Đáp án:
Bạn tham khảo lời giải ở dưới nhé!!!
Giải thích các bước giải:
12,
\(\begin{array}{l}
{n_{HCl}} = 0,1mol\\
{n_{{H_2}S{O_4}}} = 0,05mol\\
\to {n_{{H^ + }}} = {n_{HCl}} + 2{n_{{H_2}S{O_4}}} = 0,2mol\\
\to {n_{C{l^ - }}} = {n_{HCl}} = 0,1mol\\
\to {n_{S{O_4}^{2 - }}} = {n_{{H_2}S{O_4}}} = 0,05mol\\
\to C{M_{{H^ + }}} = \dfrac{{0,2}}{{0,2}} = 1M\\
\to C{M_{C{l^ - }}} = \dfrac{{0,1}}{{0,2}} = 0,5M\\
\to C{M_{S{O_4}^{2 - }}} = \dfrac{{0,05}}{{0,2}} = 0,25M\\
B{a^{2 + }} + S{O_4}^{2 - } \to BaS{O_4}\\
\to {n_{BaS{O_4}}} = {n_{S{O_4}^{2 - }}} = 0,05mol\\
\to {m_{BaS{O_4}}} = 11,65g
\end{array}\)
13,
\(\begin{array}{l}
NaOH + HCl \to NaCl + {H_2}O\\
{n_{NaOH}} = 0,02mol\\
{n_{HCl}} = 0,06mol\\
\to {n_{HCl}} > {n_{NaOH}}
\end{array}\)
Suy ra HCl dư
Vậy dung dịch A có HCl dư và NaCl
\(\begin{array}{l}
\to {n_{HCl}}dư= 0,06 - 0,02 = 0,04mol\\
\to {n_{NaCl}} = {n_{NaOH}} = 0,02mol\\
\to {n_{{H^ + }}}dư= {n_{HCl}}dư= 0,04mol\\
\to {n_{N{a^ + }}} = {n_{NaCl}} = 0,02mol\\
\to {n_{C{l^ - }}} = {n_{HCl}}dư+ {n_{NaCl}} = 0,06mol\\
\to C{M_{{H^ + }}}dư= \dfrac{{0,04}}{{0,5}} = 0,08M\\
\to C{M_{N{a^ + }}} = \dfrac{{0,02}}{{0,5}} = 0,04M\\
\to C{M_{C{l^ - }}} = \dfrac{{0,06}}{{0,5}} = 0,12M\\
pH = - \log [C{M_{{H^ + }}}dư] = 1
\end{array}\)
14,
\(\begin{array}{l}
{n_{NaOH}} = 0,01mol\\
{n_{KOH}} = 0,01mol\\
{n_{{H_2}S{O_4}}} = 0,02mol\\
\to {n_{O{H^ - }}} = {n_{NaOH}} + {n_{KOH}} = 0,02mol\\
\to {n_{{H^ + }}} = 2{n_{{H_2}S{O_4}}} = 0,04mol\\
O{H^ - } + {H^ + } \to {H_2}O\\
{n_{O{H^ - }}} < {n_{{H^ + }}}
\end{array}\)
Vậy dung dịch A có: \({H^ + }\) dư , \(N{a^ + },{K^ + },S{O_4}^{2 - }\)
\(\begin{array}{l}
\to {n_{{H^ + }}}dư= 0,04 - 0,02 = 0,02mol\\
\to {n_{N{a^ + }}} = {n_{NaOH}} = 0,01mol\\
\to {n_{{K^ + }}} = {n_{KOH}} = 0,01mol\\
\to {n_{S{O_4}^{2 - }}} = {n_{{H_2}S{O_4}}} = 0,02mol\\
\to C{M_{{H^ + }}}dư= \dfrac{{0,02}}{{0,2}} = 0,1M\\
\to C{M_{N{a^ + }}} = \dfrac{{0,01}}{{0,2}} = 0,05M\\
\to C{M_{{K^ + }}} = \dfrac{{0,01}}{{0,2}} = 0,05M\\
\to C{M_{S{O_4}^{2 - }}} = \dfrac{{0,02}}{{0,2}} = 0,1M\\
pH = - \log [C{M_{{H^ + }}}dư] = 1
\end{array}\)
