1)
Thiếu nồng độ mol của \(Ba(OH)_2\).
2)
Ta có:
\({n_{{H_2}S{O_4}}} = 0,5.0,01 = 0,005{\text{ mol}} \to {{\text{n}}_{{H^ + }}} = 2{n_{{H_2}S{O_4}}} = 0,005.2 = 0,01{\text{ mol}}\)
\({n_{NaOH}} = 0,5.0,04 = 0,02{\text{ mol = }}{{\text{n}}_{O{H^ - }}} > {n_{{H^ + }}}\)
\({H^ + } + O{H^ - }\xrightarrow{{}}{H_2}O\)
\( \to {n_{O{H^ - }{\text{ dư}}}} = 0,02 - 0,01 = 0,01{\text{ mol}}\)
\( \to {V_{dd}} = 500 + 500 = 1000{\text{ml = 1 lít}}\)
\( \to [O{H^ - }] = \frac{{0,01}}{1} = 0,01M\)
\( \to pOH = - \log [O{H^ - }] = 2 \to pH = 14 - pOH = 12\)