pH of an aqueous solution having 0.49g of ${\text{H}}_{\text{2}}{\text{SO}}_4$ in 100 ml of solution is

Given weight of  ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}=0.49\text{\hspace{0.17em}g}$

$\text{M.wt = 98}$

$\text{V = 100 ml}$

$\text{M}=\frac{0.49}{98}\times \frac{1000}{100}$

$=0.05\text{\hspace{0.17em}M}$

${\text{[H}}^{+}]=0.05\times 2=0.1={10}^{-1}$

${\text{P}}^{\text{H}}=-\log {10}^{-1}=1$