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Chemistry 12 > Unit 3 > Flashcards

Unit 3 Flashcards

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1
Q

equilibrium reaction

A

the concentrations of all reactants and products remain constant with time. it is highly dynamic as the reaction never stops. there is no limiting reactants

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2
Q

stoichiometric reaction

A

proceed to completion therefore the amount of limiting reactant the remains is negligible

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3
Q

equilibrium lies to the right

A

it favours the products, concentrations of reactants is never 0

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4
Q

equilibrium lies to the left

A

it favours the reactants

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5
Q

how can you disturb the “system”

A

temperature, change volume, changing something will stress the system and change the equilibrium

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6
Q

when is equilibrium reached

A

when rate of forward reaction equals the rate of reverse reaction

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7
Q

the equilibrium expression (K or Kc)

A

is a thermodynamic quantity
eqn is products to power of coefficients/reactants to power of coefficients
it has no units and concentrations are always in Mol/L
must always be accompanied by a balanced them eqn

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8
Q

reverse K

A

always needs a new equation! it is written as K^-1

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9
Q

when the reaction is multiplied by a factor

A

multiply the exponents by the factor.

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10
Q

equilibrium position

A

K is always the same for a certain temperature but the concentrations might change.
each equilibrium concentration is called the position and there is an infinite number of positions at a certain temperature

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11
Q

when K>1

A

the system will favour the products
equilibrium lies to the right
K»10^3, reaction will essentially be products

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12
Q

when K<1

A

the system will favour the reactants
equilibrium lies to the left
K«10^3, reaction will essentially be reactants
this would be a non-reaction but that is a lie!

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13
Q

equilibrium and kinetics

A
  • size of k and time are not directly related
  • time to reach equilibrium depends on the rate of reaction which is determined by kinetics
  • size of K is determined by thermodynamics factors such as difference in energy between products and reactants
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14
Q

heterogeneous equilibrium

A

not all reactants are in the same state
- pure solis and liquids have concentrations that don’t change and are constant so they can be ADDED to the constant equilibrium value (remove them from the eqn!)

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15
Q

decrease the temperature of equilibrium

A

that decreases the energy to be more endothermic, the equilibrium shifts right and makes the reaction more exothermic to make up for the energy lost

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16
Q

increase the temperature of equilibrium

A

that increases the energy to be more exothermic, the equilibrium shifts left and makes the reaction more endothermic to make up for the energy

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17
Q

increase the pressure

A

increase the pressure, the system respond by decreases the pressure by shifting to the side with less gas moles therefore shifting (left/right)

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18
Q

decrease the pressure

A

decrease the pressure, the system respond by increases the pressure by shifting to the side with more gas moles therefore shifting (left/right)

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19
Q

effect of concentration

A

has no effect on k,
if you increase the concentration, the system respond by decreasing the concentration of that reactant and shifting (left/right)

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20
Q

reaction quotient, Q

A

calculate the same way as K but it is used if you don’t know the concentrations at equilibrium

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21
Q

Q>K

A

too many products and shifts left to reach equilibrium

reaction will shift to reach K

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22
Q

Q

A

too many reactants

shifts right to reach equilibrium

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23
Q

Q=K

A

at equilibrium

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24
Q

ICE tables

A

initial
change
equilibrium
table can be made using moles, concentrations, pressures

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25
Q

le chatalier’s principle

A

if a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change

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26
Q

how to change the pressure of equilibrium

A
  • add or remove a gaseous reactant or product (addition of an inert gas has no effect on the concentrations or partial pressures of the reactants or products) so NO CHANGE
  • add another inert gas that does not react
  • change the volume of the container
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27
Q

when you change the volume of the container

A
  • concentrations of both reactants and products are also changed
  • when volume is reduced, pressure increases and system decreases its own pressure
  • it decreases the total number of gas moles in the system
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28
Q

by changing the temperature, you affect…

A

the value of K, and you treat energy as a reactant or product

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29
Q

adding a catalyst

A

does not change the equilibrium in any way, it just changes the time it takes to get to equilibrium

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30
Q

when a salt dissolves

A

ions get surrounded by water. its never entirely soluble so an equilibrium occurs

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31
Q

more soluble ionic compounds

A

the equilibrium lies to the right

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32
Q

insoluble ionic compound

A

the equilibrium lies to the left (more solid)

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33
Q

when writing ionic compounds

A

always write the ionic compound (solid) on the left

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34
Q

Arrhenius acid def

A

anything that has an H+

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35
Q

arrhenius base

A

donates OH-

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36
Q

bronsted-lowry acid

A

anything that donates an H+

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37
Q

bronsted-lowry base

A

accepts an H+

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38
Q

how are acids and bases defined? (water)

A

defined by their reaction with each other. H2O can’t be an acid or a base as it depends on the reactant that is associated with it

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39
Q

bronzed lowry acid base equilibrium

A

the products from the reaction can act as acids and bases themselves.

