Đáp án:
\( {C_{M{\text{ C}}{{\text{H}}_3}COOH}} = 0,4M\)
\({V_{{H_2}}} = 0,112{\text{ lít}}\)
\( \to {V_{NaOH}} {\text{ =13}}{\text{,33 ml}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(2C{H_3}COOH + Mg\xrightarrow{{}}{(C{H_3}COO)_2}Mg + {H_2}\)
Ta có:
\({n_{{{(C{H_3}COO)}_2}Mg}} = \frac{{0,71}}{{24 + 59.2}} = 0,005{\text{ mol = }}{{\text{n}}_{{H_2}}}\)
\( \to {n_{C{H_3}COOH}} = 2{n_{{H_2}}} = 0,005.2 = 0,01{\text{ mol}}\)
\( \to {C_{M{\text{ C}}{{\text{H}}_3}COOH}} = \frac{{0,01}}{{0,025}} = 0,4M\)
\({V_{{H_2}}} = 0,005.22,4 = 0,112{\text{ lít}}\)
Trung hòa axit này
\(C{H_3}COOH + NaOH\xrightarrow{{}}C{H_3}COONa + {H_2}O\)
\( \to {n_{NaOH}} = {n_{C{H_3}COOH}} = 0,01{\text{ mol}}\)
\( \to {V_{NaOH}} = \frac{{0,01}}{{0,75}} = 0,01333{\text{ lít = 13}}{\text{,33 ml}}\)