5.1, Kp, Ka, Kw ect. Flashcards
1
Q
What is the definition of an equilibrium law?
A
The equilibrium law states that for the equilibrium aA + bB ⇔ C + dD
Kc = [C]c[D]d/[A]a[B]b
2
Q
What is the definition of a homogeneous equilibrium?
A
A homogeneous equilibrium is an equilibrium in which all the species making up the reactants and products are in the same physical state.
3
Q
What is the definition of a heterogeneous equilibrium?
A
is an equilibrium in which species making up the reactants and products are in different physical states.
4
Q
How are the three meanings of K presented differently?
A
The equilibrium constant is represented by a capital K, followed by a subscript letter to indicate which equilibrium constant you are referring to a lower case k represents the rate constant and K is on its own is the symbol for potassium.
5
Q
What would Kc, and it’s units be for the reaction 2SO2 (g) + O2 (g) ⇔ 2SO3(g) ?
A
- Kc = [SO2(g)]2/ [SO2(g)]2 [O2(g)]
- Units (moldm-3)2/ (moldm-3)3 = dm3mol-1
6
Q
What needs to be known to determine concentrations at equilibrium?
A
To determine a value of Kc, the concentration of reactants and products need to be known.
Chemists determine the concentrations in a number of ways including a titration and using a colorimeter.
7
Q
What are the drawbacks of using titrations to determine concentrations at equilibrium?
A
- Titrating a reactant or product against a known concentration will show how much of the reactant or product os present.
- However the reactant or product has to be removed from the reaction mixture, and this will alter the position of equilibrium.
- Other substances may also be present, affecting the results (catalysts).
8
Q
What can monitor the concentrations of product molecules throughout reactions to determine the value of Kc?
A
A colorimeter can be used to monitor the concentrations of a reactant or product throughout.
This works because coloured substances alter the amount of light that can pass through.
9
Q
Calculating unknown equilibrium concentrations;
What can you tell from a balanced equation?
A
- the reacting quantities needed to prepare a required quantity of a product
- the quantities of products formed by reacting other known quantities of reactants.
10
Q
What are the steps involved in calculating Kc from unknown equilibrium concentrations?
A
- Find the concentrations of all three components in the equilibrium mixture
- Work out the equilibrium amounts of the reacting molecules.
- You can use the balanced equation and stiochiochemistry to find out the moles of a reactant used a product formed.
- Finally, convert the amounts to concentration.
Follow these steps on the example on page 25.
11
Q
What are the steps of calculating Kc from equilibrium concentrations?
A
- Calculate the moles at equilibrium by giving the moles at equilibrium by the volume of substance or container.
- Write the expression for Kc and work out the units. Then calculate Kc under these conditions.
Example on pg 24
12
Q
If you are given equilibrium amounts in moles and you know the total volume, what must you calculate first?
A
The concentrations.
concentrations = no. moles / volume in dm3
13
Q
What is affected by an equilibrium being heterozygous, instead of homozygous?
A
When a heterogeneous equilibrium is present, molar concentrations for solids and pure liquids do not change because their volume remains constant.
This means that the expression for Kc changes:
- the concentrations of solid substances are not included in the expression
- the concentrations of pure liquids are not included in te expression
14
Q
What is the expression for Kc for the reaction P4 (s)+ 5O2 (g) ⇔ P4O10 (s)?
A
Kc = [O2]5
15
Q
What is a mole fraction?
A
The mole fraction of a substance is a measure of how much of a given substance is present in a reaction mixture.
16
Q
What is the partial pressure?
A
The partial pressure of a substance is the pressure an individual gaseous substance would exert if it occupied a whole reaction vessel on its own.
17
Q
What symbol is given to the mole fraction?
A
It is given the symbol X.
18
Q
How is the mole fraction of a substance calculated?
A
mole fraction, XA = number of moles of substance A / total number of moles of all substances
19
Q
What is the pressure of any reaction mixture involving gases within a sealed vessel the sum of?
A
The pressure of any reaction mixture involving gases within a sealed vessel is effectively the sum of being exerted by each of the gaseous substances involved.
The amount of pressure being exerted by an individual species within a reaction vessel is called the partial pressure.
20
Q
What symbol is the partial pressure given?
A
partial pressure is given the symbol, P, followed by a subscript denoting the species to which is refers.
21
Q
How do you calculate partial pressure from the total pressure?
A
If we know the total pressure of a reaction along with the mole fraction of a given substance, we can calculate the partial pressure of that substance.
For substance A: partial pressure, PA = mole fraction x total pressure.
