Đáp án:
\(C{\% _{NaOH}} = 6,67\% \)
\(C{\% _{C{H_3}COONa}} = 6,833\% \)
Giải thích các bước giải:
Phản ứng xảy ra:
\(C{H_3}COOH + NaOH\xrightarrow{{}}C{H_3}COONa + {H_2}O\)
Ta có:
\({m_{C{H_3}COOH}} = 200.10\% = 20{\text{ gam}}\)
\( \to {n_{C{H_3}COOH}} = \frac{{20}}{{60}} = \frac{1}{3}{\text{ mol = }}{{\text{n}}_{NaOH}}\)
\( \to {m_{NaOH}} = \frac{1}{3}.40 = \frac{{40}}{3}{\text{ gam}}\)
\( \to C{\% _{NaOH}} = \frac{{\frac{{40}}{3}}}{{200}} = 6,67\% \)
BTKL:
\({m_{dd}} = {m_{dd{\text{ C}}{{\text{H}}_3}COOH}} + {m_{dd{\text{ NaOH}}}} = 200 + 200 = 400{\text{ gam}}\)
\({n_{C{H_3}COONa}} = {n_{C{H_3}COOH}} = \frac{1}{3}{\text{ mol}}\)
\( \to {m_{C{H_3}COONa}} = \frac{1}{3}.80 = \frac{{82}}{3}{\text{ gam}}\)
\( \to C{\% _{C{H_3}COONa}} = \frac{{\frac{{82}}{3}}}{{400}} = 6,833\% \)
