Em tham khảo nha:

\(\begin{array}{l}
a)\\
{\rm{[}}{H^ + }{\rm{]}} = 2{C_M}{H_2}S{O_4} = 0,004\,M\\
pH =  - \log (0,004) = 2,4\\
b)\\
{n_{{H^ + }}} = 2{n_{{H_2}S{O_4}}} = 2 \times 0,075 \times 0,003 = 4,5 \times {10^{ - 4}}\,mol\\
{n_{O{H^ - }}} = {n_{KOH}} = 0,025 \times 0,005 = 1,25 \times {10^{ - 4}}\,mol\\
{H^ + } + O{H^ - } \to {H_2}O\\
{n_{{H^ + }}} > {n_{O{H^ - }}} \Rightarrow {H^ + } \text{ dư } \\
{n_{{H^ + }}} \text{ dư }= 4,5 \times {10^{ - 4}} - 1,25 \times {10^{ - 4}} = 3,25 \times {10^{ - 4}}\,mol\\
{\rm{[}}{H^ + }{\rm{]}} \text{ dư } = \dfrac{{3,25 \times {{10}^{ - 4}}}}{{0,075 + 0,025}} = 3,25 \times {10^{ - 3}}M\\
pH =  - \log (3,25 \times {10^{ - 3}}) = 2,49
\end{array}\)