Question

$9.8g$ of $H_2{SO}_4$ is present in $500mL$ of the solution. Calculate the pH of the solution.

Answer 1

Molarity of $H_{2}S{O}_4=\frac{9.8\times 1000}{98\times 500}=0.2M$

As $H_{2}S{O}_4$ is strong acid so it dissociates completely to give $H^{+}$ ions.

$H_{2}S{O}_4\to 2H^{+}+S{O}_4^{2-}$

$0.2$

$0.2\times 2$ $0.2$

$\left[H^{+}\right]=2\times 0.2=0.4M$

$pH=-log\left(\right[H^{+}\left]\right)=0.3979$