$9.8 \mathrm{gm}$ of $\mathrm{H}_{2} \mathrm{SO}_{4}$ in $100 \mathrm{L}$ sol. $\Rightarrow 10^{-3} \mathrm{M}$ sol.

Mixture : $40 L$ of $10^{-3} \mathrm{M} \mathrm{NaOH}$ and $10 \mathrm{L}$ of

$10^{-3} \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}$ sol.

Final Conc. of $OH^-=\frac{10^{-3}(40 \times 1-10 \times 1 \times 2)}{40+10}=6 \times 10^{-4} \mathrm{M}$

$\mathrm{pOH}=-\log \left(6 \times 10^{-4}\right)$

$=4-\log 6=4-0.60=3.40$

$\mathrm{pH}=14-3.40=10.60$