22
Q
What equilibrium constant is used for reactions involving gases?
A
Equilibrium expressions can be written using partial pressures instead of concentrations.
23
Q
What would the equilibrium constant, Kp, be for the reaction aA + bB ⇔ cC + dD?
A
kp = (pC)c (PD)d / (PA)a (PB)b
24
Q
A
25
How does the expression for kp change in a heterogenous reaction?
* Solids will not be included in the expression
* pure liquids will not be included in the expression
26
What does the magnitude of the eqilibrium constants kp and kc indicate?
The extent of a chemical reaction.
A equilibrium constant, k, with a value of 1 would indicate of 1 would indicate that the position of equilibrium is halfway between reactants and products.
27
What side of the reaction does equilibrium favour when K is less than 1?
* the reaction is reactant favoured
* the reactants on the left-hand side predominate at equiibrium.
28
What side of the reaction does equilibrium favour when kp is less than one?
When kp is less than 1:
* the reaction is reactant-favoured
* the reactants on the left hand side predominate at equilibrium
29
How do changes in temperature affect the enthalpy value?
* An increase in temperature shifts the position of equilibrium in the endothermic direction (Δ*H* + ve)
* A decrease in temperature shifts the position of equilibrium in the exothermic direction (Δ*H - ve)*
30
How do the ΔH values for the forward and reverse reactions in equilibrium compare?
They have the same magnitude but have opposite signs.
31
What can cause a shift in the position of equilibrium?
Shifts in the position of equilibrium are actually controlled by the rate constant, which changes its value only with changes in temperature. When any other changes occur to conditions, K will remain constant. The way that K changes is linked to Δ*H*.
32
How would equilibrium be affected by the reaction being exothermic?
* the equilibrium yield of the products, on the right-hand side, increases.
* the equilibrium yield of the reactants, on the left-hand side, decreases.
33
How would equilibrium be affected by the reaction being endothermic?
* the equilibrium yield of product, on the right-hand side, decreases
* the equilibrium yield of the reactants, on the left-hand side, increases
34
How do changes in concentration and pressure affect K?
Although the eqilibrium constant, *k,* is altered by changes in temperature, the value of *K* i s unaffected by changes in concentration and pressure.
35
For the reaction N2O4 (aq) ⇔ 2NO(g) Kc = 12.8 mol dm-3, If the concentration of [N2O4(g)] is doubled from 2.00 mole dm-3 to 0.400 mol dm-3 how would Kc be affected?
The equilibrium position must shift to restore this ratio to the Kc value of 12.8 mol dm-3.
The system must:
* increase NO2 (g)
* decrease N2O4(g)
This causes a shift in the equilibrium poisition from left to right.
Will also be the same for Kp as it stays constant when concentrations, or the equivalent partcial pressures, are changed.
36
37
What happens to the concentrations of a gaseous reactants and products if the pressure is doubled?
If the pressure is doubled, the concentrations of both the reactants and products (g) will double because there will be the same number of partciales in half the space.
38
If the pressure of a reaction at equilibrium is doubled how will the equilibrium shift to restore the value of Kc for the reaction: N2O4 (aq) ⇔ 2NO(g) ?
* Decrease NO2(g) on the top
* Increase N2O4(g) on the bottom
This causes a shift in the equilibrium position from right to left.
the same will be true for the rate constant KP.
39
How does the presence of a catayst affect K?
It doesnt.
Catalysts affest the **rate** of a chemical recation, but not the position of equilibrium.
Cataylsts speed up both the forward and reverse reactions in the equilibrium by the same factor. Equilibrium is reached more quickly, but the equilibrium position, and hence the value of the equilibrium constant, is unchanged by the action of a catayst.
40
What is a Brønsted-Lowry acid?
What is a **Brønsted-Lowry acid** is a proton, H+, donor.
41
What is a Brønsted-Lowry base?
A Brønsted-Lowry base is any substance that can accept a proton.
42
Why are acids and bases refered to acid Brønsted-Lowry acids or bases?
Because they behave in the way described by the Brønsted-Lowry model.
43
Chemists have studied the behavior of acids and bases over many years, what discovery was made in the 1770's?
In the 1770's many chemist were investigating air and the gasses within it. Joseph Priestly and karl Scheele both reported findings that suggested the existence of the gas we now know as oxygen. Another chemist, Antoine-Laurent de Lavoisier, found, among othe rthings, that this gas was important in the rusting of metal and in 1778 he proposed that it was the source of acidity. Lavoisier as wrong, but it was an early step towards the understanding of acids.
44
Chemists have studied the behavior of acids and bases over many years, what discovery was made in the 1815's?
