Đáp án:
\({C_{M{\text{ NaOH}}}}= 1M\)
\({V_{dd\;{{\text{H}}_2}S{O_4}}} {\text{= 0}}{\text{,107456 lít}}\)
\({C_{M{\text{ N}}{{\text{a}}_2}S{O_4}}} = 0,41155M\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(N{a_2}O + {H_2}O\xrightarrow{{}}2NaOH\)
Ta có:
\({n_{N{a_2}O}} = \frac{{15,5}}{{23.2 + 16}} = 0,25{\text{ mol}}\)
\( \to {n_{NaOH}} = 2{n_{N{a_2}O}} = 0,25.2 = 0,5{\text{ mol}}\)
\( \to {C_{M{\text{ NaOH}}}} = \frac{{{n_{NaOH}}}}{{{V_{dd\;{\text{A}}}}}} = \frac{{0,5}}{{0,5}} = 1M\)
Trung hòa dung dịch \(A\)
\(2NaOH + {H_2}S{O_4}\xrightarrow{{}}N{a_2}S{O_4} + 2{H_2}O\)
\( \to {n_{{H_2}S{O_4}}} = \frac{1}{2}{n_{NaOH}} = 0,25{\text{ mol}}\)
\( \to {m_{{H_2}S{O_4}}} = 0,25.98 = 24,5{\text{ gam}}\)
\( \to {m_{dd{\text{ }}{{\text{H}}_2}S{O_4}}} = \frac{{24,5}}{{20\% }} = 122,5{\text{ gam}}\)
\( \to {V_{dd\;{{\text{H}}_2}S{O_4}}} = \frac{{122,5}}{{1,14}} = 107,456{\text{ ml = 0}}{\text{,107456 lít}}\)
Sau khi trung hòa
\({n_{N{a_2}S{O_4}}} = {n_{{H_2}S{O_4}}} = 0,25{\text{ mol}}\)
\( \to {V_{dd}} = 0,5 + 0,107456 = 0,607456{\text{ lít}}\)
\( \to {C_{M{\text{ N}}{{\text{a}}_2}S{O_4}}} = \frac{{0,25}}{{0,607456}} = 0,41155M\)
