Chemical And Ionic Equilibrium Result Question 2
2. What is the molar solubility of $\mathrm{Al}(\mathrm{OH})_3$ in 0.2 M 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. ( c )
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,
$ A B \rightleftharpoons A^{+}+B^{-} $
$ K_{\text {sp }}=\left[A^{+}\right]\left[B^{-}\right] $
$ \mathrm{NaOH} \longrightarrow \underset{0.2}{\mathrm{Na}^{+}}+\underset{0.2}{\mathrm{OH}^{-}} $
$ K_{\text {sp }} \text { of } \mathrm{Al}(\mathrm{OH})_3=2.4 \times 10^{-24} \text { (Given) } $
$ K_{\text {sp }}=\left[\mathrm{Al}^{3+}\right]\left[\mathrm{OH}^{-}\right]^3 $
$ 2.4 \times 10^{-24}=[S][3 S+0.2]^3 \quad$ $ [\because 0.2»S] $
$ 2.4 \times 10^{-24}=[S][0.008] $
$ {[S]=3 \times 10^{-22}} $
BETA