Humphrey Davy showed that some acidic substances, such as HCl, did not actually contain xoygen.
45
Chemists have studied the behaviour of acids and bases over many years, what discovery was made in the 1830's?
In 1832 Justus Liebig defined an acid as a substance containing hydrogen that could be replaced by a metal - very close to today's definition.
46
Chemists have studied the behaviour of acids and bases over many years, what discovery was made in the 1880's?
In the late 1880's Svante Arrhenius proposed that acids dissociated in water to form hydrogen ions, H+ and that bases dissociated in water to form hydroxide ions, OH-. This is true for many acids and bases, and they can be referred to as Arrhenius acids and bases. However, the model breaks down when acids and bases are not solutions in water or when bases are not soluble hydroxides.
47
When was the Brønstead-Lowry definition proposed, by who?
In 1929 the Brønsted-Lowry definition was proposed by the Danish chemist Johannes Brønstead and the British chemist Thomas Lowry.
48
What theory has been suggested as a step on from the Brønstead acid theory?
Even the Brønstead-Lowry acid can break down, such as for some solvents. The American chemist Gilbert Lewis proposed a more general theory that describes an acid (known as a Lewis acid) as an electron-pair donor. (the model isn't needed for the exam)
49
Brønstead Lowry acids;
What happens when an HCl is added to water?
HCl(g) → H+ + Cl-(aq)
The HCl has donated an H+ ion because it is a Brønstead-lowry acid.
50
How does ammonia dissociate?
NH3(aq) + H+(aq) → NH4+(aq)
The ammonia has accepted a proton (H+) - it is a Brønsted-Lowry base.
51
Write the ionic equation for HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)?
H+(aq) + OH-(aq) → H2O(l)
52
Why can different acids release different numbers of protons depending on their formulae? Give examples.
* HCl is a monobasic acid because each molecule can release one proton;
1. HCl (aq) → H+(aq) + Cl-
* H2SO4 is a dibasic acid because each molecule can release two protons, this is done in two stages.
1. H2SO4(aq) → H+(aq) + HSO4- (aq)
2. HSO4-(aq) → H+(aq) + SO42-(aq)
* H3PO4 is a tribasic acid because each molecule can release three protons, and is done in three stages.
1. H3PO4 (aq) → H+(aq) + H2PO4-(aq)
2. H2PO4- (aq) → H+(aq) + HPO42-(aq)
3. HPO42-(aq) → H+(aq) + PO43-(aq)
53
What are conjugate acid-base pairs?
An acid-base pair is a set of two species that transform into each other by gain or loss of a proton.
54
Give an example of a **conjugate acid-base pair**, for the dissociation of nitrous acid, HNO2.
55
_Conjugate acid-base pairs;_
What does acid base equilibria involve?
Acid base equilibria involves two acid-base pairs
56
State the conjugate acid-base pairs for the equilibrium of the dissociation of notrous acid, HNO2 in water:
HNO2(aq) + H2O(l) ⇔ H3O+(aq) + NO2-(aq)
HNO2 and NO2- differ by H+ and make up one acid-base pair: acid 1 and base 1.
H3O+ and H2O differ by H+ and make up a second acid-base pair: acid 2 and base 2.
57
What is the definition for an alkali?
An alkali is a base that dissloves in water forming OH-(aq) ions.
58
What is the definition for neutralisation?
Neutralisation is a chemical reaction in which an acid and a base react togther to produce a salt and a water.
59
What is the definition for a strong acid?
A strong acid is an acid that completely dissociates in solution.
60
What is the definiton for a weak acid?
A weak acid is an acid that partially dissociates in solution.
61
What is the definition for the acid dissociation constant?
*K*a, of an acid HA is defined as Ka = [H+(aq)][A-(aq)] / [HA(aq)]
62
What is the definition for *pKa* ?
*pKa* = - log10 *Ka*
63
What is the definition for Ka?
Ka = 10-pka
64
What reaction occurs when aqueous acids take part in typical acid-base reactions with carbonates, bases and alkalis.
In all of these reactions, a **neutralisation reaction** occurs and water is fomred as one of the products.
65
What is the equation and ionic equation for hydrochloric acid, HCl, reacting with solid carbonates?
_Symbol_
2HCl(aq) + CaCO3(s) → CaCl2(aq) + CO2(g) + H2O(l)
_Ionic_
2H+ (aq) + CaCO3(s) → Ca2+ (aq) + CO2(g) + H2O(l)
66
When acids release protons in water, what can they be accepted by?
