Đáp án:
1,\(pH = - \log {\rm{[}}{H^ + }{\rm{]}} = 2,4\)
2,\(pH = - \log {\rm{[}}{H^ + }{\rm{]}} = 3,5\)
19.1,
pH=11,58
\(C{M_{B{a^{2 + }}}} = \dfrac{{1,92 \times {{10}^{ - 4}}}}{{0,1}} = 1,92 \times {10^{ - 3}}M\)
\(C{M_{O{H^ - }}} = \dfrac{{3,84 \times {{10}^{ - 4}}}}{{0,1}} = 3,84 \times {10^{ - 3}}M\)
Giải thích các bước giải:
1,
\(\begin{array}{l}
{n_{HCl}} = 0,000875mol\\
{n_{{H_2}S{O_4}}} = 0,001mol\\
\to {n_{{H^ + }}} = {n_{HCl}} + 2{n_{{H_2}S{O_4}}} = 0,002875mol\\
{n_{KOH}} = 0,00075mol\\
{n_{NaOH}} = 0,000125mol\\
\to {n_{O{H^ - }}} = {n_{KOH}} + {n_{NaOH}} = 0,000875mol\\
O{H^ - } + {H^ + } \to {H_2}O\\
\to {n_{{H^ + }}} > {n_{O{H^ - }}}
\end{array}\)
Suy ra dung dịch sau phản ứng có axit dư
\(\begin{array}{l}
\to {n_{{H^ + }}}dư= 0,002875 - 0,000875 = 0,002mol\\
\to {\rm{[}}{H^ + }{\rm{]}} = \dfrac{{0,002}}{{0,5}} = 0,0004M\\
\to pH = - \log {\rm{[}}{H^ + }{\rm{]}} = 2,4
\end{array}\)
2,
\(\begin{array}{l}
{n_{HCl}} = 0,000045mol\\
{n_{HN{O_3}}} = 0,000075mol\\
\to {n_{{H^ + }}} = {n_{HCl}} + {n_{HN{O_3}}} = 0,00012mol\\
{n_{KOH}} = 0,000015mol\\
{n_{Ba{{(OH)}_2}}} = 0,0000225mol\\
\to {n_{O{H^ - }}} = {n_{KOH}} + 2{n_{Ba{{(OH)}_2}}} = 0,00006mol\\
O{H^ - } + {H^ + } \to {H_2}O\\
\to {n_{{H^ + }}} > {n_{O{H^ - }}}
\end{array}\)
Suy ra dung dịch sau phản ứng có axit dư
\(\begin{array}{l}
\to {n_{{H^ + }}}dư= 0,00012 - 0,00006 = 0,00006mol\\
\to {\rm{[}}{H^ + }{\rm{]}} = \dfrac{{0,00006}}{{0,2}} = 0,0003M\\
\to pH = - \log {\rm{[}}{H^ + }{\rm{]}} = 3,5
\end{array}\)
19.1,
\(\begin{array}{l}
{H_2}S{O_4}\\
pH = 4 \to {\rm{[}}{H^ + }{\rm{]}} = {10^{ - 4}}M\\
\to {n_{{H_2}S{O_4}}} = 8 \times {10^{ - 6}}mol\\
Ba{(OH)_2}\\
pH = 12 \to pOH = 2 \to {\rm{[}}O{H^ - }{\rm{]}} = {10^{ - 2}}M\\
\to {n_{Ba{{(OH)}_2}}} = 2 \times {10^{ - 4}}mol\\
{H_2}S{O_4} + Ba{(OH)_2} \to BaS{O_4} + 2{H_2}O\\
{n_{Ba{{(OH)}_2}}} > {n_{{H_2}S{O_4}}}
\end{array}\)
Suy ra dung dịch tạo thành có chứa bazo dư
\(\begin{array}{l}
{n_{Ba{{(OH)}_2}}}dư= 2 \times {10^{ - 4}} - 8 \times {10^{ - 6}} = 1,92 \times {10^{ - 4}}mol\\
\to {n_{B{a^{2 + }}}} = {n_{Ba{{(OH)}_2}}}dư= 1,92 \times {10^{ - 4}}mol\\
\to {n_{O{H^ - }}}dư= 2{n_{Ba{{(OH)}_2}}}dư= 3,84 \times {10^{ - 4}}mol\\
\to C{M_{B{a^{2 + }}}} = \dfrac{{1,92 \times {{10}^{ - 4}}}}{{0,1}} = 1,92 \times {10^{ - 3}}M\\
\to C{M_{O{H^ - }}} = \dfrac{{3,84 \times {{10}^{ - 4}}}}{{0,1}} = 3,84 \times {10^{ - 3}}M\\
\to pOH = - \log C{M_{O{H^ - }}} = 2,42\\
\to pH = 14 - pOH = 11,58
\end{array}\)
