Concept Explanation: Solubility of a Salt of Weak Acid in Acidic Buffer

When a salt of a weak acid (like AB, where A⁺ is the cation and B^- is the anion of the weak acid HB) dissolves in an acidic solution (pH = 3), the anion B^- can react with the H⁺ ions from the acid to form the weak acid HB. This reaction (B^- + H⁺ ⇌ HB) consumes the anion, shifting the dissolution equilibrium of the salt (AB(s) ⇌ A⁺ + B^-) to the right. This phenomenon increases the salt's solubility beyond what it would be in pure water. The total solubility (S) is the sum of the concentrations of A⁺ and all species containing the anion B (i.e., [B^-] and [HB]).

Step-by-Step Solution

Step 1: Define the Variables and Known Quantities

Let the total solubility of the salt AB be S mol L^-1.
Given:
pH = 3, so [H⁺] = 1 × 10^-3 M.
K_sp = 2 × 10^-10
K_a = 1 × 10^-8

Step 2: Write the Equilibria Involved

1. Dissolution of the salt: $\mathrm{AB}\left(s\right)\rightleftharpoons A^{+}+B^{-}$ ; K_sp = [A⁺][B^-]
2. Acid dissociation of HB: $\mathrm{HB}\rightleftharpoons H^{+}+B^{-}$ ; $K_a=\frac{\left[H^{+}\right]\left[B^{-}\right]}{\left[\mathrm{HB}\right]}$

Step 3: Relate the Concentrations

From the dissolution:
Concentration of cation A⁺ = S
The anion B^- can exist in two forms: free B^- and associated HB.
Therefore, the total solubility S is also equal to the sum of the concentrations of these two species: $S=\left[A^{+}\right]=\left[B^{-}\right]+\left[\mathrm{HB}\right]$

Step 4: Express [HB] in terms of [B^-] and [H⁺]

We can rearrange the K_a expression: $K_a=\frac{\left[H^{+}\right]\left[B^{-}\right]}{\left[\mathrm{HB}\right]}$
This gives us: $\left[\mathrm{HB}\right]=\frac{\left[H^{+}\right]\left[B^{-}\right]}{K_a}$

Step 5: Substitute [HB] back into the Solubility Equation

$S=\left[B^{-}\right]+\left[\mathrm{HB}\right]=\left[B^{-}\right]+\frac{\left[H^{+}\right]\left[B^{-}\right]}{K_a}$
Factor out [B^-]: $S=\left[B^{-}\right]\left(1+\frac{\left[H^{+}\right]}{K_a}\right)$

Step 6: Express [B^-] in terms of K_sp and S

From the K_sp expression: K_sp = [A⁺][B^-] = S [B^-]
Therefore: $\left[B^{-}\right]=\frac{K_{sp}}{S}$

Step 7: Substitute [B^-] into the Equation from Step 5

Substitute $\frac{K_{sp}}{S}$ for [B^-] in the solubility equation: $S=\left(\frac{K_{sp}}{S}\right)\left(1+\frac{\left[H^{+}\right]}{K_a}\right)$
Multiply both sides by S: $S^2=K_{sp}\left(1+\frac{\left[H^{+}\right]}{K_a}\right)$

Step 8: Insert the Known Values and Solve for S

K_sp = 2 × 10^-10
[H⁺] = 10^-pH = 1 × 10^-3
K_a = 1 × 10^-8
First, calculate the term inside the parentheses: $1+\frac{\left[H^{+}\right]}{K_a}=1+\frac{1\times {10}^{-3}}{1\times {10}^{-8}}=1+1\times {10}^5=100001\approx 1\times {10}^5$
Now, plug this value into the equation: $S^2=(2\times {10}^{-10})\times (1\times {10}^5)=2\times {10}^{-5}$
Take the square root of both sides to find S: