If 50 mL solution of pH 1.0 is mixed with 50 mL solution of pH 2.0, the pH of the mixture is

${\mathrm{V}}_1=50\mathrm{mL},{\mathrm{V}}_2=50\mathrm{mL},{\mathrm{PH}}_1=1\to {\left[{\mathrm{H}}^{+}\right]}_1={10}^{-1},{\mathrm{PH}}_2=2\to {\left[{\mathrm{H}}^{+}\right]}_2={10}^{-2}$

$\frac{{\mathrm{N}}_{\mathrm{a}}{\mathrm{V}}_{\mathrm{a}}+{\mathrm{N}}_{\mathrm{b}}{\mathrm{V}}_{\mathrm{b}}}{{\mathrm{V}}_{\mathrm{a}}+{\mathrm{V}}_{\mathrm{b}}}=\frac{{10}^{-1}\times 50+{10}^{-2}\times 50}{100}=\frac{50\left({10}^{-1}+{10}^{-2}\right)}{100}$

$\left[{\mathrm{H}}^{+}\right]=\frac{50\left({\displaystyle \frac{1}{10}}+{\displaystyle \frac{1}{10}}\right)}{100}=\frac{50\times 50}{100\times 100}=\frac{11}{200}=5.5\times {10}^{-2}=55\times {10}^{-3}$

$\mathrm{PH}=-{\log }_{10}^{\left[{\mathrm{H}}^{+}\right]}=-{\log }_{10}55\times {10}^{-3}$

$=-{\log }_{10}^{{10}^{-3}}-{\log }_{10}^{55}$

= 3 − 1.74 = 1.26