$\mathrm{pH}$ of Solution $\mathrm{A}=6$
Therefore, concentration of $\left[\mathrm{H}^{+}\right]$ion in solution $\mathrm{A}=10^{-6} \mathrm{~mol} \mathrm{~L}^{-1}$
$\mathrm{pH}$ of Solution $\mathrm{B}=4$
Therefore, Concentration of $\left[\mathrm{H}^{+}\right]$ion concentration of solution $\mathrm{B}=10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}$
On mixing one litre of each solution, total volume $=1 \mathrm{~L}+1 \mathrm{~L}=2 \mathrm{~L}$
Amount of $\mathrm{H}^{+}$ions in $1 \mathrm{~L}$ of solution $\mathrm{A}$
$\begin{aligned}
&=\text { Concentration } \times \text { volume } \mathrm{V} \\
&=10^{-6} \mathrm{~mol} \times 1 \mathrm{~L}
\end{aligned}$
Amount of $\mathrm{H}^{+}$ions in $1 \mathrm{~L}$ of solution $\mathrm{B}=10^{-4} \mathrm{~mol} \times 1 \mathrm{~L}$
$\therefore$ Total amount of $\mathrm{H}^{+}$ions in the solution formed by mixing solution $\mathrm{A}$ and $\mathrm{B}$ is $\left(10^{-6} \mathrm{~mol}+10^{-4} \mathrm{~mol}\right)$
This amount is present in $2 \mathrm{~L}$ solution.
$\begin{aligned} \therefore \text { Total }\left[\mathrm{H}^{+}\right] &=\frac{10^{-4}(1+0.01)}{2} \\ &=\frac{1.01 \times 10^{-4}}{2} \mathrm{~mol} \mathrm{~L}^{-1} \\ &=0.5 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \\ &=5 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1} \\ \mathrm{pH} &=-\log \left[\mathrm{H}^{+}\right]=-\log \left(5 \times 10^{-5}\right) \\ &=-[\log 5+(-5 \log 10)] \\ &=-\log 5+5 \\ &=5-\log 5 \\ &=5-0.6990 \\ &=4.3010=4.3 \end{aligned}$