Chemical and Ionic Equilibrium 2 Question 1
1. The molar solubility of $\mathrm{Cd}(\mathrm{OH}){2}$ is $1.84 \times 10^{-5} \mathrm{~m}$ in water. The expected solubility of $\mathrm{Cd}(\mathrm{OH}){2}$ in a buffer solution of $\mathrm{pH}=12$ is
(2019 Main, 12 April II)
(a) $1.84 \times 10^{-9} \mathrm{M}$
(b) $\frac{2.49}{1.84} \times 10^{-9} \mathrm{M}$
(c) $6.23 \times 10^{-11} \mathrm{M}$
(d) $2.49 \times 10^{-10} \mathrm{M}$
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Answer:
Correct Answer: 1. (d)
Solution:
Key Idea The concentration of substance in a saturated solution is defined as its solubility $(S)$. Its value depends upon the nature of solvent and temperature.
$$ A_{x} B_{y} \rightleftharpoons x A^{y+}+y B^{x-} K_{\mathrm{sp}}=\left[A^{y+}\right]^{x}\left[B^{x-}\right]^{y} $$
Solubility of $\mathrm{Cd}(\mathrm{OH})_{2}(\mathrm{~S})=1.84 \times 10^{-5} \mathrm{M}$
Given, $\mathrm{pH}=12$ [for $\mathrm{Cd}(\mathrm{OH})_{2}$ in buffer solution $]$
$$ \begin{array}{cl} \mathrm{So}, \mathrm{pOH}=2 & \left(\because \mathrm{pH}+\mathrm{pOH}=\mathrm{p} K_{w}\right) \ 12+\mathrm{pOH}=14 \ \mathrm{pOH}=14-12=2 \end{array} $$
$\therefore \quad\left[\mathrm{OH}^{-}\right]=10^{-2}$ in buffer solution.
$$ \begin{aligned} & \text { For reaction } \mathrm{Cd}(\mathrm{OH}){2} \longrightarrow \mathrm{Cd}{S}^{2+}+2 \mathrm{OH}{S^{2}} \ & \qquad \begin{aligned} & K{\mathrm{sp}}=\left[\mathrm{Cd}^{2+}\right]\left[\mathrm{OH}^{-}\right]^{2} \ & K_{\mathrm{sp}}=(S)(2 S)^{2}=4 S^{3}=4\left(1.84 \times 10^{-5}\right)^{3} \ & K_{\mathrm{sp}}=24.9 \times 10^{-15} \ & {\left[\mathrm{Cd}^{2+}\right] }=\frac{K_{\mathrm{sp}}}{\left[\mathrm{OH}^{-}\right]^{2}} \ & {\left[\mathrm{Cd}^{2+}\right] }=\frac{24.9 \times 10^{-15}}{\left(10^{-2}\right)^{2}}=24.9 \times 10^{-15} \times 10^{+4} \ & \Rightarrow \quad=24.9 \times 10^{-11} \mathrm{M} \ & {\left[\mathrm{Cd}^{2+}\right] \Rightarrow 2.49 \times 10^{-10} \mathrm{M} } \end{aligned} \end{aligned} $$
The expected solubility of $\mathrm{Cd}(\mathrm{OH})_{2}$ in a buffer solution of $\mathrm{pH}=12$ is $2.49 \times 10^{-10} \mathrm{M}$.
