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: 4

To determine the pH at which the metal sulfide YS will form as a precipitate, we use the solubility product constant (\(K_{sp}\)) and the dissociation constants of hydrogen sulfide (\(H_2S\)).
Given the conditions, precipitation occurs when
\([Y^{2+}][S^{2-}] \geq K_{sp}(YS)\).
An equilibrium occurs when:
\[K_{sp}(YS) = 4 \times 10^{-16}\]
The dissociation of \(H_2S\) in solution is given by:
\[H_2S \rightleftharpoons 2H^+ + S^{2-}\]
The concentration of \(S^{2-}\) at equilibrium can be calculated using the equilibrium constant:
\[K = \frac{[H^+]^2[S^{2-}]}{[H_2S]} = K_{a1} \times K_{a2} = 1.0 \times 10^{-21}\]
Rearranging gives:
\[[S^{2-}] = \frac{K}{[H^+]^2}[H_2S]\]
Substitute \( [H_2S] = 0.1 \) M into the equation:
\[[S^{2-}] = \frac{1.0 \times 10^{-21}}{[H^+]^2} \times 0.1 = 1.0 \times 10^{-22} /[H^+]^2\]
To find the point where precipitation begins, set \( [Y^{2+}] = 0.01 \) M and solve for \([H^+]\):
\[0.01 \times \frac{1.0 \times 10^{-22}}{[H^+]^2} = 4 \times 10^{-16}\]
Solving gives:
\[[H^+]^2 = \frac{1.0 \times 10^{-24}}{4 \times 10^{-16}} = 2.5 \times 10^{-9}\]
\[ [H^+] = \sqrt{2.5 \times 10^{-9}} = 5 \times 10^{-5}\]
\(pH = -\log[H^+]\), so \(pH = -\log(5 \times 10^{-5})\).
Using logarithmic properties:
\(-\log(5 \times 10^{-5}) = 5 - \log 5\)
Given \(\log 5 = 0.70\), this becomes:
\[pH = 5 - 0.70 = 4.3\]
Rounding to the nearest integer, the pH is 4.