When acids release protons in water, they are ususally accepted by the water to form a hydromium ion, H3O+ (sometimes called the oxonium ion).
67
Give an example of hydrchloric acid dissociating and forming an oxonium ion.
HCl(aq) + H2O(l) → H3O+(aq) + Cl-(aq)
But as H+ and H3O+ can be used interchangaebly when describing aqueous acid solutions, and H+ is usually used in ionic equaitions for acid-base reactons.
This is why the above equation would be simplified to: HCl(aq) → H+(aq) +Cl-(aq)
68
What are the equation and ionic equation for hydrochloric acid, HCl, reacting with aqueous carbonates?
_Symbol equation_
2HCl(aq) + Na2CO3(aq) → 2NaCl + CO2(g) + H2O(l)
_Ionic equation_
2H+(aq) + CO3(s) → CO2(g) + H2O(l)
If the carbonate is in solution, the final ionic equation simplifies beacause the carbonate is dissociated.
69
What are spectator ions?
In ionic equations species which remain unchanged are being cancelled out, these are specator ions.
70
Why is it important to include state symbols in ionic equations?
it is especially important when an acid reacts with a solid, as solids will not be dissociated into ions.
71
What do aqueous acids form when they react with bases?
Give an symbol and ionic equation.
Aqueous acids react with alkalis, forming salt and water.
**Full equation:** 2HNO3(aq) + MgO(s) → Mg(NO3)2(aq) + H2O(l)
**Ionic equations:** 2H+(aq) + MgO(s) → Mg2+(aq) + H2O(l)
72
What do aqueous acids form when they react with alkalis?
## Footnote
Give a symbol equation.
Aqueous acids react with alkalis, forming a salt and a water.
Full equation: H2SO4(aq) + 2KOH(aq) → K2SO4 (aq) + 2H2O (l)
73
What do aqueous acids form when they react with a metal?
The reaction of an acid with a metal is a redox reaction, so does not fit in with the acid-base model. however, they are common acid reaction.
The general equation is;
acid + metal → salt + hydogen
74
Give the full, all ions and ionic equation for the recation of hydrochloric acid and magnesium to show the process of cancelling out unchanged species to create a ionic equation.
75
What may be an acception to acids reacting with metals to form a salt and hydrogen?
Some acids such as sulphuric an nitric acids, are powerful oxidising agents and other reactions may also take place, especially when acids are concentrated. It is best to sterr clear of clear of H2SO4 and HNO3 when giving examples of acid reactions with metals.
76
What are the common strong acids?
These acids 100% dissociate in an aqueous solution.
77
Show the dissociation of ethanoic acid in water, and describe how a weak acid dissociates.
CH3COOH-(aq) ⇔ H+(aq) + CH3COO-(aq)
* The equilibrium position lies well over to the left
* There are only small concentrations of dissociated ions, H+(aq) undissociated CH3COO-(aq) compared with the concentration of undissociated CH3COOH(aq)
We can actually say that CH3COO- is a very good base - it is very good at accepting the dissociated H+ back.
78
What is the actual extent of acid dissociation meausred by?
An equilibrium constant calle dthe acid dissociation constant, *Ka.*
79
How do the terms 'strong' and 'weak' describe the extent of dissociation?
* a large *K*a value indiactes a large extent of dissociation - the acid is strong.
* a small *K*a value indicates a small extent of dissociation - the acid is weak.
80
How is pKa related to ka?
The range of values of Ka are vast, for this reason a logarithmic scale is used to decribe them. pKa is a more manageable number than Ka, and the two quamtities are related as follows;
* pka = - log10 Ka
* Ka = 10-pka
* A low value of Ka matches a high value of pKa.
* A high value of Ka matches a matches a low value of pKa.
* The smaller the pKa value, the stornger the acid.
81
What is the defintion for pH?
-log[H+(aq)]
82
What is the definition for [H+(aq)]?
10-pH
83
Who came up with the pH scale?
Søren Sørensen, a Danish chemist.
84
Describe the work of Søren Sørensen, on creating the pH scale.
Søren Sørensen, a Danish chemist, carried out research that involved measuring hyrogen ion concentrations. These numbers could be cumbersome to measure and record.
Aqueous solutions have concentrations of H+(aq) ions in the range:
* from about 10 mol dm-3 (101 mol dm-3)
* to about 0.0000000000000001 mol dm-3 (10-15 mol dm-3).
Sørensen came up with an easier way to measure and denote these concentrations - a logarithmic scale called the pH scale.
85
What are are [H+] values and corrosponding pH values that make up the pH scale.
