QJEEAdv23Chcd77: JEE Advanced 2023 Chemistry - Class 11 JEEAdv23Chcd77 Solution: On decreasing the from 7 to 2 , the solubility of a sparingly…

Question

Verified Solution • 2025–26 Session moderate

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

Accepted Answer

Relationship between solubility,

\mathrm{H}^{+}

and

\mathrm{K}_{\mathrm{a}}

is given by

\begin{aligned}\mathrm{S} &amp; =\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 &amp; \Rightarrow\left(\mathrm{H}^{+}\right)=10^{-7} \\\\\mathrm{~S} &amp; =10^{-4} \mathrm{~mol} / \mathrm{L}\end{aligned}

\begin{aligned}\Rightarrow 10^{-4} &amp; =\sqrt{\frac{\mathrm{K}_{\mathrm{SP}}\left(10^{-7}+\mathrm{K}_{\mathrm{a}}\right)}{\mathrm{K}_{\mathrm{a}}}} ...........(i) \\\\10^{-3} &amp; =\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}&amp; \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)} \\\\&amp; \Rightarrow 10^{-2}=\frac{10^{-7}+\mathrm{K}_{\mathrm{a}}}{10^{-2}+\mathrm{K}_{\mathrm{a}}} \\\\&amp; \Rightarrow 10^{-4}+10^{-2} \cdot \mathrm{K}_{\mathrm{a}}=10^{-7}+\mathrm{K}_{\mathrm{a}} \\\\&amp; \therefore \mathrm{K}_{\mathrm{a}} \simeq 10^{-4} \\\\&amp; \Rightarrow \mathrm{pK}_{\mathrm{a}}=4\end{aligned}

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