The pH of solution obtained by mixing $500 ml$ of $0.15 M$  $H_{2}S{O}_4$ with 500 ml of 0.1 M  $NaOH$ is

Millimole of  $H^{+}=0.3\times 500=150$
Millimole of  $O{H}^{-}=0.1\times 500=50$
After neutralization Millimole of $H^{+}$ left $150-50=100$
So,  $\left[H^{+}\right]=\frac{100}{500+500}=\frac{1}{10}$

$pH=-\log \left(\frac{1}{10}\right)=1$