Chemical and Ionic Equilibrium 2 Question 2
2. What is the molar solubility of $\mathrm{Al}(\mathrm{OH})_{3}$ in $0.2 \mathrm{M} \mathrm{NaOH}$ solution? Given that, solubility product of
$\mathrm{Al}(\mathrm{OH})_{3}=2.4 \times 10^{-24}$
(2019 Main, 12 April II)
(a) $3 \times 10^{-19}$
(b) $12 \times 10^{-21}$
(c) $3 \times 10^{-22}$
(d) $12 \times 10^{-23}$
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Answer:
Correct Answer: 2. (b)
Solution:
- Key Idea Concentration of substance in a saturated solution is defined as its solubility $(S)$. Its value depends upon the nature of solvent and temperature. For reaction,
$$ \begin{gathered} A B \rightleftharpoons A^{+}+B^{-} \ K_{\mathrm{sp}}=\left[A^{+}\right]\left[B^{-}\right] \end{gathered} $$
$\mathrm{NaOH} \longrightarrow \underset{0.2}{\mathrm{Na}^{+}}+\underset{0.2}{\mathrm{OH}^{-}}$
$K_{\text {sp }}$ of $\mathrm{Al}(\mathrm{OH})_{3}=2.4 \times 10^{-24}$ (Given)
$$ K_{\mathrm{sp}}=\left[\mathrm{Al}^{3+}\right]\left[\mathrm{OH}^{-}\right]^{3} $$
$2.4 \times 10^{-24}=[S][3 S+0.2]^{3}$
$[\because 0.2»S]$
$2.4 \times 10^{-24}=[S][0.008]$
$[S]=3 \times 10^{-22}$
