The dilution processes of different aqueous solutions with water are given in LIST-I. The effects of…

1. a-q;b-t;c-p;d-r;
2. a-p;b-t;c-s;d-p;
3. a-s;b-t;c-p;d-r;
4. a-s;b-q;c-p;d-t;

Solution

$\left(\mathrm{A}\right)$

$\mathrm{pH}=\mathrm{pKa}\Rightarrow \left[{\mathrm{H}}^{+}\right]$ will not change on dilution.
$\left(\mathrm{B}\right)$

$\left[{\mathrm{OH}}^{-}\right]=\sqrt{{\mathrm{K}}_{\mathrm{H}}\mathrm{C}}=\sqrt{\left(\frac{{\mathrm{k}}_{\mathrm{w}}}{{\mathrm{k}}_{\mathrm{a}}}\mathrm{C}\right)}$
${\left[{\mathrm{H}}^{+}\right]}_1=\sqrt{\frac{{\mathrm{k}}_{\mathrm{w}}{\mathrm{k}}_{\mathrm{a}}}{\mathrm{C}}}$
$\frac{{\left[{\mathrm{H}}^{+}\right]}_2}{{\left[{\mathrm{H}}^{+}\right]}_1}=\sqrt{\frac{{\mathrm{C}}_1}{{\mathrm{C}}_2}}=\sqrt{\frac{0.05}{0.025}}=\sqrt{2}$
$\left(\mathrm{C}\right)$

$\left[{\mathrm{H}}^{+}\right]=\sqrt{{\mathrm{K}}_{\mathrm{H}}\mathrm{C}}$
$\frac{{\left[{\mathrm{H}}^{+}\right]}_2}{{\left[{\mathrm{H}}^{+}\right]}_1}=\sqrt{\frac{{\mathrm{C}}_2}{{\mathrm{C}}_1}}=\frac{1}{\sqrt{2}}$
$\left(\mathrm{D}\right)$ Because of dilution, the solubility does not change. So, $\left[{\mathrm{H}}^{+}\right]=\mathrm{constant}$.

Asked in: JEE Advanced 2018 (Paper 2)

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