+ + 2NH4+

[Cu(H2O)6]2+ + 4NH3  [Cu(NH3)4(H2O)2]2+ + 4H2O

Question 559

Equation for addition of HCl to [Cu(H2O)6]2+

Answer 559

[Cu(H2O)6]2+ + 4Cl- [CuCl4]2- + 6H2O

Question 560

Equation for addition of ammonia to [Cr(H2O)6]3+

Answer 560

[Cr(H2O)6]3+ + 6NH3  [Cr(NH3)6]3+ + 6H2O

Question 561

What is the stability constant?

Answer 561

Kstab – the equilibrium constant existing between a transition metal ion surrounded by water ligands and the complex formed when the same ion has undergone a ligand substitution reaction

Like KC but for equilibrias including complex ions

Question 562

Colour of Sc3+

Answer 562

Colourless

Question 563

Colour of Ti2+

Answer 563

colourless

Question 564

Colour of Ti3+

Answer 564

lilac

Question 565

Colour of Ti+4

Answer 565

colourless

Question 566

Colour of Ti5+

Answer 566

colourless

Question 567

Colour of V2+

Answer 567

lilac

Question 568

Colour of V3+

Answer 568

green

Question 569

Colour of V4+

Answer 569

blue

Question 570

Colour of V5+

Answer 570

yellow

Question 571

Colour of Cr2+

Answer 571

blue

Question 572

Colour of Cr3+

Answer 572

green

Question 573

Colour of Cr4+

Answer 573

colourless

Question 574

Colour of Cr5+

Answer 574

colourless

Question 575

Colour of Cr6+

Answer 575

orange

Question 576

Colour of Mn2+

Answer 576

pale pink

Question 577

Colour of Mn+3

Answer 577

colourless

Question 578

Colour of Mn4+

Answer 578

dark purple

Question 579

Colour of Mn5+

Answer 579

colourless

Question 580

Colour of Mn6+

Answer 580

green

Question 581

Colour of Mn7+

Answer 581

lilac

Question 582

Colour of Fe2+

Answer 582

pale green

Question 583

Colour of Fe3+

Answer 583

pale yellow

Question 584

Colour of Fe4+

Answer 584

colourless

Question 585

Colour of Fe5+

Answer 585

colourless

Question 586

Colour of Fe6+

Answer 586

colourless

Question 587

Colour of Co2+

Answer 587

Pink

Question 588

Colour of Co3+

Answer 588

green

Question 589

Colour of Co4+

Answer 589

colourless

Question 590

Colour of Co5+

Answer 590

colourless

Question 591

Colour of Ni2+

Answer 591

green

Question 592

Colour of Ni3+

Answer 592

colorless

Question 593

Colour of Ni4+

Answer 593

colourless

Question 594

Colour of Cu1+

Answer 594

colourless

Question 595

Colour of Cu2+

Answer 595

blue

Question 596

Colour of Cu3+

Answer 596

colourless

Question 597

Colour of Zn2+

Answer 597

colourless

Question 598

[Cu(H2O)6]2+ colour

Answer 598

blue solution

Question 599

Cu(OH)2 colour

Answer 599

Blue precipitate

Question 600

[Cu(NH3)4(H2O)2]2+ colour

Answer 600

Deep blue solution

Question 601

[CuCl4]2- colour

Answer 601

Yellow solution

Question 602

HCl + Cu2+(aq) reaction colour and why

Answer 602

the complex can look green as the reaction is reversible so both blue and yellow species present

Question 603

[Fe(H2O)6]2+ colour

Answer 603

pale green solution

Question 604

Fe(OH)2 colour

Answer 604

Green precipitate, if left in air a reddy brown colour appears (Fe2+ oxidizes to Fe3+)

Question 605

[Fe(H2O)6]3+ colour

Answer 605

Yellow solution

Question 606

Fe(OH)3 colour

Answer 606

Reddy brown precipitate

Question 607

[Mn(H2O)6]2+ colour

Answer 607

Very pale solution

Question 608

Mn(OH)2 colour

Answer 608

Light brown precipitate which darkens in air

Question 609

[Cr(H2O)6]3+ colour

Answer 609

Violet solution

Question 610

Cr(OH)3 colour

Answer 610

Grey/green precipitate

Question 611

[Cr(OH)6]3- colour

Answer 611

Green solution

Question 612

[Cr(NH3)6]3+

Answer 612

Purple solution

Question 613

[Cu(H2O)6]2+ + 4NH3

Answer 613

[Cu(H2O)6]2+ + 4NH3  [Cu(NH3)4(H2O)2]2+ + 4H2O

Blue solution Deep blue solution

Question 614

[Co(H¬2O)6]2+ + 6NH3

Answer 614

[Co(H¬2O)6]2+ + 6NH3  [Co(NH3)6]2+ + 6H2O

Question 615

[Cu(H2O)6]2+ + 4Cl-

Answer 615

[Cu(H2O)6]2+ + 4Cl-  [CuCl4]2- + 6H2O

Blue solution Yellow/green solution

Question 616

[Co(H2O)6]2+ + 4Cl-

Answer 616

[Co(H2O)6]2+ + 4Cl-  [CoCl4]2- + 6H2O

Pink solution Blue solution

Question 617

Cu2+ + 2OH-  Cu(OH)2 colours

Answer 617

Cu2+ + 2OH-  Cu(OH)2

Blue solution  blue precipitate

Question 618

Mn2+ + 2OH-  Mn(OH)2 colours

Answer 618

Mn2+ + 2OH-  Mn(OH)2

Very pale pink solution  pale brown precipitate

Question 619

Fe2+ + 2OH-  Fe(OH)2 COLOURs

Answer 619

Fe2+ + 2OH-  Fe(OH)2

Green solution  green precipitate

Question 620

Fe3+ + 3OH-  Fe(OH)3 colours

Answer 620

Fe3+ + 3OH-  Fe(OH)3

Yellow/brown solution  brown precipitate

Question 621

Cr3+ + 3OH-  Cr(OH)3 colours

Answer 621

Cr3+ + 3OH-  Cr(OH)3

Green solution  green precipitate

Question 622

[Mn(H2O)6]2+ + 2NH3

Answer 622

[Mn(H2O)6]2+ + 2NH3  Mn(H2O)4(OH)2 + 2NH4+

Question 623

[Fe(H2O)6]3+ + 3NH3

Answer 623

[Fe(H2O)6]3+ + 3NH3  Mn(H2O)3(OH)3 + 3NH4+

Question 624

Cr(H2O)3(OH)3 + 3OH-

Answer 624

Cr(H2O)3(OH)3 + 3OH-  [Cr(OH)6]3- + 3H2O

Green precipitate  green solution

Question 625

Cr(H2O)3(OH)3 + 3H+

Answer 625

Cr(H2O)3(OH)3 + 3H+  [Cr(H2O)6]3+

Green precipitate  green solution

Question 626

Cr(OH)3(H2O)3 + 6NH3

Answer 626

Cr(OH)3(H2O)3 + 6NH3  [Cr(NH3)6]3+ + 3H2O + 3OH-

Green precipitate  purple solution

Question 627

Cu(OH)2(H2O)4 + 4NH3

Answer 627

Cu(OH)2(H2O)4 + 4NH3  [Cu(NH3)4(H2O)2]2+ + 2H2O + 2OH-

Blue precipitate  deep blue solution