Đáp án:
\(\begin{array}{l}
a)\\
{\rm{[}}N{a^ + }{\rm{]}} = 0,67M\\
{\rm{[}}{K^ + }{\rm{]}} = 0,33M\\
{\rm{[}}O{H^ - }{\rm{]}} = 1M\\
b)\\
{\rm{[}}{H^ + }{\rm{]}} = 1,5M\\
{\rm{[}}N{a^ + }{\rm{]}} = 0,33M\\
{\rm{[}}{K^ + }{\rm{]}} = 0,167M\\
{\rm{[}}S{O_4}^{2 - }{\rm{]}} = 1M
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
{n_{NaOH}} = 0,1 \times 2 = 0,2\,mol\\
{n_{KOH}} = 0,2 \times 0,5 = 0,1\,mol\\
{n_{N{a^ + }}} = {n_{NaOH}} = 0,2\,mol\\
{n_{{K^ + }}} = {n_{KOH}} = 0,1\,mol\\
{n_{O{H^ - }}} = 0,2 + 0,1 = 0,3\,mol\\
{\rm{[}}N{a^ + }{\rm{]}} = \dfrac{{0,2}}{{0,3}} = 0,67M\\
{\rm{[}}{K^ + }{\rm{]}} = \dfrac{{0,1}}{{0,3}} = 0,33M\\
{\rm{[}}O{H^ - }{\rm{]}} = \dfrac{{0,3}}{{0,3}} = 1M\\
b)\\
{n_{{H_2}S{O_4}}} = 0,3 \times 2 = 0,6\,mol\\
{n_{{H^ + }}} = 2{n_{{H_2}S{O_4}}} = 1,2\,mol\\
O{H^ - } + {H^ + } \to {H_2}O\\
\dfrac{{0,3}}{1} < \dfrac{{1,2}}{1} \Rightarrow\text{ H+ dư} \\
{n_{{H^ + }}} \text{ dư}= 1,2 - 0,3 = 0,9\,mol\\
{n_{S{O_4}^{2 - }}} = {n_{{H_2}S{O_4}}} = 0,6\,mol\\
{\rm{[}}{H^ + }{\rm{]}} = \dfrac{{0,9}}{{0,3 + 0,3}} = 1,5M\\
{\rm{[}}N{a^ + }{\rm{]}} = \dfrac{{0,2}}{{0,6}} = 0,33M\\
{\rm{[}}{K^ + }{\rm{]}} = \dfrac{{0,1}}{{0,6}} = 0,167M\\
{\rm{[}}S{O_4}^{2 - }{\rm{]}} = \dfrac{{0,6}}{{0,6}} = 1M
\end{array}\)