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40
Q

conjugate

A

results from the transfer of H+

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41
Q

Acid to conjugate….

A

base

take away an H+ ion from acid to form the conjugate base

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42
Q

base to conjugate…

A

acid

add an H+ ion from base to form conjugate acid

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43
Q

remember… charges

A

loosing an H, leaves the thing with a -ve charge

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44
Q

amphiprotic/amphoteric

A

bronstead lowry term to mean it can act as an acid or a base
ex.
water and HCO3-

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45
Q

hydronium

A

H3O+

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46
Q

pH

A

measures the concentration of H3O+ in a solution
usually positive and unitless
change by one is a 10 fold change in H+

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47
Q

pH eqns

A

pH= -log[H3o+]

[H+]=10^-pH

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48
Q

pOH eqns

A

1.0x10^-14 = [H+][OH-] at 25 celsius
pOH= -log[OH-]
pH + pOH = 14

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49
Q

significant figures in log

A

only the stuff after the number (AKA decimals) are significant

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50
Q

strong acids

A

completely dissociate in water, not in equilibrium

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51
Q

strong acids you should know

A

HCL, HBr, HI, H2SO4, HNO3

HF has a small radius and too strong intermolecular force to ionize!

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52
Q

weak acid

A

partially dissociates, it is in equilibrium

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53
Q

K a

A

acid dissociation constant, a stronger acid will have a larger Ka
more H3O+ because H+ is dissociated
the equilibrium lies more towards the right compared to a weaker acid

54
Q

ka eqn

A

= [H3O+][A-]/[HA]

HA is the acid

55
Q

strong bases

A

completely ionize

56
Q

strong bases to know

A

group 1 and 2 + OH-

57
Q

Weak bases

A

partially ionize, in equilibrium

58
Q

Kb

A

a stronger base will have a larger Kb

a stronger base will have an equilibrium which lies more towards the right

59
Q

auto ionization of water

A

water can react with itself to make hydronium
kw at 25 degrees celsius= 1.0x10^-14

This is an endothermic reaction in the forward direction. increasing the temperature decreases the pH more H30+

60
Q

kb eqn

A

[HA][OH-]/[A-]

61
Q

KaKb relationship

A

Ka times Kb= [H3O+][OH-]= kw=1.0 x 10^-14

that means that if ka is large then kb is small

62
Q

in a neutral solution

A

hydronium concentration equals OH- concentration

63
Q

in an acidic solution

A

hydronium concentration is more than OH-

64
Q

in a basic solution

A

hydronium concentration is less than OH-

65
Q

-logkw

A

pH+pOH

66
Q

in calculations, if the concentration of HA is much greater (>100) than Ka then

A

you can make the approximation that the concentration of HA initial is equal to the concentration of HA at equilibrium. must be stated each time you use this

67
Q

if b initial> 100Kb

A

assume B initial equals B equilibrium

68
Q

ionic compounds and acidic/ basic properties

A

determining whether acid/base is weak acid/natural salt/weak base

69
Q

hydrolysis (salts)

A

certain ions react with water to change pH
salts that are made from the conjugate base of a strong acid and the cation of a strong base do NOT react with water (no change)

70
Q

strong bases that don’t react with water

A

group 1 and 2 cations come from strong bases

71
Q

conjugate bases that don’t react with water

A

BR-, I-, Cl-, NO3-, CLO4-, CLO3- come from strong acids

72
Q

ions and pH

A

conjugate base of weak acid will react which adds OH+ and changes the pH (basic)

conjugate acids of a weak base will react with water to change its pH (acidic)

73
Q

two ions that hydrolyse

A

a salt can be made from the anion of a weak acid and the cation of a weak acid
this is ammonium acetate and ammonium formate
compare relative strength of their ka and kb

