a) Ta có:
\([{H^ + }] = {C_{M{\text{ HCl}}}} = 0,01M \to pH = - \log [{H^ + }] = 2\)
b)
Ta có:
\({n_{{H_2}S{O_4}}} = \frac{{3,92}}{{98}} = 0,04{\text{ mol}} \to {{\text{n}}_{{H^ + }}} = 2{n_{{H_2}S{O_4}}} = 0,08{\text{ mol}} \to [{{\text{H}}^ + }] = \frac{{0,08}}{2} = 0,04M \to pH = - \log [{H^ + }] = 1,398\)
c)
Ta có:
\([O{H^ - }] = 2{C_{M{\text{ Ba(OH}}{{\text{)}}_2}}} = 0,05.2 = 0,1M \to pOH = - \log [O{H^ - }] = 1 \to pH = 14 - pOH = 13\)
d)
Ta có:
\({n_{NaOH}} = \frac{4}{{40}} = 0,1{\text{ mol;}}{{\text{n}}_{KOH}} = \frac{{16,8}}{{56}} = 0,3{\text{ mol}} \to {{\text{n}}_{O{H^ - }}} = {n_{NaOH}} + {n_{KOH}} = 0,4{\text{ mol}} \to {\text{[O}}{{\text{H}}^ - }{\text{] = }}\frac{{0,4}}{4} = 0,1\)
\( \to pOH = - \log [O{H^ - }] = 1 \to pH = 14 - pOH = 13\)
