Equilibria

Every objective in this topic is printed under "A Level subject content" in the 9701 syllabus and carries the tier "A Level". None of it is AS, and it is a separate topic from the AS topic 7 of the same name. It is examined on Paper 4. Paper 1 is the AS multiple-choice paper and the whole 9701 bank on this site comes from it, so no practice here is tagged to this topic.

Syllabus points

25.1 Acids and bases

• Understand and use the terms conjugate acid and conjugate base.
• Define conjugate acid-base pairs, identifying such pairs in reactions.
• Define mathematically the terms pH, K_a, pK_a and K_w and use them in calculations (K_b and the equation K_w = K_a × K_b will not be tested).
• Calculate [H⁺(aq)] and pH values for strong acids, strong alkalis and weak acids.
• Define a buffer solution; explain how a buffer solution can be made; explain how buffer solutions control pH, using chemical equations in these explanations; describe and explain the uses of buffer solutions, including the role of HCO₃⁻ in controlling pH in blood.
• Calculate the pH of buffer solutions, given appropriate data.
• Understand and use the term solubility product, K_sp.
• Write an expression for K_sp.
• Calculate K_sp from concentrations and vice versa.
• Understand and use the common ion effect to explain the different solubility of a compound in a solution containing a common ion, and perform calculations using K_sp values and concentration of a common ion.

25.2 Partition coefficients

• State what is meant by the term partition coefficient, K_pc.
• Calculate and use a partition coefficient for a system in which the solute is in the same physical state in the two solvents.
• Understand the factors affecting the numerical value of a partition coefficient in terms of the polarities of the solute and the solvents used.

Conjugate acid-base pairs

On the Brønsted-Lowry definition, an acid is a proton donor and a base is a proton acceptor.

When an acid donates its proton, what is left is its conjugate base. When a base accepts a proton, what it becomes is its conjugate acid. The two members of a pair therefore differ by exactly one H⁺, and that is the test to apply.

For the reaction

CH₃COOH + H₂O ⇌ CH₃COO⁻ + H₃O⁺

there are two pairs:

• CH₃COOH and CH₃COO⁻, acid and its conjugate base.
• H₂O and H₃O⁺, base and its conjugate acid.

Two points that questions rely on. A strong acid has a weak conjugate base, and vice versa, because a species that gives up its proton readily has little tendency to take it back. And water is amphoteric: it acts as a base with an acid and as an acid with a base, so which role it plays depends on its partner.

pH and the ionic product of water

pH = -log[H⁺]

and running it back,

[H⁺] = 10^(-pH)

Water ionises very slightly, and the equilibrium constant for that is the ionic product of water:

K_w = [H⁺][OH⁻] = 1.00 × 10⁻¹⁴ mol² dm⁻⁶ at 298 K

Note the units, and note the temperature. K_w is an equilibrium constant, so it changes with temperature: the ionisation of water is endothermic, so warming water increases K_w and therefore lowers the pH of pure water below 7. Pure water at 50 °C has a pH under 7 and is still neutral, because [H⁺] still equals [OH⁻]. Neutrality means the two concentrations are equal, not that the pH is 7.

Strong acids

A strong acid is fully dissociated, so [H⁺] equals the acid concentration, multiplied by the number of protons the acid supplies.

Worked example. Find the pH of 0.0500 mol dm⁻³ hydrochloric acid.

[H⁺] = 0.0500, and pH = -log(0.0500) = 1.30.

Worked example, a diprotic acid. For 0.0500 mol dm⁻³ sulfuric acid, treated as fully dissociating twice:

0.0500 × 2 = 0.1

so [H⁺] = 0.100 and pH = 1.00.

Strong alkalis

A strong alkali is fully dissociated, so [OH⁻] is known, and K_w gives [H⁺].

Worked example. Find the pH of 0.0200 mol dm⁻³ sodium hydroxide.

[H⁺] = 1.00 × 10⁻¹⁴ / 0.0200 = 5.0 × 10⁻¹³

pH = -log(5.0 × 10⁻¹³) = 12.30

Going through K_w is the step people forget; taking -log of the hydroxide concentration gives the pOH, not the pH.

Weak acids

A weak acid is partially dissociated, and the position of that equilibrium is measured by the acid dissociation constant:

HA ⇌ H⁺ + A⁻

K_a = [H⁺][A⁻] / [HA]

with units mol dm⁻³, and

pK_a = -log K_a

so a larger K_a, and therefore a smaller pK_a, means a stronger acid. The inversion of the pK_a scale is a reliable source of wrong answers.

Two approximations are made, and both should be stated when a question asks for assumptions:

• The dissociation of the acid produces equal amounts of H⁺ and A⁻, and the contribution of water to [H⁺] is negligible, so [H⁺] = [A⁻].
• The acid is only slightly dissociated, so the equilibrium concentration of HA is effectively the initial concentration.

