JEE Challenger
More from Equilibrium
ChemMedAdv 2018 • Paper 1

Solubility of Weak Acid Salt at Specific pH

The solubility of a sparingly soluble salt of a weak acid (denoted as AB\mathbf{AB}) in a solution of pH=3\text{pH} = 3 is measured to be Y×10−3 mol L−1\mathbf{Y} \times 10^{-3} \text{ mol L}^{-1}. Determine the numerical value of Y\mathbf{Y}.\n\n(Given: The solubility product constant of AB\mathbf{AB} is Ksp=2×10−10K_{sp} = 2 \times 10^{-10} and the acid dissociation constant of HB\mathbf{HB} is Ka=1×10−8K_a = 1 \times 10^{-8})

Official Numerical Answer4.47

Topics & Concepts

EquilibriumSolubility ProductAcid-Base Equilibrium

Step-by-Step Solution

To find the solubility of the sparingly soluble salt AB\mathbf{AB} in a buffered solution of pH=3\text{pH} = 3, we consider the simultaneous equilibria of the salt dissolution and the protonation of the anion B−\mathbf{B^-}.

1. Equilibrium Equations

When AB\mathbf{AB} dissolves in water, it dissociates as: AB(s)⇌A+(aq)+B−(aq)\mathbf{AB(s)} \rightleftharpoons \mathbf{A^+}(aq) + \mathbf{B^-}(aq) The solubility product expression is: Ksp=[A+][B−]K_{sp} = [\mathbf{A^+}][\mathbf{B^-}]

Since HB\mathbf{HB} is a weak acid, the anion B−\mathbf{B^-} hydrolyzes in the presence of H+\mathbf{H^+}: B−(aq)+H+(aq)⇌HB(aq)\mathbf{B^-}(aq) + \mathbf{H^+}(aq) \rightleftharpoons \mathbf{HB}(aq) The acid dissociation constant expression for HB\mathbf{HB} is: Ka=[H+][B−][HB]K_a = \frac{[\mathbf{H^+}][\mathbf{B^-}]}{[\mathbf{HB}]}

2. Mass Balance and Speciation

Let the molar solubility of AB\mathbf{AB} in this solution be S mol L−1S\text{ mol L}^{-1}.

From the stoichiometry of the dissolution: [A+]=S[\mathbf{A^+}] = S

The total concentration of all dissolved species containing B\mathbf{B} is equal to the solubility SS: S=[B−]+[HB]S = [\mathbf{B^-}] + [\mathbf{HB}]

Using the expression for KaK_a, we can express [HB][\mathbf{HB}] in terms of [B−][\mathbf{B^-}]: [HB]=[H+][B−]Ka[\mathbf{HB}] = \frac{[\mathbf{H^+}][\mathbf{B^-}]}{K_a}

Substitute [HB][\mathbf{HB}] into the mass balance relation: S=[B−]+[H+][B−]Ka=[B−](1+[H+]Ka)S = [\mathbf{B^-}] + \frac{[\mathbf{H^+}][\mathbf{B^-}]}{K_a} = [\mathbf{B^-}] \left(1 + \frac{[\mathbf{H^+}]}{K_a}\right)

Therefore, the equilibrium concentration of the free anion B−\mathbf{B^-} is: [B−]=S1+[H+]Ka[\mathbf{B^-}] = \frac{S}{1 + \frac{[\mathbf{H^+}]}{K_a}}

3. Relation with Solubility Product

Substitute [A+][\mathbf{A^+}] and [B−][\mathbf{B^-}] into the KspK_{sp} expression: Ksp=[A+][B−]=S⋅S1+[H+]Ka=S21+[H+]KaK_{sp} = [\mathbf{A^+}][\mathbf{B^-}] = S \cdot \frac{S}{1 + \frac{[\mathbf{H^+}]}{K_a}} = \frac{S^2}{1 + \frac{[\mathbf{H^+}]}{K_a}}

Rearranging gives: S2=Ksp(1+[H+]Ka)S^2 = K_{sp} \left(1 + \frac{[\mathbf{H^+}]}{K_a}\right)

4. Calculation

Given:

  • pH=3  ⟹  [H+]=10−3 M\text{pH} = 3 \implies [\mathbf{H^+}] = 10^{-3}\text{ M}
  • Ksp=2×10−10K_{sp} = 2 \times 10^{-10}
  • Ka=1×10−8K_a = 1 \times 10^{-8}

Calculate the ratio: [H+]Ka=10−31×10−8=105\frac{[\mathbf{H^+}]}{K_a} = \frac{10^{-3}}{1 \times 10^{-8}} = 10^5

Since 105≫110^5 \gg 1: 1+[H+]Ka≈1051 + \frac{[\mathbf{H^+}]}{K_a} \approx 10^5

Now, calculate S2S^2: S2=(2×10−10)×105=2×10−5=20×10−6 mol2 L−2S^2 = (2 \times 10^{-10}) \times 10^5 = 2 \times 10^{-5} = 20 \times 10^{-6}\text{ mol}^2\text{ L}^{-2}

Taking the square root: S=20×10−6=20×10−3≈4.47×10−3 mol L−1S = \sqrt{20 \times 10^{-6}} = \sqrt{20} \times 10^{-3} \approx 4.47 \times 10^{-3}\text{ mol L}^{-1}

Given that the solubility is expressed as Y×10−3 mol L−1\mathbf{Y} \times 10^{-3}\text{ mol L}^{-1}: Y=4.47\mathbf{Y} = 4.47

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