2)
Ta có:
\({n_{NaOH}} = 0,05.0,2 = 0,01{\text{ mol = }}{{\text{n}}_{O{H^ - }}}\)
\({V_{dd}} = 50 + 150 = 200ml = 0,2{\text{ lít}}\)
\( \to [O{H^ - }] = \frac{{0,01}}{{0,2}} = 0,05M \to pOH = - \log [O{H^ - }] = 1,3 \to pH = 14 - pOH = 12,7\)
3)
\({m_{HN{O_3}}} = 0,2.0,2 = 0,04{\text{ mol;}}{{\text{n}}_{NaOH}} = 0,3.0,3 = 0,09{\text{ mol}}\)
\(NaOH + HN{O_3}\xrightarrow{{}}NaN{O_3} + {H_2}O\)
\( \to {n_{NaOH{\text{ dư}}}} = 0,09 - 0,04 = 0,05{\text{ mol = }}{{\text{n}}_{O{H^ - }}}\)
\({V_{dd}} = 200 + 300 = 500ml = 0,5{\text{ lít}}\)
\( \to [O{H^ - }] = \frac{{0,05}}{{0,5}} = 0,1 \to pOH = - \log [O{H^ - }] = 1 \to pH = 14 - pOH = 13\)