86
[H+(aq)] What do the [] stand for?
[] are shorthand for 'the concentration of'
87
Describe the relationship between pH and [H+(aq)].
* A low pH value means a large [H+(aq)].
* A high pH value means a small [H+(aq)].
* A pH change of 1 changes [H+(aq)] by 10 times.
* An acid with a pH of 2 contains 1000 times the [H+(aq)] of an acid with a pH of 5.
88
How do you calculate the pH of strong acids?
A strong monobasic acid, HA, has virtually complete dissociation in water:
HA(aq) → H+(aq) + A-(aq)
This means the [H+(aq)] of a strong acid is approximately equal to the concentration of the acid, [HA(aq)]:
[H+(aq)] = [HA(aq)]
The pH can then be calculated using pH = -log[H+(aq)]
89
Show the working out required to answer the question;
## Footnote
A sample of HCl has a concentration of 1.22 x 10-3 mol dm-3, what is it, pH?
HCl is a strong acid, and therfore completely dissociates:
HCl → H+ + Cl-
Therefore [H+] = [HCl] = 1.22 x 10-3 mol dm-3
pH = -log[H+]
= -log(1.22 x 10-3)
=2.91
90
How do you work out the pH of a weak acid?
* Firts we find the H+,
Ka = [H+] [A-] / [HA]
* If we assume that the H+ and A- ions only come from the dissociation of [HA, then this approximates to [H+] / [A-], therefore;
[H+]2 = Ka x [HA]
[H+] = (the square root of) Ka x [HA]
pH = - log [H+]
91
When calculating the pH of a weak acid why do you have to calculate Ka?
In aqueous solution, a weak monobasic acid, HA, partially dissociates, setting up the equilibrium: HA⇔ H+ + A-
Because you can no longer assume [H+] is equal to the concentration of the acid, we need to work out the value for Ka for the reaction - to find the extent of dissociation that has occurred, as this will dictate the value of [H+].
92
When calculating the pH of weak acids, how do you work out the value of Ka?
Ka = [H+][A-] / [HA] is simplified to Ka = [H+]2 / [HA]
93
Why can the equaltion for Ka be simplified to Ka = [H+]2 / [HA]?
* For weak acids, [H+] is much less than [HA] becuase the extent of dissociation is so small.
* [H+] and [A-] can be considered equal as when HA molecules dissociate H+ and A- ions are formed equally (any water present is expected to have disaccoiated into a negligable amount that will not affect the concentration of H+)
* This means the expression [H+] [A-] in the ka calculation becomes [H+]2.
* As some HA molecule have dissociated;
[HA] undissociated - [H+]
* Finally we make an approximation to simplifiy the pH calculation. We assume that such a small amount of the acid has dissociated that:
[HA]undissociated - [H+] = [HA] undissociated
* The Ka expression now becomes:
* Ka = [H+]2 / [HA]
94
When calcualting the pH of a weak acid how do we rearrange the equation to find [H+]?
The equation Ka = [H+]2 / [HA] is rearranged to give: [H+] = -/ Ka x [HA].
Thw pH can then be calculated using pH = - log[H+]
95
Recap; what is the calculation for H+ for strong and weak acids?
For strong acids, HA, [H+] = [HA]
For weak acids, HA, [H+] = -/Ka x [HA]
96
When can you not use the approximation - used in the calculating the pH of a weak acid - that so little of the original acid has dissociated that the concentration of the acid at qulibrium is effectively the same as the concentration of the orginal amount of acid?
It is actually the case that there will be some acid which does dissociate, even if it is a small amount. If this amount is less than 5% of the total of [HA] undissociated then it is deemed safe to use for the above approximation. This will b the case for very weak acids, with very low Ka values.
If, however, more than 5% of [HA] undissociated dissociates, then the above approximation cannot be use. This would occur with stronger examples of weak acids, with higher Ka values.
In this instance Ka is solved by a quadratic equation, which we dont need to know.
97
How do you determine ka for a weak acid?
To determine Ka for a weak acid, we needs to measure the pH of a solution of the weak acid using a pH meter. We also need the concentration of the weak acid.
1. First we find the [H+] from the pH: [H+] = 10-pH
2. Now we can calculate Ka: Ka = [H+][A-]
3. which approximates to ka = [H+]2 / [HA]
98
What is the definition for the ionic product of water?
Kw, is defined as Kw = [H+][OH-]. At 25oC, Kw = 1.00 x 10-14mol2dm-6
99
Describe teh ionisation of water.
