The specific gravity of $H_{2}S{O}_4$  is 1.8 g/cc and this solution is found to contain 98% $H_{2}S{O}_4$  by weight. 10 cc of this solution is mixed with 350 cc of pure water. 25 ml of this dil. $H_{2}S{O}_4$  solution neutralizes 500 ml of $NaOH$  solution. Then the $P^H$  of $NaOH$  solution is

$N_{H_{2}S{O}_4}=\frac{\%\times Spgr\times 10}{GEW}$

              $=\frac{98\times 1.8\times 10}{49}$

              $=36N$

        $N_1{V}_1=N_2{V}_2$

    $36\times 10=N_2\times 360$

     $\Rightarrow N_2=1$

       $N_a{V}_a=N_b{V}_b$

       $1\times 25=N_b\times 500$

$\Rightarrow N_b=\left[O{H}^{-}\right]=5\times {10}^{-2}$

         $\left[H^{+}\right]=\frac{{10}^{-14}}{5\times {10}^{-2}}$

                     $=0.2\times {10}^{-12}$

                     $=2\times {10}^{-13}$

             $\therefore P^H=13-\log 2$

                        $=13-0.3010$

                        $=12.6990$