Chemical And Ionic Equilibrium Result 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.
$A_x B_y \rightleftharpoons x A^{y+}+y B^{x-} K_{\text {sp }}=\left[A^{y+}\right]^x\left[B^{x-}\right]^y$
Solubility of $\mathrm{Cd}(\mathrm{OH})_2(S)=1.84 \times 10^{-5} \mathrm{M}$
Given, $\mathrm{pH}=12$ [for $\mathrm{Cd}(\mathrm{OH})_2$ in buffer solution]
$\begin{gathered} \mathrm{So}, \mathrm{pOH}=2 \quad\left(\because \mathrm{pH}+\mathrm{pOH}=\mathrm{p} K_w\right) \\ 12+\mathrm{pOH}=14 \\ \mathrm{pOH}=14-12=2 \end{gathered}$
$\therefore \quad\left[\mathrm{OH}^{-}\right]=10^{-2}$ in buffer solution.
$\begin{aligned} & \text { For reaction } \mathrm{Cd}(\mathrm{OH})2 \longrightarrow \mathrm{Cd}S^{2+}+2 \mathrm{OH}^{-} \\ & \qquad \begin{aligned} & K{\mathrm{sp}}=\left[\mathrm{Cd}^{2+}\right]\left[\mathrm{OH}^{-}\right]^2 \\ & K{\mathrm{sp}}=(S)(2 S)^2=4 S^3=4\left(1.84 \times 10^{-5}\right)^3 \\ & K_{\mathrm{sp}}=24.9 \times 10^{-15} \\ & {\left[\mathrm{Cd}^{2+}\right] }=\frac{K_{\mathrm{sp}}}{\left[\mathrm{OH}^{-}\right]^2} \\ & {\left[\mathrm{Cd}^{2+}\right] }=\frac{24.9 \times 10^{-15}}{\left(10^{-2}\right)^2}=24.9 \times 10^{-15} \times 10^{+4} \\ &=24.9 \times 10^{-11} \mathrm{M} \\ & {\left[\mathrm{Cd}^{2+}\right] \Rightarrow 2.49 \times 10^{-10} \mathrm{M} } \end{aligned} \end{aligned}$
The expected solubility of $\mathrm{Cd}(\mathrm{OH})_2$ in a buffer solution of $\mathrm{pH}=12$ is $2.49 \times 10^{-10} \mathrm{M}$
BETA