74
Q

buffer

A

resist pH after the addition of OH- or protons

75
Q

buffered solutions contain

A 
a weak acid and its salt (its conjugate)
 HF and NaF
OR 
a weak base and its salt (its conjugate)
NH3 and NH4CL
76
Q

buffers in word problems

A

the concentration is on the other side of the equilibrium as well, not 0

77
Q

buffer mechanism

A

buffer solutions needs a large quantity of weak acid and conjugate base (or other way) compared to the amount of strong acid or base added

78
Q

buffer mechanism for hydroxide ions

A

hydroxide ions will not accumulate as they will react with the weak acid or conjugate base
OH- + HA –> A- + H2O
OH- +BH+ –> B + H20
*stoichiometry arrows

79
Q

buffer mechanism for protons

A

protons will react with conjugate base or weak base
H+ + A—> HA
H+ + B –> BH+

80
Q

pH and buffers

A

will be determined by the ratio of HA to A-
if the amounts of HA and A- originally present is large then any small changes shouldn’t affect the ratio by much therefore shouldn’t affect the pH by much

81
Q

how to create a buffer

A

1) combine a weak acid and its salt (or base ad its salt)
- should be about equal moles
- ex. NH3 and NH4Cl (source of NH4+)
2) taking a weak acid and combing it with a lower amount of stronger base (or weak base and strong acid)
- this is more realistic
- take about half amount of strong acid to weak base

82
Q

titration

A

used to determine the amount of acid or base in a solution of unknown concentration

83
Q

end point

A

indicator changes colour

84
Q

equivalence point

A

when the moles of acid = moles of base if monoprotic

85
Q

titration curve

A

a graph of pH as a function of volume of titrant added
partly sigmoidal
equivalence point is the POI of the curve

86
Q

how can end point show equivalence point

A

because the tangent of the equivalence point and end point are essentially the same as the tangent is vertical (same x)

87
Q

drawing a titration curve of strong acid and strong base

A

strong acid starts at pH 1-2
weak acid is at 3-4 pH
weak base is 9-11 pH
strong base is 11-12

curve is symmetrical around equivalence point
pH rises sharply at equivalence point
equivalence point is 7 pH

88
Q

weak acid and strong base half equivalence

A

ka= {h30+] at half equivalence (1/2 moles required for equivalence) and creates a buffer with two curves on graph

89
Q

weak acid and strong base initial

A

HA

90
Q

weak acid and strong base equivalence

A

A- so basic pH

91
Q

weak acid and strong base after equivalence

A

OH- basic pH

92
Q

titration of weak acid and strong base graph

A

initial pH is much higher (around 3)
pH equivalence point is greater than 7
curve is not symmetrical around equivalence point
after equivalence the curve flattens out at a high value (pH of strong base)

buffer region:
- will be flat until the equivalence point
half of the equivalence point volume (x six) is the half equivalence point

93
Q

weak base and strong acid initial

A

B

94
Q

weak base and strong acid half equivalence

A

B + BH+ –> buffer

95
Q

weak base and strong acid equivalence

A

BH+ so acidic

96
Q

weak base and strong acid after equivalence

A

H+

97
Q

weak acid and weak acid curve

A

initial pH is high (3) from weak acid
addition of base makes pH rise gradually
rise around the equivalence point is less sharp
after equivalence point, the curve flattens out at relatively low pH (around 9ish)
hard to find the equivalence point (POI)

98
Q

le chatelier statement

A

if I _(increase/decrease)___ the system responds by ___(doing opposite)__ by shifting __(left/right)___

99
Q

indicator

A

a weak base that changes colour. they have a Ka

when its colour A, its an acid
Colour b is the dissociated side with the H+ ion and is the base

the Ka/H+ ratio will show the indicator colour at a pH

100
Q

when to see a indicator colour change

A

tipping point of colour a and b is ka= [H+]
visible change is noticed when ratio of Ln-/ HLN = 0.1 to 10
colour change also is pKA = +/_ 1 because at half equivalence, Ka is equal to the [H+]???

101
Q

indicator for strong acid and strong bas

A

pretty much anything

102
Q

indicator for weak acid/ strong base

A

any indicator that is basic

103
Q

indicator for weak base and strong acid

A

acid indicator

104
Q

molar solubility

A
  • is the amount of moles that dissolve

- it is the x value in the ICE table!

