Đáp án:

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Giải thích các bước giải:

3,

\(\begin{array}{l}
HCl \to {H^ + } + C{l^ - }\\
HN{O_3} \to {H^ + } + N{O_3}^ - \\
 \to C{M_{{H^ + }}} = C{M_{HCl}} + C{M_{HN{O_3}}} = 0,01M\\
 \to pH =  - \log ({H^ + }) = 2\\
 \to pOH = 14 - 2 = 12 \to C{M_{O{H^ - }}} = {10^{ - 12}}M
\end{array}\)

4,

\(\begin{array}{l}
NaOH \to N{a^ + } + O{H^ - }\\
KOH \to {K^ + } + O{H^ - }\\
 \to C{M_{O{H^ - }}} = C{M_{NaOH}} + C{M_{KOH}} = 0,1M\\
 \to pOH =  - \log (O{H^ - }) = 1\\
 \to pH = 14 - 1 = 13 \to C{M_{{H^ + }}} = {10^{ - 13}}M
\end{array}\)

5,

a, 

\(\begin{array}{l}
NaHC{O_3} + HCl \to NaCl + C{O_2} + {H_2}O\\
HC{O_3}^ -  + {H^ + } \to C{O_2} + {H_2}O
\end{array}\)

b,

\(\begin{array}{l}
NaHC{O_3} + HCl \to NaCl + C{O_2} + {H_2}O\\
{n_{NaHC{O_3}}} = 0,1mol\\
 \to {n_{HCl}} = {n_{NaHC{O_3}}} = 0,1mol\\
 \to {V_{HCl}} = \dfrac{{0,1}}{{0,05}} = 2l\\
 \to {n_{C{O_2}}} = {n_{NaHC{O_3}}} = 0,1mol\\
 \to {V_{C{O_2}}} = 2,24l
\end{array}\)

6,

a,

\(\begin{array}{l}
{(N{H_4})_2}S{O_4} + 2NaOH \to N{a_2}S{O_4} + 2N{H_3} + 2{H_2}O\\
N{H_4}^ +  + O{H^ - } \to N{H_3} + {H_2}O
\end{array}\)

b,

\(\begin{array}{l}
{(N{H_4})_2}S{O_4} + 2NaOH \to N{a_2}S{O_4} + 2N{H_3} + 2{H_2}O\\
{n_{NaOH}} = 0,2mol\\
 \to {n_{{{(N{H_4})}_2}S{O_4}}} = \dfrac{1}{2}{n_{NaOH}} = 0,1mol\\
 \to {m_{{{(N{H_4})}_2}S{O_4}}} = 13,2g\\
 \to {n_{N{H_3}}} = {n_{NaOH}} = 0,2mol\\
 \to {V_{N{H_3}}} = 4,48l
\end{array}\)