3)
Ta có:
\({n_{HN{O_3}}} = 0,2.0,2 = 0,04{\text{ mol = }}{{\text{n}}_{{H^ + }}}\)
\({n_{NaOH}} = 0,3.0,3 = 0,09{\text{ mol = }}{{\text{n}}_{O{H^ - }}}\)
\({H^ + } + O{H^ - }\xrightarrow{{}}{H_2}O\)
\({n_{{H^ + }}} < {n_{O{H^ - }}} \to {n_{O{H^ - }dư}} = 0,09 - 0,04 = 0,05{\text{ mol}}\)
\({V_{dd}} = 200 + 300 = 500ml = 0,5{\text{ lít}}\)
\( \to [O{H^ - }] = \frac{{0,05}}{{0,5}} = 0,1M \to pOH = - \log [O{H^ - }] = 1 \to pH = 13\)
4)
\({n_{{H_2}S{O_4}}} = 0,1.3 = 0,3{\text{ mol}} \to {{\text{n}}_{{H^ + }}} = 2{n_{{H_2}S{O_4}}} = 0,6{\text{ mol}}\)
\({n_{KOH}} = 0,3.1,5 = 0,45{\text{ mol = }}{{\text{n}}_{O{H^ - }}}\)
\({H^ + } + O{H^ - }\xrightarrow{{}}{H_2}O\)
\({n_{{H^ + }}} > {n_{O{H^ - }}} \to {n_{{H^ + }{\text{ dư}}}} = 0,6 - 0,45 = 0,15{\text{ mol}}\)
\({V_{dd}} = 100 + 300 = 400ml = 0,4{\text{ lít}}\)
\( \to [{H^ + }] = \frac{{0,15}}{{0,4}} = 0,375M \to pH = - \log [{H^ + }] = 0,4256\)
