Đáp án:
a, \(C{M_{{H_2}S{O_4}}} = \dfrac{{0,02}}{{0,08}} = 0,25M\)
b,
\(\begin{array}{l}
C{M_{N{a^ + }}} = \dfrac{{0,04}}{{(0,02 + 0,08)}} = 0,4M\\
C{M_{S{O_4}^{2 - }}} = \dfrac{{0,02}}{{(0,02 + 0,08)}} = 0,2M
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
2NaOH + {H_2}S{O_4} \to N{a_2}S{O_4} + 2{H_2}O\\
{n_{NaOH}} = 0,04mol\\
\to {n_{{H_2}S{O_4}}} = \dfrac{1}{2}{n_{NaOH}} = 0,02mol\\
\to C{M_{{H_2}S{O_4}}} = \dfrac{{0,02}}{{0,08}} = 0,25M\\
{n_{N{a_2}S{O_4}}} = \dfrac{1}{2}{n_{NaOH}} = 0,02mol\\
\to {n_{N{a^ + }}} = 2{n_{N{a_2}S{O_4}}} = 0,04mol\\
\to {n_{S{O_4}^{2 - }}} = {n_{N{a_2}S{O_4}}} = 0,02mol\\
\to C{M_{N{a^ + }}} = \dfrac{{0,04}}{{(0,02 + 0,08)}} = 0,4M\\
\to C{M_{S{O_4}^{2 - }}} = \dfrac{{0,02}}{{(0,02 + 0,08)}} = 0,2M
\end{array}\)