Equal volume of two solutions, one having pH 6 and the other having pH 4 are mixed. The pH of the resulting solution would be

$$\begin{gathered} \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right] = \frac{{10}^{-6} +{10}^{-4} }{2} = 5.05 \times {10}^{-5} \\ \mathrm{pH} = -\log \left(5.05 \times {10}^{-5}\right) = 4.3 \\ \mathrm{Alternately} \\ \mathrm{When} \mathrm{equal} \mathrm{volumes} \mathrm{of} \mathrm{two} \mathrm{acidic} \mathrm{solutions} \mathrm{are} \mathrm{mixed} \mathrm{resulting} {\mathrm{p}}^{\mathrm{H} }=\mathrm{Lower} {\mathrm{p}}^{\mathrm{H}}+0.3 \end{gathered}$$