a)
Phản ứng xảy ra:
\(2Na + 2{H_2}O\xrightarrow{{}}2NaOH + {H_2}\)
Ta có:
\({n_{Na}} = \frac{{4,6}}{{23}} = 0,2{\text{ mol = }}{{\text{n}}_{NaOH}} = {n_{O{H^ - }}} \to {\text{[O}}{{\text{H}}^ - }{\text{]}} = \frac{{0,2}}{{0,2}} = 1M \to pOH = - \log [O{H^ - }{\text{]}} = 0 \to pH = 14 - pOH = 14\)
b)
Cân bằng:
\(C{H_3}COOH\overset {} \leftrightarrows C{H_3}COO{^ - } + {H^ + }\)
Ta có:
\({\text{[}}{H^ + }{\text{]}} = {\text{[}}C{H_3}COO{H_{phân{\text{ ly}}}}{\text{]}} = 0,01.\alpha = 0,01.12,5\% = 0,00125M \to pH = - \log [{H^ + }{\text{]}} = 2,9\)
c) \(HClO\overset {} \leftrightarrows {H^ + } + Cl{O^ - }\)
\({\text{[}}{H^ + }{\text{]}} = {\text{[}}HCl{O_{phân{\text{ ly}}}}{\text{]}} = {10^{ - 3}}.\alpha = {10^{ - 3}}.0,13 = 0,00013M \to pH = - \log [{H^ + }] = 3,886\)
