Đáp án:
\({{\text{C}}_{M{\text{ HCl}}}} = 0,15;{C_{M{\text{ }}{{\text{H}}_2}S{O_4}}} = 0,05\)
Giải thích các bước giải:
Gọi \({n_{{H_2}S{O_4}}} = a \to {n_{HCl}} = 3a\)
Phản ứng xảy ra:
\(HCl + NaOH\xrightarrow{{}}NaCl + {H_2}O\)
\({H_2}S{O_4} + 2NaOH\xrightarrow{{}}N{a_2}S{O_4} + 2{H_2}O\)
Ta có:
\({m_{NaOH}} = 20.\frac{{50}}{{1000}} = 1{\text{ gam}} \to {{\text{n}}_{NaOH}} = \frac{1}{{40}} = 0,025{\text{ mol = a}}{\text{.2 + 3a}} \to {\text{a = 0}}{\text{,005}} \to {{\text{C}}_{M{\text{ HCl}}}} = \frac{{3a}}{{0,1}} = 0,15;{C_{M{\text{ }}{{\text{H}}_2}S{O_4}}} = \frac{a}{{0,1}} = 0,05\)
200 ml A chứa
\({n_{HCl}} = 0,15.0,2 = 0,03{\text{ mol;}}{{\text{n}}_{{H_2}S{O_4}}} = 0,05.0,2 = 0,01{\text{ mol}} \to {{\text{n}}_{H{\text{ axit}}}} = {n_{HCl}} + 2{n_{{H_2}S{O_4}}} = 0,03 + 0,01.2 = 0,05{\text{ mol = }}{{\text{n}}_{OH{\text{ bazo}}}}\)
\( \to {V_B}.0,2 + {V_B}.0,1.2 = {n_{OH}} = 0,05 \to {V_B} = 0,125{\text{ lít = 12}}{\text{5ml}}\)
