Potential of hydrogen

We will use here : $\large\color{#20A900}{ pH= - log_{10}[H^+]} \hspace{1cm} \ldots (1)$ Now here , $\ce{H_2SO_4}$ is a dibasic acid and strong acid(100% dissociation) , i.e. , it gives $\color{#D61F06}{\text{2 moles}}$ of $\ce{H^+}$ $\color{#D61F06}{\text{ per mole}}$ of the acid in aqueous solution. Therefore : $\large [H^+] = \color{#3D99F6}{2} \times 0.005 = \text{0.01 M}$ Now from eq.(1) , $\large pH = - log_{10}(10^{-2})= \color{#EC7300}{\boxed{2}}$

$H_2SO_4 \rightarrow 2H^+ + SO_4^-$ . As $H_2SO4$ is a strong acid, it will dissolve completely and $0.005 \text{M }H_2SO4$ will become $0.01 \text{M }H^+$ and $0.005 \text{M }SO_4^-$ . We can ignore the effect of $SO_4$ on pH as it is a negligible base. Therefore the pH is $pH = -\log{[H^+]} = -\log{0.01} = -(-2) = 2$ .