Giải thích các bước giải:
\(\begin{array}{l}
Ba{(OH)_2} + 2HCl \to BaC{l_2} + 2{H_2}O\\
Ba{(OH)_2} + {H_2}S{O_4} \to BaS{O_4} + 2{H_2}O\\
{n_{Ba{{(OH)}_2}}} = 0,03mol\\
{n_{BaS{O_4}}} = 0,02mol\\
\to {n_{Ba{{(OH)}_2}(2)}} = {n_{{H_2}S{O_4}}} = {n_{BaS{O_4}}} = 0,02mol\\
\to {n_{Ba{{(OH)}_2}(1)}} = 0,01mol\\
\to {n_{HCl}} = 2{n_{Ba{{(OH)}_2}(1)}} = 0,02mol\\
\to C{M_{HCl}} = C{M_{{H_2}S{O_4}}} = \dfrac{{0,02}}{{0,02}} = 1M
\end{array}\)