JEE Advance - Chemistry (2023 - Paper 1 Online - No. 6)
On decreasing the $p \mathrm{H}$ from 7 to 2 , the solubility of a sparingly soluble salt (MX) of a weak acid (HX) increased from $10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}$ to $10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}$. The $p \mathrm{~K}_{\mathrm{a}}$ of $\mathrm{HX}$ is
3
4
5
2
Explanation
Relationship between solubility, $\mathrm{H}^{+}$and $\mathrm{K}_{\mathrm{a}}$ is given by
$$ \begin{aligned} \mathrm{S} & =\sqrt{\frac{\left(\mathrm{K}_{\mathrm{SP}}\left[\mathrm{H}^{+}\right]+\mathrm{K}_{\mathrm{a}}\right)}{\mathrm{K}_{\mathrm{a}}}} \\\\ \text { If } \mathrm{pH}=7 & \Rightarrow\left(\mathrm{H}^{+}\right)=10^{-7} \\\\ \mathrm{~S} & =10^{-4} \mathrm{~mol} / \mathrm{L} \end{aligned} $$
$$ \begin{aligned} \Rightarrow 10^{-4} & =\sqrt{\frac{\mathrm{K}_{\mathrm{SP}}\left(10^{-7}+\mathrm{K}_{\mathrm{a}}\right)}{\mathrm{K}_{\mathrm{a}}}} ...........(i) \\\\ 10^{-3} & =\sqrt{\frac{\mathrm{K}_{\mathrm{SP}}\left(10^{-2}+\mathrm{K}_{\mathrm{a}}\right)}{\mathrm{K}_{\mathrm{a}}}} ...........(ii) \end{aligned} $$
Dividing and squaring equation (i) by equation (ii),
$$ \begin{aligned} & \frac{\left(10^{-4}\right)^2}{\left(10^{-3}\right)^2}=\frac{K_{SP}\left(10^{-7}+K_a\right)}{K_a} \times \frac{K_a}{K_{SP}\left(10^{-2}+K_a\right)} \\\\ & \Rightarrow 10^{-2}=\frac{10^{-7}+\mathrm{K}_{\mathrm{a}}}{10^{-2}+\mathrm{K}_{\mathrm{a}}} \\\\ & \Rightarrow 10^{-4}+10^{-2} \cdot \mathrm{K}_{\mathrm{a}}=10^{-7}+\mathrm{K}_{\mathrm{a}} \\\\ & \therefore \mathrm{K}_{\mathrm{a}} \simeq 10^{-4} \\\\ & \Rightarrow \mathrm{pK}_{\mathrm{a}}=4 \end{aligned} $$
$$ \begin{aligned} \mathrm{S} & =\sqrt{\frac{\left(\mathrm{K}_{\mathrm{SP}}\left[\mathrm{H}^{+}\right]+\mathrm{K}_{\mathrm{a}}\right)}{\mathrm{K}_{\mathrm{a}}}} \\\\ \text { If } \mathrm{pH}=7 & \Rightarrow\left(\mathrm{H}^{+}\right)=10^{-7} \\\\ \mathrm{~S} & =10^{-4} \mathrm{~mol} / \mathrm{L} \end{aligned} $$
$$ \begin{aligned} \Rightarrow 10^{-4} & =\sqrt{\frac{\mathrm{K}_{\mathrm{SP}}\left(10^{-7}+\mathrm{K}_{\mathrm{a}}\right)}{\mathrm{K}_{\mathrm{a}}}} ...........(i) \\\\ 10^{-3} & =\sqrt{\frac{\mathrm{K}_{\mathrm{SP}}\left(10^{-2}+\mathrm{K}_{\mathrm{a}}\right)}{\mathrm{K}_{\mathrm{a}}}} ...........(ii) \end{aligned} $$
Dividing and squaring equation (i) by equation (ii),
$$ \begin{aligned} & \frac{\left(10^{-4}\right)^2}{\left(10^{-3}\right)^2}=\frac{K_{SP}\left(10^{-7}+K_a\right)}{K_a} \times \frac{K_a}{K_{SP}\left(10^{-2}+K_a\right)} \\\\ & \Rightarrow 10^{-2}=\frac{10^{-7}+\mathrm{K}_{\mathrm{a}}}{10^{-2}+\mathrm{K}_{\mathrm{a}}} \\\\ & \Rightarrow 10^{-4}+10^{-2} \cdot \mathrm{K}_{\mathrm{a}}=10^{-7}+\mathrm{K}_{\mathrm{a}} \\\\ & \therefore \mathrm{K}_{\mathrm{a}} \simeq 10^{-4} \\\\ & \Rightarrow \mathrm{pK}_{\mathrm{a}}=4 \end{aligned} $$
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