Those give

K_a = [H⁺]² / [HA], so [H⁺] = √(K_a × [HA])

Worked example. Find the pH of 0.100 mol dm⁻³ ethanoic acid, K_a = 1.74 × 10⁻⁵ mol dm⁻³.

K_a × [HA]:

1.74 × 10⁻⁵ × 0.100 = 1.74 × 10⁻⁶

Taking the square root gives [H⁺] = 1.319 × 10⁻³ mol dm⁻³, so

pH = -log(1.319 × 10⁻³) = 2.88

Compare that with a strong acid at the same concentration, which would have pH 1.00. The difference is the whole meaning of "weak": same concentration, far fewer free protons.

Buffer solutions

A buffer solution is one that resists changes in pH when small amounts of acid or alkali are added, or when it is diluted. The words "small amounts" and "resists" both matter: a buffer minimises change, it does not prevent it, and it can be exhausted.

How to make one

• An acidic buffer: a weak acid together with its salt (its conjugate base). Ethanoic acid with sodium ethanoate. Or make it by adding a limited amount of strong alkali to an excess of weak acid, so the alkali is used up in converting some of the acid into its salt.
• An alkaline buffer: a weak base together with its salt. Ammonia with ammonium chloride.

The essential feature is a reservoir of both the weak acid and its conjugate base in appreciable amounts, so that there is something present to react with added H⁺ and something to react with added OH⁻.

How it works

Take the ethanoic acid and ethanoate buffer.

• Add acid. The added H⁺ reacts with the large reservoir of ethanoate ions:

CH₃COO⁻ + H⁺ → CH₃COOH

so the H⁺ is removed and the pH barely changes.

• Add alkali. The added OH⁻ reacts with the large reservoir of undissociated ethanoic acid:

CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O

so the OH⁻ is removed and the pH barely changes.

The mark scheme wants the equations, and it wants the word reservoir or its equivalent: the buffer works because both species are present in large amounts compared with what is added.

Calculating buffer pH

Rearranging the K_a expression, with [A⁻] as the salt concentration and [HA] as the acid concentration:

[H⁺] = K_a × [HA] / [A⁻]

Worked example. A buffer contains 0.200 mol dm⁻³ ethanoic acid and 0.150 mol dm⁻³ sodium ethanoate. K_a = 1.74 × 10⁻⁵.

The ratio:

0.200 / 0.150 = 1.333

[H⁺]:

1.74 × 10⁻⁵ × 1.333 = 2.32 × 10⁻⁵

pH = -log(2.32 × 10⁻⁵) = 4.63

Notice that only the ratio of acid to salt appears, so diluting a buffer does not change its pH, since both concentrations fall by the same factor. That is a favourite question and the answer surprises people.

It also follows that a buffer works best when the ratio is close to 1, where pH equals pK_a, because then it has equal capacity to absorb acid and alkali. Choosing a buffer means choosing a weak acid whose pK_a is close to the required pH.

Buffers in blood

Blood is held between pH 7.35 and 7.45, and the main system is carbonic acid and hydrogencarbonate:

CO₂ + H₂O ⇌ H₂CO₃ ⇌ H⁺ + HCO₃⁻

• If acid enters the blood, H⁺ reacts with HCO₃⁻ to form H₂CO₃, which decomposes to CO₂ and water, and the extra CO₂ is breathed out.
• If alkali enters, OH⁻ reacts with H₂CO₃, and more CO₂ dissolves to replace it.

The link to breathing is what makes this system so effective: one of the products can be removed from the body entirely, so the buffer is continuously renewed rather than being used up. A fall below about pH 7.0 or a rise above about 7.8 is fatal, which is why the control is so tight.

Other uses worth naming: shampoos and toiletries buffered near the pH of skin, buffered food products to control spoilage, and standard buffer solutions used to calibrate a pH meter.

Solubility product

For a sparingly soluble ionic solid in contact with its saturated solution:

A_xB_y(s) ⇌ xA^y+(aq) + yB^x-(aq)

K_sp = [A^y+]^x [B^x-]^y

The solid does not appear, because its concentration is constant. Examples:

• AgCl: K_sp = [Ag⁺][Cl⁻], units mol² dm⁻⁶
• Mg(OH)₂: K_sp = [Mg²⁺][OH⁻]², units mol³ dm⁻⁹
• Ca₃(PO₄)₂: K_sp = [Ca²⁺]³[PO₄³⁻]², units mol⁵ dm⁻¹⁵

The units change with the expression, so work them out rather than recalling them.

Worked example, K_sp from solubility. Silver chloride has a solubility of 1.46 × 10⁻³ g dm⁻³. M_r = 143.4.

Solubility in mol dm⁻³:

1.46 × 10⁻³ / 143.4 = 1.018 × 10⁻⁵

Each formula unit gives one Ag⁺ and one Cl⁻, so both are 1.018 × 10⁻⁵:

K_sp = 1.018 × 10⁻⁵ × 1.018 × 10⁻⁵ = 1.036 × 10⁻¹⁰

so K_sp is about 1.04 × 10⁻¹⁰ mol² dm⁻⁶.