Water can act as an acid by donating a proton
H2O → H+ + OH-
Water can act as a base by accepting a proton
H2O + H+ → H3O+
This back and fourth transfer of protns happens continuosly in water. In fact, water exisits a equilibrium: H2O ⇔ H+ + OH-
100
Why is [H2O] considered a constant, just like Kc.
* The equilbrium of the dissociation of water lies well to the left and only an extremenly small amount of water is dissociated at any given time.
* Because the amount of disociation, or ionisation, of water is so small, [H2O] is considered constant, just like Kc. We rearrange the equation as follows to combine the two constants.
* Kc = [H+][OH-] / [H2O] ⇒ Kc x [H2O] = [H+][OH-]
* This can be further simplified by reffering to the expression Kc x [H2O] as the new constant, Kw:
* Kw = Kc x [H2O] = Kc = [H+][OH-] and is simplified to Kw = [H+][OH]
101
What are the units for Kw?
The units for Kw are always mol-2dm-6 because [H+] mol dm-3 x [OH-3] mol dm-3 = mol2dm-6
102
Why does Kw = 1.00 x 10-14 at 25oC? The pH of water is 7, and the [H+] = 10-7 mol dm-3.
* At 25 oC, the measured pH of water is 7 and [H+] = 10-7 mol dm-3.When pure water ionises, te smae number of OH- and H+ ions are released:
[H+] = [OH-]
* This means that the concentration of OH- and H+ ions are released:
[H+] = [OH-]
* This means that the concentration of OH- is also equal to 10-7 mol dm-3
* so the calculation for Kw becomes :
Kw = [H+][OH-] = 1.0 x 10-7 x 1.0 x 10-7 = 10-14mol2dm-6
* Therefore, at 25oC, Kw = 1.00 x 10-14mol2dm-6
103
Why would the ionisation of water shift?
The balance between [H+] and [OH-] can be affected by the addition of extra H+ or OH- ions that cause the equilibrium to shift. (the equilibrium will shift to restore change)
104
What will cause Kw to change?
Kw changes only if the temperature is altered.
105
When are the concentrations of OH- and H+ balanced in water?
At 25oC, the pH value of 7 is the neutral point at which H+ and OH- concentrations are the same equal to 10-7 mol dm-3 . This applies to neutral solutions.
106
At 25oC what must Kw always equal for water?
Kw = [H+] x [OH-] = 1.00 x 10-14 mol2dm-6
107
Describe in terms of the dissociation of water, why rainwater becomes more acidic with the addition of carbon doixide.
At 25oC rainwater is acididc, CO2 dissolves forming a weak acid called carbonic acid, H2CO3, which dissociates to release H+ions into the water.
Because [H+] has increased [OH-] must fall until:
[H+][OH-] = 1.00 x 10-14mol2dm-6.
The overall effect is the pH decreases. Typical rainwater has a pH value of about 5.6.
108
How does the pH of mineral water differ from normal rainwater?
Mineral water contains dissolved ions such as carbonate, CO32-. These ions react with acids and remove H+ ions and lower the [H+].
Because [H+] has decreased, [OH-] must increase untill:
[H+][OH-] = 1.00 x 10-14 mol-2dm-6.
pH increases (7.0 - 8.0)
109
How can you find the concentrations of H+ and OH- iions in solution using the link between [H+] and [OH-]?
* using [H+][OH-] = 1.00 x 10-14mol-2dm-6.
* The indicies of [H+] and [OH-] will always add to -14.
110
What are the most common strong bases?
* Strong bases tend to be hydroxides of the metals in groups 1 and 2 in the periodic table.
* NaOH, KOH, and Ca(OH)2
111
How do strong bases dissociate? Use NaOH as an example.
* In aqueous solution strong bases dissociates completely:
* NaOH(aq) → Na+(aq) + OH-(aq)
* Bases that dissociate in water to release hydroxide ions are called alkalis.
* Strong bases are alkalis - they are 100% dissociated in aqueous solution.
112
How does a weak base dissociate?
Ammonia, NH3, is a weak base. In aqueous solution, an equilibrium is set up. The equilibrium position lies well to the left hand side:
NH3 + H2O ⇔ NH4+ + OH-
113
To work out the pH of a strong base we need to know [H+] and this depends on:
* the concentration of the base
* the ionic product of water, Kw = 1.00 x 10-14 mol2dm-6 (because this links [H+] with [OH-]).
114
Show the calculations for finding the pH of a strong monobasic alkali such as NaOH.
