$\begin{array}{*{20}{l}}
\begin{array}{l}
a/\\
{n_{N{a_2}S{O_4}}} = 0,05{\mkern 1mu} {\kern 1pt} {\mkern 1mu} {\kern 1pt} {\mkern 1mu} {\kern 1pt} {\mkern 1mu} {\kern 1pt} {\mkern 1mu} {\kern 1pt} {\mkern 1mu} {\kern 1pt} {\mkern 1mu} {\kern 1pt} {\mkern 1mu} {\kern 1pt} {\mkern 1mu} {\kern 1pt} {\mkern 1mu} {\kern 1pt} {\mkern 1mu} {\kern 1pt} {\mkern 1mu} {\kern 1pt} {\mkern 1mu} {\kern 1pt} {\mkern 1mu} {\kern 1pt} {\mkern 1mu} {\kern 1pt} {\mkern 1mu} {\kern 1pt} {\mkern 1mu} {\kern 1pt} {n_{KCl}} = 0,1{\mkern 1mu} {\kern 1pt} {\mkern 1mu} {\kern 1pt} {\mkern 1mu} {\kern 1pt} {\mkern 1mu} {\kern 1pt} {\mkern 1mu} {\kern 1pt} {\mkern 1mu} {\kern 1pt} {\mkern 1mu} {\kern 1pt} {\mkern 1mu} {\kern 1pt} {\mkern 1mu} {\kern 1pt} {\mkern 1mu} {\kern 1pt} {\mkern 1mu} {\kern 1pt} {\mkern 1mu} {\kern 1pt} {\mkern 1mu} {\kern 1pt} {\mkern 1mu} {\kern 1pt} {\mkern 1mu} {\kern 1pt} {\mkern 1mu} {\kern 1pt} {n_{NaCl}} = 0,05
\end{array}\\
{}\\
{\left[ {N{a^ + }} \right] = 2.0,05 + 0,05 = 0,15M}\\
{\left[ {SO_4^{2 - }} \right] = 0,05M}\\
{\left[ {{K^ + }} \right] = 0,1M}\\
{\left[ {C{l^ - }} \right] = 0,1 + 0,05 = 0,15M}\\
{}\\
{b/}\\
\begin{array}{l}
\left[ {C{l^ - }} \right] = \left[ {N{a^ + }} \right] = \frac{{{n_{NaCl}}}}{{0,4}} = 0,15 \to {n_{NaCl}} = 0,06\\
\left[ {{K^ + }} \right] = 2\left[ {SO_4^{2 - }} \right] = \frac{{2{n_{{K_2}S{O_4}}}}}{{0,4}} = 0,1 \to {n_{{K_2}S{O_4}}} = 0,02\\
\\
c/\\
\left[ {C{l^ - }} \right] \ne \left[ {{K^{ + |}}} \right]
\end{array}
\end{array}$
Mà KCl tuân theo tỉ lệ $\left[ {{K^ + }} \right] = \left[ {C{l^ - }} \right]$
Suy ra không thể pha được