a)
Ta có:
\({n_{HCl}} = 0,1.1 = 0,1{\text{ mol;}}{{\text{n}}_{{H_2}S{O_4}}} = 0,1.0,5 = 0,05{\text{ mol}} \to {{\text{n}}_{{H^ + }}} = {n_{HCl}} + 2{n_{{H_2}S{O_4}}} = 0,1 + 0,05.2 = 0,2{\text{ mol;}}{{\text{V}}_{dd}} = 100 + 100 = 200{\text{ ml}} \to {\text{[}}{{\text{H}}^ + }] = \frac{{0,2}}{{0,2}} = 1\)
\( \to pH = - \log [{H^ + }] = 0\)
b)
\({n_{NaOH}} = 0,05.2 = 0,1{\text{ mol;}}{{\text{n}}_{Ba{{(OH)}_2}}} = 0,15.2 = 0,3{\text{ mol}} \to {{\text{n}}_{O{H^ - }}} = {n_{NaOH}} + 2{n_{Ba{{(OH)}_2}}} = 0,1 + 0,3.2 = 0,7{\text{ mol;}}{{\text{V}}_{dd}} = 0,05 + 0,15 = 0,2{\text{ lit}} \to {\text{[O}}{{\text{H}}^ - }] = \frac{{0,7}}{{0,2}} = 0,35M \to pOH = - \log [O{H^ - }] = 0,456 \to pH = 14 - 0,456 = 13,544\)
c)
\({n_{HCl}} = 0,05.0,12 = 0,006{\text{ mol;}}{{\text{n}}_{NaOH}} = 0,05.0,1 = 0,005{\text{ mol}}\)
Phản ứng xảy ra:
\(NaOH + HCl\xrightarrow{{}}NaCl + {H_2}O\)
\( \to {n_{HCl}} > {n_{NaOH}} \to {n_{HCl{\text{ dư}}}} = 0,006 - 0,005 = 0,001{\text{ mol;}}{{\text{V}}_{dd}} = 0,05 + 0,05 = 0,1{\text{ lít}} \to {\text{[}}{{\text{H}}^ + }] = \frac{{0,001}}{{0,1}} = 0,01 \to pH = - \log [{H^ + }] = 2\)
d)
\({n_{{H_2}S{O_4}}} = 0,2.0,05 = 0,01{\text{ mol;}}{{\text{n}}_{NaOH}} = 0,3.0,06 = 0,018{\text{ mol}}\)
Phản ứng xảy ra:
\(2NaOH + {H_2}S{O_4}\xrightarrow{{}}N{a_2}S{O_4} + 2{H_2}O\)
Vì
\({n_{{H_2}S{O_4}}} > \frac{1}{2}{n_{NaOH}} \to {n_{{H_2}S{O_4}{\text{ dư}}}} = 0,01 - \frac{{0,018}}{2} = 0,001{\text{ mol}} \to {{\text{n}}_{{H^ + }}} = 0,002{\text{ mol}}{{\text{V}}_{dd}} = 0,2 + 0,3 = 0,5{\text{ lít}} \to {\text{[}}{{\text{H}}^ + }] = \frac{{0,002}}{{0,5}} = 0,004M\)
\( \to pH = - \log [{H^ + }] = 2,398\)