1. The [OH-] of a strong base is equal to the concentration of the base.
2. so [OH-] = [NaOH]
3. we can find [H+] from Kw and [OH-] Kw = [H+] [OH-]
4. [H+] = Kw / [OH-]
5. The pH can then be calculated using pH = -log10[H+].
(worked example on pg 41)
115
Recap: what can we use Kw to find?
[H+] = Kw / [OH-]
We can use Kw to find either the concentraion of H+ or OH-, as long as we know the value for Kw and the other concentration.
116
What is the definition of a buffer solution.
A buffer solution is a mixture that minimises pH changes on addition of small amounts of acid or base.
The word 'minimses' is essential to this definition.
117
How can a buffer solution be made?
118
Describe the carbonic acid buffer system.
* The weak acid, CH3COOH, dissociates partially;
CH3COOH ⇔ H+ + CH3COO-
* The salt dissociates completely, generating the conjugate base, CH3COO-:
CH3COO-Na+ → CH3COO- + Na+
The equilibrium mixture formed contains a high concentration of the undissociated weak acid, CH3COOH, and its conjugate base, CH3COO-. The high concentration of the conjugate base pushes the equilibrium to the left, so the concentration of H+ ions is very small.
The resulting buffer solution contains large reservoirs of the weak acid and its conugate base.
119
What is an alternative method of producing a buffer than from a weak acid and a salt?
Buffers can be made from a weak acid and a strong alkali. In this situation, a solution containing a mixture of the salt and the excess of weak acid is formed. For example, a weak acid such as methanoic acid, HCOOH, can be partially neutralised by an aqueous alkali, such as NaOH.
120
How does a buffer solution react to the addition of an acid?
* [H+] is increased
* The conjugate base, A- reacts with H+ ions
* the equilibrium shifts to the left, removing most of the added H+ ions.
121
How does a buffer solution react to the addition of an alkali to the buffer mixture
* [OH-] is increased
* The small concentration of H+ ions reacts with the OH- ions.
* HA dissociates, shifting the equilibrium to the right to restore most of the H+ ions that have reacted.
122
When calculating pH of buffer solutions what does the pH of a buffer solution depend on?
The pH of a buffer solution depends on:
* the acid dissociation constant, Ka, of a buffer system.
* the concentration ratio of the weak acid and its conjugate base.
123
What equation do we use for calculating the pH of buffer solutions?
Ka = [H+] [A-] / [HA] therefore [H+] = Ka x [HA] / [A-]
We need to check on the concentrations of HA and A-. Only a very small proportion of HA dissociates so we can assume that [HA]equilibrium = [HA]undissociated
The salt of the weak acid is ionic and dissociated completely in aqueous solution, for example:
CH3COO-Na+ → CH3COO- + Na+
so effectively [CH3COO-] = [CH3COO-Na+]
(example on page 43)
124
What buffer system controls the pH of blood?
The carbonic acid- hydrogen carbonate buffer system.
125
What is the importance of the carbonic acid - hydrogen carbonate buffer system?
Healthy human blood plasma needs to have a pH between 7.35 and 7.45. If the pH falls below 7.35, a condition called acidosis occurs. If the pH rises above 7.45, then the condition is called alkalosis.
The pH of blood is controlled by the mixture of buffers. The carbonic acid-hydrogen carbonate ion buffer, which is present in blood plasma, is the most important buffer system in the blood:
126
How does the carbonic acid - hydrogen carbonate buffer system work?
* carbonic acid, H2CO3, acts as the weak acid
* the hydrogencarbonate ion, HCO3- , acts as the conjugate base.
127
How does the carbonic acid - hydrogencarbonate buffer system react to an increase in [H+] ions?
Any increase in [H+] ions in the blood is removed by the conjugate base, HCO3-.The equilibrium shifts to the left, removing most of the H+ ions: H2CO3 ⇔ H+ + HCO3-
128
How does the carbonic acid - hydrogencarbonate buffer system react to an increase in [OH-] ions?
* Any increase in [OH-] ions in the blood is removed by the weak acid, H2CO3. The small concentration of H+ ions reacts with the OH- ions: H+ + OH- → H2O
* H2CO3 dissociates, shifting the equilibrium to the right to restore most of the H+ ions:
129
What is the acid dissociation constant, Ka, for the carbonic acid - hydrogencarbonate ion equilibrium?
4.3 x 10-7 mol dm-3
130
In the carbonic acid - hydrogencarbonate ion buffer system what happens to the H2CO3 when acid is added?
The carbonic acid is converted into aqueous carbon dioxide through the action of an enzyme. In the lungs, the dissolved carbon dioxide is converted into carbon dioxide gas, which is then exhaled.
131
What is the definition of the equivalence point?
