Em tham khảo nha:
\(\begin{array}{l}
\text{ Gọi a là thể tích của NaOH }\\
{V_{{H_2}S{O_4}}} = 2{V_{NaOH}} = 2a\,(l)\\
{n_{{H_2}S{O_4}}} = 0,01 \times 2a = 0,02a\,mol\\
{n_{NaOH}} = 0,01 \times a = 0,01a\,mol\\
2NaOH + {H_2}S{O_4} \to N{a_2}S{O_4} + 2{H_2}O\\
\dfrac{{{n_{NaOH}}}}{2} < {n_{{H_2}S{O_4}}} \Rightarrow {H_2}S{O_4} \text{ dư }\\
{n_{{H_2}S{O_4}}} \text{ dư }= 0,02a - \dfrac{{0,01a}}{2} = 0,015a\,mol\\
{n_{N{a_2}S{O_4}}} = \dfrac{{0,01a}}{2} = 0,005a\,mol\\
{n_{{H^ + }}} = 2{n_{{H_2}S{O_4}}}\text{ dư } = 0,03a\,mol\\
{n_{N{a^{2 + }}}} = 2{n_{N{a_2}S{O_4}}} = 0,01a\,mol\\
{n_{S{O_4}^{2 - }}} = 0,015a + 0,005a = 0,02a\,mol\\
{\rm{[}}{H^ + }{\rm{]}} = \dfrac{{0,03a}}{{2a + a}} = 0,01M\\
{\rm{[}}N{a^{2 + }}{\rm{]}} = \dfrac{{0,01a}}{{2a + a}} \approx 0,0033M\\
{\rm{[}}S{O_4}^{2 - }{\rm{]}} = \dfrac{{0,02a}}{{2a + a}} \approx 0,0067M
\end{array}\)