On mixing equal volumes of two buffer solutions of pH value 3 and 5, the pH of the resultant solution will be

$$\begin{gathered} \left[{\mathrm{H}}^{+}\right]\text{for the solution of }\mathrm{pH}3=1\times {10}^{-3} \\ \left[{\mathrm{H}}^{+}\right]\text{for the solution of }\mathrm{pH}5=1\times {10}^{-5} \\ \mathrm{Let} \mathrm{V} \mathrm{volumes} \mathrm{of} \mathrm{both} \mathrm{the} \mathrm{solutions} \mathrm{are} \mathrm{added}, \mathrm{then} \mathrm{concentration} \mathrm{of} \mathrm{H}+ \mathrm{in} \mathrm{final} \mathrm{mixture} \\ =\frac{1\times {10}^{-3}\times \mathrm{V}+1\times {10}^{-5}\times \mathrm{V}}{\mathrm{V}+\mathrm{V}} \\ =\frac{\mathrm{V}\left(1\times {10}^{-3}+1\times {10}^{-5}\right)}{2\mathrm{V}} \\ =\frac{1.01\times {10}^{-3}}{2}=5.05\times {10}^{-4} \\ \mathrm{pH}\text{ of resultant solution }=-\log \left[{\mathrm{H}}^{+}\right] \\ =-\log \left(5.05\times {10}^{-4}\right)\approx 3.3 \end{gathered}$$