The equivalence point is the point in a titration at which the volume of one solution has reacted exactly with the volume of the second solution. This matches the stoichiometry of the reaction taking place.
132
What is the definition of the end point?
The end point is the point in a titration at which there are equal concentrations of the weak acid and its conjugate base forms of the indicator.the colour at the endpoint is midway between the colours of the acid and conjugate base forms.
133
Describe the first section of this titration curve.
Section 1 - a slight increase in pH occurs as the base is added. The increase is only slight because the acid is in such excess. It is important to note that this section of the curve is not horizontal - pH is rising slightly.
134
Describe the middle section of the titration curve.
Section 2 - a sharp rise in pH occurs. The acid is no longer in excess so any base added has a large impact on the pH.
The equivalence point is the centre of the verticle section of the titration curve.
135
Describe the last section of the titration curve.
Section 3 - a slight increase in pH occurs as further base is added. The increase is only slight because the base is in excess now and an extra base has little impact on the pH. It is important to note however that this section is not horizontal - pH is rising slightly.
136
How can the data that is used to plot a titration curve be recorded?
pH meters or data loggers can be used to measure the pH of the reaction mixture as a solution from the burette is added over time. The varying pH values can then be plotted as a graph - known as an acid-base titration curve. The curve may be plotted by hand or by a computer attached to a pH data logger.
137
What symbols are often used to represent an indicator in an acid-base titration?
An acid-base titration is a weak acid, often represented as HIn. An indicator has one colour in its acid form (HIn) and a different colour in its conjugate base form (In-).
138
What are the advantages of using a pH meter?
An indicator such as universal indicator can only give a general indication of the pH of a solution. A pH meter can give an accurate reading for pH, usually to two decimal places.
139
What does a pH meter contain?
pH meters contain a protected electrode (the pH probe), which is placed into the solution in question, and a small computer display, which gives the pH reading.
140
When using a pH probe, it is important to calibrate it first do that the results are accurate. How do you calibrate a pH probe?
Usually, this is done in a two or three step calibration;
* The probe is removed from its storage solution and rinsed with deionized water.
* The probe is blotted dry and then placed into a solution of a known pH, often starting with a solution of pH4. The pH reading is allowed top settle before checking that a pH of 4 has been registered.
* This process is repeated with other solutions of known pH, often using prepared solutions of pH 7 and pH 10.
141
The ranges of the indicators are not actually on the equivalence points on most titration curves, so how can they show the reaction has reached this point?
Becuase the equivalence points occurs halfway through an incredibly sharp rise (or fall) in pH, the moment at which the indicator passes through its end point is effectively the same as the moment that the equivalence point is effectively the same moment that the equivalence point takes place.
142
What do you notice from the graph? The indicators used are phenolphthalein pink and methyl orange.
* The verticle section of the graph covers a large change in pH, starting around pH 3 and ending around pH 11, with an equivalence point at pH7.
* Both indicators, methyl orange and phenolphthalein, have end points that fall within its pH range.
* Either indicator would be suitable to use in a titration between a strong acid and a strong base.
143
Draw a strong acid - strong base titration curve.
144
Draw a strong acid- weak base titration curve.
145
What do you notice from the graph? The indicators used are phenolphthalein pink and methyl orange.
* The verticle section of the graph covers a smaller change in pH, starting around pH 3 and ending around pH 7.5, with the equivalence point occurring at a more acidic value, i.e. at pH lower than 7.
* Methyl orange has an end point that falls within its pH range.
* Phenolphthalein does not have an end point that falls within this pH range.
* Only methyl orange would be a suitable indicator for a titration between a strong acid and a weak base.
146
Draw a titration curve for a weak acid - strong base titration.
147
What do you notice from the graph? The indicators used are phenolphthalein pink and methyl orange.
* The vertical section id the graph covers a smaller change in pH, and occurs further up towards the higher pH values, starting around pH 6.5 and ending up around 11.5.
* The equivalence point occurs at a more basic pH, i.e. a pH value above 7.
* Methyl orange has an end point that falls outside this pH range.
* Phenolphthalein has an end point that falls within this pH range.
* Only phenolphthalein would be a suitable indicator for a titration between a strong acid and a weak base.
148
Draw a titration curve for the reaction between a weak acid and a weak base.
149
What do you notice from the graph? The indicators used are phenolphthalein pink and methyl orange.
* There is no real vertical section.
* Neither indicator has an end point near the equivalence point.
* Neither indicator is suitable.
* An indicator would change colour gradually over a few cm3 of base added.
* No indicators are really suitable for a weak acid - weak base titration.
150
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