Đáp án:
a)
\({p_H} = 2,079\)
b)
\(pH = 10\)
c)
\(pH = 13\)
d)
\(pH = 1\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
{n_{{H_2}S{O_4}}} = \dfrac{{4,9}}{{98}} = 0,05mol\\
{n_H} = 0,05 \times 2 = 0,1mol\\
{\rm{[}}{H^ + }{\rm{]}} = \dfrac{{0,1}}{{12}} = \dfrac{1}{{120}}\\
{p_H} = - \log (\dfrac{1}{{120}}) = 2,079\\
b)\\
{\rm{[}}O{H^ - }{\rm{]}} = 2 \times {C_{{M_{Ba{{(OH)}_2}}}}} = 2 \times 0,00005 = 0,0001\\
pOH = - \log (0,0001) = 4\\
pH = 14 - 4 = 10\\
c)\\
{n_{NaOH}} = \dfrac{{0,4}}{{40}} = 0,01mol\\
{\rm{[}}O{H^ - }{\rm{]}} = {C_{{M_{NaOH}}}} = \dfrac{{0,01}}{{0,1}} = 0,1M\\
pOH = - \log (0,1) = 1\\
pH = 14 - 1 = 13\\
d)\\
{n_{HCl}} = \dfrac{{1,46}}{{36,5}} = 0,04mol\\
{\rm{[}}{H^ + }{\rm{]}} = {C_{{M_{HCl}}}} = \dfrac{{0,04}}{{0,4}} = 0,1M\\
pH = - \log (0,1) = 1
\end{array}\)
