Đáp án:
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Giải thích các bước giải:
1,
\(\begin{array}{l}
2NaOH + {H_2}S{O_4} \to N{a_2}S{O_4} + 2{H_2}O\\
{n_{NaOH}} = 0,03mol\\
{n_{{H_2}S{O_4}}} = 0,025mol\\
\to \dfrac{{{n_{NaOH}}}}{2} < {n_{{H_2}S{O_4}}}
\end{array}\)
Suy ra \({H_2}S{O_4}\) dư
\(\begin{array}{l}
\to {n_{{H_2}S{O_4}}}dư= 0,025 - \dfrac{1}{2} \times 0,03 = 0,01mol\\
\to {n_{N{a_2}S{O_4}}} = 2{n_{NaOH}} = 0,06mol\\
\to {n_{{H^ + }}} = 2{n_{{H_2}S{O_4}}}dư= 0,02mol\\
\to {n_{S{O_4}^{2 - }}} = {n_{{H_2}S{O_4}}}dư+ {n_{N{a_2}S{O_4}}} = 0,07mol\\
\to {n_{N{a^ + }}} = 2{n_{N{a_2}S{O_4}}} = 0,12mol\\
C{M_{{H^ + }}} = \dfrac{{0,02}}{{0,015 + 0,05}} = 0,3M\\
C{M_{S{O_4}^{2 - }}} = \dfrac{{0,07}}{{0,015 + 0,05}} = 1,1M\\
C{M_{N{a^ + }}} = \dfrac{{0,12}}{{0,015 + 0,05}} = 1,8M\\
\to pH = - \log (C{M_{{H^ + }}}) = 0,52
\end{array}\)
2,
\(\begin{array}{l}
HCl + KOH \to KCl + {H_2}O\\
{n_{HCl}} = 0,1mol\\
{n_{KOH}} = 0,05mol\\
\to {n_{HCl}} > {n_{KOH}}
\end{array}\)
Suy ra HCl dư
\(\begin{array}{l}
\to {n_{HCl}}dư= 0,1 - 0,05 = 0,05mol\\
\to {n_{KCl}} = {n_{KOH}} = 0,05mol\\
\to {n_{{H^ + }}} = {n_{HCl}}dư= 0,05mol\\
\to {n_{C{l^ - }}} = {n_{HCl}}dư+ {n_{KCl}} = 0,1mol\\
\to {n_{{K^ + }}} = {n_{KCl}} = 0,05mol\\
\to C{M_{{H^ + }}} = C{M_{{K^ + }}} = \dfrac{{0,05}}{{0,2}} = 0,25M\\
\to C{M_{C{l^ - }}} = \dfrac{{0,1}}{{0,2}} = 0,5M
\end{array}\)
Dung dịch D gồm HCl dư và KCl
\(\begin{array}{l}
NaOH + HCl \to NaCl + {H_2}O\\
\to {n_{NaOH}} = {n_{HCl}}dư= 0,05mol\\
\to {V_{NaOH}} = \dfrac{{0,05}}{1} = 0,05l
\end{array}\)