105
Q

for le chatilier

A

think about what is in K, temp, volume moles and stuff

106
Q

KSP is super small

A

so anything that has an initial volume the isn’t zero is insignificant in the table

107
Q

acid to conjugate base

A

loose an H+

108
Q

conjugate base to acid

A

add an H+

109
Q

base to conjugate acid

A

add an H+

110
Q

conjugate acid to base

A

loose an H+

111
Q

weak acids to kno

A

CH3COOH, HF, HNO2, NH4+

112
Q

calculating pH of acid/base strong

A

its just stoichiometry of the concentration as its monoprotic 1-1

113
Q

Basic anions in water

A

CH3CO2-, F-, CO3 2-, S 2- (conjugate bases)

114
Q

acidic cations in water

A

Al3+, NH4+ (conjugate acid)

115
Q

salt can be made from anion of weak acid and cation of weak base

A

just determine the strength of their relative Ka and Kb

whichever is larger

116
Q

titration of strong acid and strong base

A

1) initial is just -log of thing being titrated
2) eon with H+ and OH- makes water (stoichiometry) then divide by the full volume after subtracting the LR
3) equivalence point is pH 0
4) same as 2

117
Q

titration of weak acid and strong base

A

1) weak acid so its in equilibrium with water, use Ka to get H3O+ concentration
2) half equivalence point is a buffer. first to the titrated thingy with strong base (OH-) in eon, then get concentration of ion, then do that ion concentration in equilibrium which is now a buffer and is Ka= [H3O+]
3) equivalence. first do the weak acid and OH- stoichiometry, then get ion concentration then do it in equilibrium, and get pH
4) just stoichiometry as OH- is in excess (subtract LR then do pH)

118
Q

titration of a weak base and strong acid

A

1) equilibrium of weak base, then use Kb
2) half equivalence so stoichiometry, ion concentration, equilibrium to get OH- then pOH then pH as pOH= pKb at half equivalence
3) stoichiometry to get ion, the use equilibrium to get pH
4) stoichiometry as H+ is in excess (- Lr)

119
Q

titration of weak base and strong acid at half equivalence

A

pOH= pKb

120
Q

when doing acidic/basic salt calculations

A

decide first if anion is going to react, then cation, then do equilibrium with only that iON AND DO THE PH

121
Q

when asked about rate and equilibrium constant

A

Equilibrium and rate are measures of two different things and are not related. The ____reaction____ occurs to a __large/small__ extent and so the reaction has a _same as above (large/small)__ equilibrium constant. It is quite a __slow/fast__ reaction under normal conditions of temperature and pressure and so rate of reaction is __same as above (slow/fast)__.

122
Q

when you change the volume of vessel or increase/decrease temperature (something that changes the whole equilibrium)

A

all concentrations drop immediately, then proceed to increase/decrease as you would predict for the new equilibrium and then flatten out

when volume is suddenly decreased, it gives a greater concentration of the solution and temporarily darkens the colour. then, the solution lightens up in favour of equilibrium as concentrations level out

123
Q

when asked about ions and equilibrium le chatelier

A

as you add water, it dilutes all species, and the reaction will proceed to side with the greater number of moles (as it will be ‘diluted’ to the greatest extent).

remove water will shift to more solid form

124
Q

auto ionization of water

A

transfer of proton from one water molecule to another to produce hydroxide ion and hydronium ion

can happen in liquid ammonia as well, it turns to NH4+ and NH2-

125
Q

water properties

A

water is a stronger base than the conjugate base of a strong acid but a weaker base than the conjugate base of a weak acid

126
Q

to check if a reaction is endo or exo based on K value changes over temperature

A

see if K increases or decrease with increase or decrease of temperature. then think of le chatelier and which way equilibrium will shift when you change it and think of ratio of products to reactants to find if its endo or exo

127
Q

when the concentration of strong acid is super super low (like 10^-10)

A

its just pH 7

128
Q

when H+ is added to a buffered solution

A

reacts with the weak base

H+ + A- –> HA
or H+ + B –> BH+

129
Q

when OH- is added to a buffered solution

A

reacts with the weak acid

OH- + HA –> A- + H20
OH- + BH+ –> B + H20

130
Q

when there is a titration and indicator changes colour quickly

A

that means that titration solution (in burette) is very concentrated and no concentration can be determined

131
Q

h2CO3 and sO3

A

dissociates!! to water and carbon dioxide!

132
Q

when asked about the indicator and changing colour pH

A

look at the Ka of the indicator, the do the ratio of HIn/Ln- the top ratio should be 10 times greater than the bottom so multiply the Ka by 10 and find pH

Chemistry 12 (6 decks)

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