Giải bài 9 trang 20 - SGK môn Hóa học lớp 11 nâng cao

\(\begin{aligned} & pH=10\Rightarrow \text{ }\!\![\!\!\text{ }{{H}^{+}}\text{ }\!\!]\!\!\text{ }={{10}^{-10}}\Rightarrow \text{ }\!\![\!\!\text{ }O{{H}^{-}}\text{ }\!\!]\!\!\text{ }=\dfrac{{{10}^{-14}}}{{{10}^{-10}}}={{10}^{-4}}\Rightarrow {{n}_{O{{H}^{-}}}}={{10}^{-4}}.0,3={{3.10}^{-3}}\,mol \\ & NaOH\to N{{a}^{+}}+O{{H}^{-}} \\ & {{3.10}^{-3}}\,\,\,\,\,\,\,\,{{3.10}^{-3}} \\ & \Rightarrow {{m}_{NaOH}}=0,12\,gam \\ \end{aligned} \)

Ghi nhớ:

\(pH=-log[H^+]\)

\([H^+].[OH^-]=10^{-14}\)