Đáp án:
\(\begin{array}{l}
7)\\
a)\\
{\rm{[}}O{H^ - }{\rm{]}} = {\rm{[}}N{a^ + }{\rm{]}} = \frac{{0,5}}{{0,5}} = 1M\\
b)\\
{V_{HCl}} = 0,25\,l\\
8)\\
a)\\
{\rm{[}}O{H^ - }{\rm{]}} = 1M\\
{\rm{[}}N{a^ + }{\rm{]}} = 0,67M\\
{\rm{[}}{K^ + }{\rm{]}} = 0,33M\\
b)\\
{C_M}{H_2}S{O_4} = 0,5M
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
7)\\
a)\\
{n_{NaOH}} = \dfrac{{20}}{{40}} = 0,5\,mol\\
\Rightarrow {n_{N{a^ + }}} = {n_{O{H^ - }}} = 0,5\,mol\\
\Rightarrow {\rm{[}}O{H^ - }{\rm{]}} = {\rm{[}}N{a^ + }{\rm{]}} = \dfrac{{0,5}}{{0,5}} = 1M\\
b)\\
NaOH + HCl \to NaCl + {H_2}O\\
{n_{HCl}} = {n_{NaOH}} = 0,5\,mol\\
{V_{HCl}} = \dfrac{{0,5}}{2} = 0,25\,l\\
8)\\
a)\\
{n_{NaOH}} = 0,1 \times 2 = 0,2\,mol\\
{n_{KOH}} = 0,2 \times 0,5 = 0,1\,mol\\
{n_{O{H^ - }}} = 0,2 + 0,1 = 0,3mol\\
{n_{N{a^ + }}} = {n_{NaOH}} = 0,2\,mol\\
{n_{{K^ + }}} = {n_{KOH}} = 0,1\,mol\\
{V_{{\rm{dd}}}} = 0,1 + 0,2 = 0,3l\\
{\rm{[}}O{H^ - }{\rm{]}} = \dfrac{{0,3}}{{0,3}} = 1M\\
{\rm{[}}N{a^ + }{\rm{]}} = \dfrac{{0,2}}{{0,3}} = 0,67M\\
{\rm{[}}{K^ + }{\rm{]}} = \dfrac{{0,1}}{{0,3}} = 0,33M\\
b)\\
O{H^ - } + {H^ + } \to {H_2}O\\
{n_{{H^ + }}} = {n_{O{H^ - }}} = 0,3\,mol\\
\Rightarrow {n_{{H_2}S{O_4}}} = \dfrac{{0,3}}{2} = 0,15\,mol\\
{C_M}{H_2}S{O_4} = \dfrac{{0,15}}{{0,3}} = 0,5M
\end{array}\)
