solubility $=\mathrm{S}_1=\sqrt{\mathrm{K}_{>p}}$
At $\mathrm{pH}=2$
$\Rightarrow \mathrm{MX}(\mathrm{s})+\mathrm{aq} \stackrel{\mathrm{K}_{\mathrm{sp}}}{\rightleftharpoons} \mathrm{M}^{+}(\mathrm{aq})+\mathrm{X}^{-}(\mathrm{aq})$
$5 \quad 5-\mathrm{x}$
$\mathrm{X}^{-}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq}) \stackrel{1 / \mathrm{K},}{\rightleftharpoons} \mathrm{HX}(\mathrm{aq})$
$5-x \quad 10^{-2} \quad x=5$
Approximation : $5-x=0\left[\mathrm{X}^{-}\right.$is limiting reagent]
$\Rightarrow 5=x$
$\Rightarrow s(5-x)=K_{\infty}$ $. . . . . . .(1)$
$\frac{5}{(s-x)\left(10^{-2}\right)}=\frac{1}{K_2}$ $. . . . . . .(2)$
Multiply (1) $\times(2) \Rightarrow \frac{s^2}{10^{-2}}=\frac{\mathrm{K}_{\mathrm{up}}}{\mathrm{K}_{\mathrm{a}}}$
$\Rightarrow s=\frac{\sqrt{\mathrm{K}_{s p}}}{10 \sqrt{\mathrm{K}_v}}$
Now given : $\frac{5}{s_1}=\frac{10^{-3}}{10^{-4}}$
$\frac{\frac{\sqrt{\mathrm{K}_{\Delta p}}}{10 \sqrt{\mathrm{K}_a}}}{\sqrt{\mathrm{K}_{\varphi}}}=10 \Rightarrow \frac{1}{10 \sqrt{\mathrm{K}_{\mathrm{a}}}}=10$
$\Rightarrow \sqrt{\mathrm{K}_2}=10^{-2}$
$\Rightarrow \mathrm{K}_{\mathrm{s}}=10^{-4}$
$\Rightarrow p \mathrm{~K}_a=4$
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(Molecular Weight of $\mathrm{HNO}_{3}=63$ )
${C_6}{H_5} - H\mathop {\xrightarrow{{[X]}}}\limits_{AlC{l_3}} $ $\begin{matrix}
O\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\
||\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\
{{C}_{6}}{{H}_{5}}-C-C{{H}_{2}}-{{C}_{6}}{{H}_{5}} \\
\end{matrix}$
$[X]$ will be :