\(NaN{O_2} \to N{a^ + } + NO_2^ - \)
1 \( \to \) 1 \( \to \) 1
\(NO_2^ - + {H_2}O \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over {\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} HN{O_2} + O{H^ - }\)
Trước thủy phân: 1
Thủy phân: x \( \to \) x \( \to \) x
Cân bằng: (1 - x) x x
Ta có \({K_b} = {{\left[ {HN{O_2}} \right]\left[ {O{H^ - }} \right]} \over {\left[ {NO_2^ - } \right]}} = 2,{5.10^{ - 11}}\)
\(\Rightarrow {{xx} \over {\left( {1 - x} \right)}} = 2,{5.10^{ - 11}}\)
Vì x << 1 \( \Rightarrow (1 - x) \approx 1 \)
\(\Rightarrow x.x = 2,{5.10^{ - 11}} = {25.10^{ - 12}}\)
\( \Rightarrow x = {5.10^{ - 6}}\)
Ta có \(\left[ {O{H^ - }} \right].\left[ {{H^ + }} \right] = {10^{ - 14}}\)
\(\Rightarrow \left[ {{H^ + }} \right] = {{{{10}^{ - 14}}} \over {{{5.10}^{-6}}}} = {2.10^{ - 9}}\,\,mol/l\)
