Đáp án:

\({C_{M{\text{ Ba(OH}}{{\text{)}}_2}}} = 0,1{\text{ mol;}}{{\text{C}}_{M{\text{NaOH}}}} = 4M\)

Giải thích các bước giải:

 Ta có: 

\({n_{{H_2}S{O_4}}} = 0,1.2 = 0,2{\text{ mol;}}{{\text{n}}_{HCl}} = 0,1.2 = 0,2{\text{ mol}} \to {{\text{n}}_{{H^ + }}} = 2{n_{{H_2}S{O_4}}} + {n_{HCl}} = 0,6{\text{ mol}}\)

Phản ứng xảy ra:

\({H^ + } + O{H^ - }\xrightarrow{{}}{H_2}O\)

\(B{a^{2 + }} + S{O_4}^{2 - }\xrightarrow{{}}BaS{O_4}\)

Ta có: 

\({n_{O{H^ - }}} = {n_{{H^ + }}} = 0,6{\text{ mol = }}{{\text{n}}_{NaOH}} + 2{n_{Ba{{(OH)}_2}}}\)

\({n_{BaS{O_4}}} = \frac{{23,3}}{{233}} = 0,1{\text{ mol < }}{{\text{n}}_{{H_2}S{O_4}}} \to S{O_4}^{2 - }\) dư.

\( \to {n_{B{a^{2 + }}}} = {n_{BaS{O_4}}} = 0,1{\text{ mol = }}{{\text{n}}_{Ba{{(OH)}_2}}}\)

\( \to {n_{NaOH}} = 0,6 - 0,1.2 = 0,4{\text{ mol}}\)

\( \to {C_{M{\text{ Ba(OH}}{{\text{)}}_2}}} = \frac{{0,1}}{{0,1}} = 0,1{\text{ mol;}}{{\text{C}}_{M{\text{NaOH}}}} = \frac{{0,4}}{{0,1}} = 4M\)