The pH of a solution is 5.0. To a 10 ml of this solution, 990 ml of water is added. Then the pH of the resulting solution is

$$\begin{gathered} {\mathrm{V}}_1{\mathrm{N}}_1={\mathrm{V}}_2{\mathrm{N}}_2; \\ 10\left({10}^{-5}\right)=\left(10+990\right){\mathrm{N}}_2; \\ {\mathrm{N}}_2={10}^{-7} \mathrm{M} ; \end{gathered}$$

When the concentration of the acid or base is less than equal to 10^-7 M, then [ H⁺] or [OH^-] furnished from water is also taken into accout.

$$\begin{gathered} \left[{\mathrm{H}}^{+}\right]={10}^{-7}\left(\mathrm{acid}\right)+{10}^{-7}\left(\mathrm{water}\right) ; \\ \left[{\mathrm{H}}^{+}\right]=2\times {10}^{-7}; \\ {\mathrm{p}}^{\mathrm{H}}=7-\log 2=6.7 ; \end{gathered}$$