Giải bài 10 trang 31- SGK môn Hóa học lớp 11 nâng cao

$\begin{aligned} & NaN{{O}_{2}}\to N{{a}^{+}}+N{{O}_{2}}^{-} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,N{{O}_{2}}^{-}+{{H}_{2}}O\rightleftharpoons HN{{O}_{2}}+O{{H}^{-}} \\ & \text{phân li:}\,\,\,\,\,\,\,\,\,\,\,\,\text{x}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,x \\ & \text{cân bằng }\,\,\,\,\,\,\,\,\text{1-x}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,x \\ & {{K}_{b}}=2,{{5.10}^{-11}}\Leftrightarrow \dfrac{{{x}^{2}}}{1-x}=2,{{5.10}^{-11}} \\ & \Leftrightarrow x\approx {{5.10}^{-6}}\Leftrightarrow [O{{H}^{-}}]\approx {{5.10}^{-6}}\Leftrightarrow [{{H}^{+}}]=\dfrac{{{10}^{-14}}}{[O{{H}^{-}}]}={{2.10}^{-9}} \\ \end{aligned}$

Ghi nhớ:

Tích số ion của nước : $[OH^-].[H^+]=10^{-14}$