The \(\mathrm{pH}\) of the solution obtained by mixing \(100 \mathrm{ml}\) of a solution of \(\mathrm{pH}=3\) with \(400 \mathrm{~mL}\) of a solution of \(\mathrm{pH}=4\) is

For solution \(\mathrm{I}, \mathrm{pH}=3\)
\[
\Rightarrow \quad\left[\mathrm{H}^{+}\right]=10^{-3}
\]
\(\Rightarrow\) Concentration of solution \(\mathrm{M}_{1}=10^{-3} \mathrm{M}\)
\[
\mathrm{V}_{1}=100 \mathrm{~mL}
\]
For solution II, pH = 4
\(\Rightarrow \quad\left[\mathrm{H}^{+}\right]=10^{-4}\) \(\Rightarrow\) Concentration of solution, \(\mathrm{M}_{2}=10^{-4} \mathrm{M}\) \(\mathrm{V}_{2}=400 \mathrm{~mL}\)
Concentration of resulting solution
\[
\begin{aligned}
\mathrm{M} &=\frac{\mathrm{M}_{1} \mathrm{~V}_{1}+\mathrm{M}_{2} \mathrm{~V}_{2}}{\mathrm{~V}_{1}+\mathrm{V}_{2}} \\
&=\frac{10^{-3} \times 100+10^{-4} \times 400}{100+400} \\
&=\frac{0.14}{500} \\
\mathrm{M} &=0.00028=2.8 \times 10^{-4} \\
{\left[\quad\left[\mathrm{H}^{+}\right]\right.} &=2.8 \times 10^{-4} \\
\therefore \mathrm{pH} \text { of resulting solution } \\
&=-\log \left[\mathrm{H}^{+}\right] \\
&=-\log \left(2.8 \times 10^{-4}\right) \\
&=-\log 2.8-\log 10^{-4} \\
&=4-\log 2.8
\end{aligned}
\]