Worked example, solubility from K_sp, with a 1 : 2 salt. Magnesium hydroxide has K_sp = 1.10 × 10⁻¹¹ mol³ dm⁻⁹.

If the solubility is s, then [Mg²⁺] = s and [OH⁻] = 2s, so

K_sp = s × (2s)² = 4s³

Rearranging:

1.10 × 10⁻¹¹ / 4 = 2.75 × 10⁻¹²

and the cube root of that is s = 1.40 × 10⁻⁴ mol dm⁻³.

Getting the 2s and the cube right is what this question tests. Writing K_sp = s² for a 1 : 2 salt is the standard error.

Predicting precipitation. Calculate the ionic product, the same expression using the actual concentrations after mixing, and compare:

• Ionic product greater than K_sp: a precipitate forms.
• Ionic product less than K_sp: no precipitate, and the solution is unsaturated.

Remember to allow for the dilution that occurs on mixing two solutions, which halves each concentration if equal volumes are used.

The common ion effect

A sparingly soluble salt is less soluble in a solution that already contains one of its ions.

The explanation is Le Chatelier applied to the solubility equilibrium. Adding extra Cl⁻ to a saturated solution of silver chloride shifts

AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)

to the left, so more solid precipitates and the concentration of dissolved Ag⁺ falls. Note that K_sp itself does not change; only the concentrations adjust to satisfy it.

Worked example. Find the solubility of silver chloride in 0.100 mol dm⁻³ sodium chloride, with K_sp = 1.04 × 10⁻¹⁰.

Here [Cl⁻] is dominated by the sodium chloride, so take it as 0.100. Then:

[Ag⁺] = 1.04 × 10⁻¹⁰ / 0.100 = 1.04 × 10⁻⁹

so the solubility is 1.04 × 10⁻⁹ mol dm⁻³, against 1.02 × 10⁻⁵ in pure water. It is about ten thousand times less soluble, which is why a precipitate is washed with a solution containing the common ion rather than with pure water.

Partition coefficients

When a solute is shaken with two immiscible solvents, it distributes itself between them, and at equilibrium the ratio of its concentrations is a constant at a given temperature:

K_pc = [X in solvent 1] / [X in solvent 2]

The partition coefficient has no units when the solute is in the same physical state in both solvents, because it is a ratio of two concentrations.

Worked example. 2.00 g of a solute is shaken with 100 cm³ of water and 50.0 cm³ of an organic solvent. At equilibrium 0.400 g remains in the water.

Mass in the organic layer:

2.00 - 0.400 = 1.6

Concentration in the organic layer, in g cm⁻³:

1.6 / 50.0 = 0.032

Concentration in water:

0.400 / 100 = 0.004

K_pc, organic over aqueous:

0.032 / 0.004 = 8

so K_pc is 8. Note that both concentrations must be expressed per unit volume of their own layer, and the volumes are usually different on purpose.

What determines the value. "Like dissolves like". A solute distributes itself in favour of the solvent whose polarity resembles its own:

• A polar solute, able to hydrogen bond, favours the aqueous layer.
• A non-polar solute favours the organic layer, because the induced dipole attractions it can form there are stronger than the hydrogen bonds it would have to break in water.

That principle is the basis of solvent extraction, and it explains a practical result worth knowing: extracting with several small portions of solvent removes more solute than one large portion of the same total volume, because equilibrium is re-established each time.

Common mistakes

• Identifying a conjugate pair that differs by more than one proton, such as H₂SO₄ and SO₄²⁻.
• Saying pure water always has pH 7. It does at 298 K, and it is neutral at every temperature, but the pH changes with temperature.
• Taking -log of a hydroxide concentration and calling it the pH.
• Forgetting to double the concentration for a strong diprotic acid.
• Treating a weak acid as fully dissociated, so [H⁺] is taken as the acid concentration.
• Forgetting the square root in [H⁺] = √(K_a[HA]).
• Reading a larger pK_a as a stronger acid. Larger K_a means stronger, so larger pK_a means weaker.
• Describing a buffer as a solution that keeps the pH constant. It resists change.
• Saying a buffer is a mixture of a strong acid and its salt.
• Explaining a buffer without equations, or without saying that both species are present in large amounts.
• Saying diluting a buffer raises its pH. The ratio is unchanged, so the pH is unchanged.
• Writing K_sp = s² for a salt of formula MX₂, where it is 4s³.
• Quoting K_sp units without deriving them from the expression.
• Saying the common ion effect changes the value of K_sp.
• Forgetting the dilution when two solutions are mixed before comparing the ionic product with K_sp.
• Dividing masses rather than concentrations when calculating a partition coefficient, so the different volumes of the two layers are ignored.