Đáp án:

\(\begin{array}{l}
1)\\
a)\\
pH = 1\\
b)\\
pH = 13,3\\
c)\\
pH = 3\\
d)\\
pH = 2\\
e)\\
pH = 11\\
f)\\
pH = 11\\
2)\\
a)\\
pH = 2\\
b)\\
pH = 12\\
c)\\
pH = 12\\
d)\\
pH = 0,4
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
1)\\
a)\\
{n_{HCl}} = \dfrac{{1,46}}{{36,5}} = 0,04\,mol \Rightarrow {n_{{H^ + }}} = 0,04\,mol\\
{\rm{[}}{H^ + }{\rm{]}} = \dfrac{{0,04}}{{0,4}} = 0,1(M)\\
pH =  - \log (0,1) = 1\\
b)\\
{n_{NaOH}} = \dfrac{{1,6}}{{40}} = 0,04\,mol \Rightarrow {n_{O{H^ - }}} = 0,04\,mol\\
{\rm{[}}O{H^ - }{\rm{]}} = \dfrac{{0,04}}{{0,2}} = 0,2(M)\\
pH = 14 - {\rm{[}} - \log (0,2){\rm{]}} = 13,3\\
c)\\
{\rm{[}}{H^ + }{\rm{]}} = {C_M}HCl = 0,001M\\
pH =  - \log (0,001) = 3\\
d)\\
{\rm{[}}{H^ + }{\rm{]}} = 2{C_M}{H_2}S{O_4} = 0,005 \times 2 = 0,01M\\
pH =  - \log (0,01) = 2\\
e)\\
{\rm{[O}}{{\rm{H}}^ - }{\rm{]}} = {C_M}NaOH = 0,001M\\
pH = 14 - {\rm{[}} - \log (0,001){\rm{]}} = 11\\
f)\\
{\rm{[}}O{H^ - }{\rm{]}} = 2{C_M}Ba{(OH)_2} = 0,0005 \times 2 = 0,001M\\
pH = 14 - {\rm{[}} - \log (0,001){\rm{]}} = 11\\
2)\\
a)\\
{n_{{H^ + }}} = 0,1 \times 0,01 + 2 \times 0,05 \times 0,005 = 0,0015\,mol\\
{\rm{[}}{H^ + }{\rm{]}} = \dfrac{{0,0015}}{{0,15}} = 0,01M\\
pH =  - \log (0,01) = 2\\
b)\\
{n_{O{H^ - }}} = 0,5 \times 0,005 + 0,25 \times 0,02 = 0,0075\,mol\\
{\rm{[}}O{H^ - }{\rm{]}} = \dfrac{{0,0075}}{{0,75}} = 0,01M\\
pH = 14 - {\rm{[}} - \log (0,01){\rm{]}} = 12\\
c)\\
{n_{NaOH}} = 0,05 \times 0,52 = 0,026\,mol\\
{n_{HCl}} = 0,05 \times 0,5 = 0,025\,mol\\
NaOH + HCl \to NaCl + {H_2}O\\
\dfrac{{0,026}}{1} > \dfrac{{0,025}}{1} \Rightarrow\text{ NaOH dư} \\
{n_{NaOH}} = 0,026 - 0,025 = 0,001\,mol\\
{\rm{[}}O{H^ - }{\rm{]}} = \dfrac{{0,001}}{{0,1}} = 0,01\\
pH = 14 - {\rm{[}} - \log (0,01){\rm{]}} = 12\\
d)\\
{n_{{H^ + }}} = 2{n_{{H_2}S{O_4}}} = 0,2 \times 0,5 \times 2 = 0,2\,mol\\
{n_{O{H^ - }}} = {n_{KOH}} = 0,05 \times 2 = 0,1\,mol\\
O{H^ - } + {H^ + } \to {H_2}O\\
\dfrac{{0,2}}{1} > \dfrac{{0,1}}{1} \Rightarrow {H^ + }\text{ dư}\\
{n_{{H^ + }}} = 0,2 - 0,1 = 0,1\,mol\\
{\rm{[}}{H^ + }{\rm{]}} = \dfrac{{0,1}}{{0,25}} = 0,4\\
pH =  - \log (0,4) = 0,4
\end{array}\)