JEE Chemistry Equilibrium Notes Set 03

Polyprotic Acids and Polyacidic Bases

A polyprotic acid \( \ce{H_nA} \) ionises in steps, each with its own constant, and \( K_{a1} \gg K_{a2} \gg K_{a3} \) because it is harder to remove a proton from an ion that is already negatively charged. For a solution of a weak polyprotic acid of concentration C, almost all the H\(^+\) comes from the first step: \( [H^+] = \sqrt{K_{a1}C} \) (valid when the degree of ionisation is below 10 percent), so \( pH = \frac12(pK_{a1} - \log C) \). The concentration of the ion formed in the second step is approximately \( K_{a2} \) when \( [HS^-] = [H^+] \), as in a pure solution of H\(_2\)S, where \( [S^{2-}] \approx K_{a2} \).
In general \( [S^{2-}] = K_{a1}K_{a2}[H_2S]/[H^+]^2 \). A strong acid added to H\(_2\)S raises [H\(^+\)], and the sulphide ion concentration falls with the square of [H\(^+\)]. The same relations hold for a polyacidic base with \( K_b \) and pOH.

Mixtures of Acids and Bases

Weak acid with a strong acid: the strong acid fixes [H\(^+\)] (ignore the weak acid) and the weak acid ionises to \( \alpha = K_a/[H^+] \). Weak base with a strong base: [OH\(^-\)] is fixed by the strong base. Two weak monoprotic acids: \( [H^+] = \sqrt{C_1K_{a1} + C_2K_{a2}} \), and the conjugate anion concentrations are in the ratio of their K\(_a\) values. When the acid or base is extremely weak, the ionisation of water must be added: \( [OH^-] = \sqrt{K_bC + K_w} \). If the degree of ionisation is not small, solve the quadratic equation instead of using the square-root formula.

Salt Hydrolysis

Salt of a strong base and a weak acid: \( K_h = K_w/K_a \), \( h = \sqrt{K_h/C} \), \( pH = \frac12(pK_w + pK_a + \log C) \). Salt of a strong acid and a weak base: \( pH = \frac12(pK_w - pK_b - \log C) \). Salt of a weak acid and a weak base: \( pH = \frac12(pK_w + pK_a - pK_b) \), independent of concentration. For the anion of a polyprotic acid \( A^{x-} \), the first hydrolysis uses the last dissociation constant: \( K_{h1} = K_w/K_{a,x} \). For a cation \( BH_y^{y+} \) of a polyacidic base, \( K_{h1} = K_w/K_{b,y} \).

These values do not depend on the concentration of the salt.

Buffer Solutions and Buffer Capacity

An acidic buffer (weak acid plus its salt) has \( pH = pK_a + \log([\text{salt}]/[\text{acid}]) \) and a basic buffer has \( pOH = pK_b + \log([\text{salt}]/[\text{base}]) \). A polyprotic acid gives a buffer for each adjacent pair of species, and the relevant pK\(_a\) is that of the pair involved. A buffer is most effective when the acid and salt are present in equal amounts (pH = pK\(_a\)) and its range is about pK\(_a\) plus or minus 1. The buffer capacity is the amount of strong acid or base needed to change the pH of one litre of buffer by one unit; for a buffer with a mol of acid and b mol of salt the capacity is \( 2.303\,ab/(a + b) \), which is largest when a = b.

Acid-Base Titrations

To find the pH at any point, carry out the neutralisation by mole counting and then decide what is left. For a weak acid titrated with a strong base: at half equivalence pH = pK\(_a\); at x percent neutralisation \( pH = pK_a + \log[x/(100 - x)] \); at equivalence the solution is the salt of a weak acid, so the pH is above 7. For a weak base titrated with a strong acid, at half equivalence pOH = pK\(_b\) and at equivalence the pH is below 7.
For a polyprotic acid, the pH at the first half equivalence point is pK\(_{a1}\), at the first equivalence point it is \( \frac12(pK_{a1} + pK_{a2}) \), at the second half equivalence point it is pK\(_{a2}\), and so on. The last equivalence point is the pure salt of the last anion. A weak acid titrated with a weak base shows no sharp jump, which is why such titrations are not used.

Worked Examples

Example 1: The pH of 0.1 M H\(_2\)CO\(_3\) with \( K_{a1} = 4 \times 10^{-7} \): \( [H^+] = \sqrt{4 \times 10^{-8}} = 2 \times 10^{-4} \), so pH = 3.7.
Example 2: 100 mL of 0.1 M acetic acid is mixed with 50 mL of 0.1 M NaOH. Acid left 5 mmol and acetate 5 mmol, so pH = pK\(_a\) = 4.74.
Example 3: The pH of 0.1 M sodium acetate is \( \frac12(14 + 4.74 - 1) = 8.87 \). The degree of hydrolysis is \( \sqrt{5.6 \times 10^{-10}/0.1} = 7.5 \times 10^{-5} \).