Đáp án
\({{\text{C}}_{M{\text{ N}}{{\text{a}}_2}S{O_4}}} = 0,45M;{C_{M{\text{ NaOH dư}}}} = 0,3M\)
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:
\({n_{NaOH}} = 0,1.3 = 0,3{\text{ mol; }}{{\text{n}}_{{H_2}S{O_4}}}\)
\(= 0,75.0,15 = 0,1125{\text{ mol < }}\frac{1}{2}{n_{NaOH}}\)nên NaOH dư.
A chứa
\({n_{N{a_2}S{O_4}}} = {n_{{H_2}S{O_4}}} = 0,1125{\text{ mol; }} \)
\({{\text{n}}_{NaOH dư}}= 0,3 - 0,1125.2 = 0,075{\text{ mol}}\)
\({V_{dd{\text{ A}}}} = 100 + 150 = 250ml = 0,25{\text{ lít}} \)
\(\to {{\text{C}}_{M{\text{ N}}{{\text{a}}_2}S{O_4}}} = \frac{{0,1125}}{{0,25}} = 0,45M;{C_{M{\text{ NaOH dư}}}} = \frac{{0,075}}{{0,25}} = 0,3M\)
