Đáp án:
\( C{\% _{NaCl}} = 1,95\% ;C{\% _{NaOH}}= 4\% \)
\( {C_{M{\text{ NaCl}}}} = 0,3616M;{C_{M{\text{ NaOH}}}} = 1,0847M\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(NaOH + HCl\xrightarrow{{}}NaCl + {H_2}O\)
Ta có:
\({m_{HCl}} = 150.3,65\% = 5,475{\text{ gam}}\)
\( \to {n_{HCl}} = \frac{{5,475}}{{36,5}} = 0,15{\text{ mol}}\)
\({m_{NaOH}} = 300.8\% = 24{\text{ gam}}\)
\( \to {n_{NaOH}} = \frac{{24}}{{40}} = 0,6{\text{ mol > }}{{\text{n}}_{HCl}}\)
Vậy sau phản ứng dung dịch chứa
\({n_{NaCl}} = {n_{HCl}} = 0,15{\text{ mol;}}{{\text{n}}_{NaOH}} = 0,6 - 0,15 = 0,45{\text{ mol}}\)
\({m_{NaCl}} = 0,15.58,5 = 8,775{\text{ gam;}}{{\text{m}}_{NaOH}} = 0,45.40 = 18{\text{ gam}}\)
\({m_{dd}} = 150 + 300 = 450{\text{ gam}}\)
\( \to C{\% _{NaCl}} = \frac{{8,775}}{{450}} = 1,95\% ;C{\% _{NaOH}} = \frac{{18}}{{450}} = 4\% \)
\({V_{dd{\text{HCl}}}} = \frac{{150}}{{1,02}} = 147{\text{ ml;}}{{\text{V}}_{dd{\text{ NaOH}}}} = \frac{{300}}{{1,12}} = 267,86{\text{ ml}}\)
\( \to {V_{dd}} = 147 + 267,86 = 414,86{\text{ ml = 0}}{\text{,41486 lít}}\)
\( \to {C_{M{\text{ NaCl}}}} = \frac{{0,15}}{{0,41486}} = 0,3616M;{C_{M{\text{ NaOH}}}} = \frac{{0,45}}{{0,41486}} = 1,0847M\)
