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Calculate the molar solubility of calcium hydroxide $\mathrm{Ca}(\mathrm{OH})_2$ in $0.10 \mathrm{M} \mathrm{NaOH}$ solution. The ionic product of calcium hydroxide is $5.5 \times 10^{-6}$.
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Verified Answer
The correct answer is:
$5.5 \times 10^{-4}$
Given,
Ionic product of $\mathrm{Ca}(\mathrm{OH})_2=5.5 \times 10^{-6}$
Concentration of $\mathrm{NaOH}$ solution $=0.10 \mathrm{M}$
$\because$ Ionic product of $\mathrm{Ca}(\mathrm{OH})_2=5.5 \times 10^{-6}$
or, $\mathrm{Ca}(\mathrm{OH})_2=\left[\mathrm{Ca}^{2+}\right]+2\left[\mathrm{OH}^{-}\right]$
Let, solubility $=S$
$\begin{gathered}\mathrm{Ca}^{2+}+2\left[\mathrm{OH}^{-}\right] \\ S \quad(2 S)^2\end{gathered}$
Also, $K_{\mathrm{sp}}=4 S^3$
where, $K_{\mathrm{sp}}=$ solubility product
$\therefore$ Due to common ion effect,
Molar solubility of $\mathrm{Ca}(\mathrm{OH})_2=\frac{5.5 \times 10^{-6}}{0.1 \times 0.1}$
$=5.5 \times 10^{-4}$
Ionic product of $\mathrm{Ca}(\mathrm{OH})_2=5.5 \times 10^{-6}$
Concentration of $\mathrm{NaOH}$ solution $=0.10 \mathrm{M}$
$\because$ Ionic product of $\mathrm{Ca}(\mathrm{OH})_2=5.5 \times 10^{-6}$
or, $\mathrm{Ca}(\mathrm{OH})_2=\left[\mathrm{Ca}^{2+}\right]+2\left[\mathrm{OH}^{-}\right]$
Let, solubility $=S$
$\begin{gathered}\mathrm{Ca}^{2+}+2\left[\mathrm{OH}^{-}\right] \\ S \quad(2 S)^2\end{gathered}$
Also, $K_{\mathrm{sp}}=4 S^3$
where, $K_{\mathrm{sp}}=$ solubility product
$\therefore$ Due to common ion effect,
Molar solubility of $\mathrm{Ca}(\mathrm{OH})_2=\frac{5.5 \times 10^{-6}}{0.1 \times 0.1}$
$=5.5 \times 10^{-4}$
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