Đáp án:
\( {C_{M{\text{ N}}{{\text{a}}_2}S{O_4}}} = 0,02857M;{C_{M{\text{ NaOH}}}} = 1,3714M\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(2NaOH + {H_2}S{O_4}\xrightarrow{{}}N{a_2}S{O_4} + 2{H_2}O\)
Ta có:
\({n_{NaOH}} = 0,5.2 = 1{\text{ mol;}}{{\text{n}}_{{H_2}S{O_4}}} = 0,2.0,1 = 0,02{\text{ mol}}\)
\( \to {n_{NaOH}} > 2{n_{{H_2}S{O_4}}}\) nên \(NaOH\) dư
\( \to {n_{N{a_2}S{O_4}}} = {n_{{H_2}S{O_4}}} = 0,02{\text{ mol;}}{{\text{n}}_{NaOH{\text{ dư}}}} = 1 - 0,02.2 = 0,96{\text{ mol}}\)
\( \to {V_{dd}} = 500 + 200 = 700{\text{ ml = 0}}{\text{,7 lít}}\)
\( \to {C_{M{\text{ N}}{{\text{a}}_2}S{O_4}}} = \frac{{0,02}}{{0,7}} = 0,02857M;{C_{M{\text{ NaOH}}}} = \frac{{0,96}}{{0,7}} = 1,3714M\)
