Đáp án:
\({C_{{M_{HCl}}}} = 0,93M\)
Giải thích các bước giải:
\(\begin{array}{l}
NaOH + HCl \to NaCl + {H_2}O\\
Ba{(OH)_2} + 2HCl \to BaC{l_2} + 2{H_2}O\\
{m_{NaOH}} = \dfrac{{300 \times 10}}{{100}} = 30g\\
{n_{NaOH}} = \dfrac{{30}}{{40}} = 0,75mol\\
{m_{{\rm{dd}}Ba{{(OH)}_2}}} = 125 \times 1,28 = 160g\\
{m_{Ba{{(OH)}_2}}} = \dfrac{{160 \times 10}}{{100}} = 16g\\
{n_{Ba{{(OH)}_2}}} = \dfrac{{16}}{{171}} = 0,09mol\\
{n_{HCl}} = {n_{NaOH}} + 2{n_{Ba{{(OH)}_2}}} = 0,93mol\\
{C_{{M_{HCl}}}} = \dfrac{{0,93}}{1} = 0,93M
\end{array}\)