10 mL of ${10}^{-6}$ M HCl solution is mixed with 90 mL ${\mathrm{H}}_2\mathrm{O}$. pH will change approximately

pH of ${10}^{-6}$ M HCl = 6 

$$\begin{gathered} \mathrm{on} \mathrm{dilution} \\ \left[{\mathrm{H}}^{+}\right] \mathrm{obtained} \mathrm{by} \mathrm{using} \\ {\mathrm{M}}_1{\mathrm{V}}_1={\mathrm{M}}_2{\mathrm{V}}_2 ; {\mathrm{M}}_1={{10}^{-6}}_{ }{\mathrm{M}}_2=[\mathrm{H}+] \\ {\mathrm{V}}_1=10\mathrm{ml} {\mathrm{V}}_2=100\mathrm{ml} \end{gathered}$$

$\left[{\mathrm{H}}^{+}\right] = \frac{{10}^{-6} \times 10}{100} = {10}^{-7}$

Thus new pH of HCl is not 7 because it is acid. pH new solution:

$$\begin{gathered} \left[{\mathrm{H}}^{+}\right] = {10}^{-7} + {10}^{-7} (\mathrm{from} {\mathrm{H}}_2\mathrm{O}) = 2 \times {10}^{-7} \\ \therefore \mathrm{pH} = 6.7 \left(\mathrm{approximately}\right) \end{gathered}$$