Question
In a qualitative analysis, ${Bi}^{3 +}$ is detected by appearance of precipitate of $BiO(OH)(s)$. Calculate pH when the following equilibrium exists at 298 K.
$$BiO(OH)(s) \rightleftharpoons {BiO}^{+}(aq) + {OH}^{-}(aq),K = 4 \times 10^{- 10} $$(Given: $log2 = 0.3010$ )
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Solution
Sol. [Key Concept: Solubility product Ksp from ionic concentrations]
$BiO(OH)(s) \rightleftharpoons {BiO}_{s}^{+}(aq) + {OH}_{s}^{-}(aq)$
$$\begin{matrix} & K = \frac{\left\lbrack {BiO}^{+} \right\rbrack\left\lbrack {OH}^{-} \right\rbrack}{\left\lbrack {BiO}^{-}(OH)(s) \right\rbrack} \\ & K = \frac{\left\lbrack {BiO}^{+} \right\rbrack\left\lbrack {OH}^{-} \right\rbrack}{1} \\ & \text{ }K = s \times s = s^{2} \\ & \text{ }s = \sqrt{K} = \sqrt{4 \times 10^{- 10}} = 2 \times 10^{- 5}M \\ & \left\lbrack H^{+} \right\rbrack = \frac{K_{w}}{\left\lbrack {OH}^{-} \right\rbrack} = \frac{1}{2} \times 10^{- 9},pH = - log\left\lbrack H^{+} \right\rbrack = 9 + log2 = 9.301 \end{matrix}$$
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