One litre of $\text{1}\text{M}$ solution of an acid $\text{HA}\left({\text{K}}_{\text{a}}\text{=}{\text{10}}^{-\text{4}}\text{ at 25}{}^{\text{o}}\text{C}\right)$ has $\text{pH}\text{=}\text{2}\text{.}$ It is diluted by water so the new $\mathrm{p}\mathrm{H}$ becomes double. The solution was diluted to $\mathrm{y}\times 10^{\mathrm{z}}\mathrm{ }\mathrm{m}\mathrm{l}$ . The value of $\frac{\mathrm{y}+\mathrm{z}}{2}$ is:

Solution

$\mathrm{pH}=2\Rightarrow {\left[{\mathrm{H}}^{+}\right]}_1={10}^{-2}={\mathrm{c}}_1{\mathrm{\alpha }}_1=1\times {\mathrm{\alpha }}_1$

${\mathrm{\alpha }}_1={10}^{-2}$

$\mathrm{pH}=4\Rightarrow {\left[{\mathrm{H}}^{+}\right]}_2={10}^{-4}=\frac{1\times 1000}{\mathrm{V}}\times {\mathrm{\alpha }}_2$

${\mathrm{k}}_{\mathrm{a}}={10}^{-4}=\frac{{\mathrm{\alpha }}_2^2{\mathrm{C}}_2}{1-{\mathrm{\alpha }}_2}=\frac{{\mathrm{\alpha }}_2\times {\mathrm{\alpha }}_2\times {\mathrm{C}}_2}{1-{\mathrm{\alpha }}_2}$

${10}^{-4}=\frac{{\mathrm{\alpha }}_2\times {10}^{-4}}{1-{\mathrm{\alpha }}_2}\Rightarrow {\mathrm{\alpha }}_2=0.5$

${\left[{\mathrm{H}}^{+}\right]}_2=\frac{1\times 1000}{\mathrm{V}}\times {\mathrm{\alpha }}_2={10}^{-4}$

$\frac{1\times 1000}{\mathrm{V}}\times 0.5={10}^{-4}$

$\mathrm{V}=5\times {10}^6 \mathrm{ml}$

$\Rightarrow \frac{\mathrm{x}+\mathrm{y}}{2}=\frac{5+6}{2}=\frac{11}{2}=5.5$

Hence, the correct answer is 5.5.