Question

Consider two Group IV metal ions X$^{2+}$ and Y$^{2+}$. A solution containing 0.01 M X$^{2+}$ and 0.01 M Y$^{2+}$ is saturated with H$_2$S. The pH at which the metal sulphide YS will form as a precipitate is ___. (Nearest integer)
Given:
$K_{sp}(\mathrm{XS}) = 1 \times 10^{-22}$ at 25$^\circ$C
$K_{sp}(\mathrm{YS}) = 4 \times 10^{-16}$ at 25$^\circ$C
$[\mathrm{H_2S}] = 0.1$ M
$K_{a1} \times K_{a2} (\mathrm{H_2S}) = 1.0 \times 10^{-21}$
$\log 2 = 0.30,\ \log 3 = 0.48,\ \log 5 = 0.70$

Hint:

Selective precipitation depends on the relative values of $K_{sp}$ and sulphide ion concentration controlled by pH.

Answer 1

Correct Answer: 2

Step 1: Write expression for sulphide ion concentration.
\[ [\mathrm{S^{2-}}] = \frac{K_{a1} K_{a2} [\mathrm{H_2S}]}{[\mathrm{H^+}]^2} \]
Step 2: Use precipitation condition for YS.
\[ K_{sp}(\mathrm{YS}) = [\mathrm{Y^{2+}}][\mathrm{S^{2-}}] \] \[ 4 \times 10^{-16} = 0.01 \times [\mathrm{S^{2-}}] \] \[ [\mathrm{S^{2-}}] = 4 \times 10^{-14} \]
Step 3: Substitute values.
\[ 4 \times 10^{-14} = \frac{(1 \times 10^{-21})(0.1)}{[\mathrm{H^+}]^2} \] \[ [\mathrm{H^+}]^2 = 2.5 \times 10^{-9} \] \[ [\mathrm{H^+}] = 5 \times 10^{-5} \]
Step 4: Calculate pH.
\[ \text{pH} = -\log (5 \times 10^{-5}) = 5 - 0.70 = 4.3 \] Nearest integer pH $\approx$ 2.