Đáp án:
Giải thích các bước giải:
$\begin{array}{l}
{n_{HCl}} = 0,1.0,1 = 0,01\\
HCl \to {H^ + } + C{l^ - }\\
0,01\,\,\,\,\,\,\,\,0,01\,\,\\
n{H_2}S{O_4} = 0,02.0,1 = 0,002\\
{H_2}S{O_4} \to 2{H^ + } + S{O_4}^{2 - }\\
0,002\,\,\,\,\,\,\,0,004\\
\Rightarrow {n_{{H^ + }}} = 0,014\\
{n_{NaOH}} = 0,01.0,1 = 0,001\\
NaOH \to N{a^ + } + O{H^ - }\\
0,001\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,001\\
{n_{Ba{{(OH)}_2}}} = 0,02.0,1 = 0,002\\
Ba{(OH)_2} \to B{a^{2 + }} + 2O{H^ - }\\
0,002\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,004\\
{n_{KOH}} = 0,02.0,1 = 0,002\\
KOH \to {K^ + } + O{H^ - }\\
0,002\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,002\\
\Rightarrow {n_{O{H^ - }}} = 0,001 + 0,004 + 0,002 = 0,007\\
PTHH:\,\,\,{H^ + } + O{H^ - } \to {H_2}O\\
\,\,\,bd\,\,\,\,\,\,\,\,\,\,0,014\,\,\,\,0,007\\
\,\,pu\,\,\,\,\,\,\,\,\,\,\,0,007\,\,\,0,007\\
\,\,du\,\,\,\,\,\,\,\,\,\,\,\,0,007\,\,\,0\\
{n_{{H^ + }du}} = 0,007 \to {{\rm{[}}{H^ + }{\rm{]}}_{du}} = \frac{{0,007}}{{0,2}} = 0,035\\
pH = - \log [{H^ + }{\rm{]}} = 1,45
\end{array}$
