The pH of the solution obtained by mixing 100 mL of a solution of pH = 3 with 400 mL of a solution of pH = 4 is

$$\begin{gathered} \text{ For solution }\mathrm{I},\mathrm{pH}=3 \\ \Rightarrow \left[{\mathrm{H}}^{+}\right]={10}^{-3} \\ \text{ Concentration of solution, }{\mathrm{M}}_1={10}^{-3}\mathrm{M} \\ {\mathrm{V}}_1=100\mathrm{mL} \\ \text{ For solution II, }\mathrm{pH}=4 \\ \left[{\mathrm{H}}^{+}\right]={10}^{-4} \\ \text{ Concentration of solution, }{\mathrm{M}}_2={10}^{-4}\mathrm{M} \\ {\mathrm{V}}_2=400\mathrm{mL} \\ \mathrm{Concenmation} \mathrm{of} \mathrm{resulting} \mathrm{solution} \\ \mathrm{M}=\frac{{\mathrm{M}}_1{\mathrm{V}}_1+{\mathrm{M}}_2{\mathrm{V}}_2}{{\mathrm{V}}_1+{\mathrm{V}}_2} \\ =\frac{{10}^{-3}\times 100+{10}^{-4}\times 400}{100+400}=\frac{0.14}{500} \\ =0.00028=2.8\times {10}^{-4} \\ \therefore \left[{\mathrm{H}}^{+}\right]=2.8\times {10}^{-4} \\ \therefore \mathrm{pH}\text{ of resulting solution }=-\log \left[{\mathrm{H}}^{+}\right]=-\log \left(2.8\times {10}^{-4}\right) \\ =-\log 2.8-\log {10}^{-4} \\ =4-\log 2.8 \end{gathered}$$