Giải bài 10 trang 16 - SGK môn Hóa học lớp 11 nâng cao
Có 2 dung dịch sau:
a. \(C{{H}_{3}}COOH\,0,1M\,({{K}_{a}}=1,{{75.10}^{-5}}) \). Tính nồng độ mol của ion \({{H}^{+}} \)
b. \(N{{H}_{3}}\,0,1M\,({{K}_{b}}=1,{{8.10}^{-5}}) \). Tính nồng độ mol của ion \(O{{H}^{-}} \)
a. Gọi nồng độ \({{H}^{+}} \) là x
\(\begin{align} & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{{H}_{3}}COOH\rightleftharpoons C{{H}_{3}}CO{{O}^{-}}+{{H}^{+}} \\ & pl\text{:}\,\,\,\,\,\,\,\,\,\,\,\text{x}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x \\ & cb\text{:}\,\,\,\,\,\,\,\,0,1-\text{x}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x \\ & {{K}_{a}}=1,{{75.10}^{-5}}\Leftrightarrow \frac{{{x}^{2}}}{0,1-x}=1,{{75.10}^{-5}}\Leftrightarrow x\approx 1,{{3.10}^{-3}}\Leftrightarrow [{{H}^{+}}]\approx 1,{{3.10}^{-3}} \\ \end{align} \)
b. Gọi nồng độ \(O{{H}^{-}}\) là x
\(\begin{align} & \,\,\,\,\,\,\,\,\,\,\,\,\,\,N{{H}_{3}}+{{H}_{2}}O\rightleftharpoons N{{H}_{4}}^{+}+O{{H}^{-}} \\ & pl\text{:}\,\,\,\,\,\,\,\,\,\,\text{x}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x \\ & cb\text{:}\,\,\,\,\,\,\,\,\,0,1-\text{x}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x \\ & {{K}_{b}}=1,{{810}^{-5}}\Leftrightarrow \frac{{{x}^{2}}}{0,1-x}=1,{{8.10}^{-5}}\Leftrightarrow x\approx 1,{{3.10}^{-3}}\Leftrightarrow [O{{H}^{-}}]\approx 1,{{3.10}^{-3}} \\ \end{align} \)
Ghi nhớ:
Hằng số phân ki của axit, bazơ không thay đổi và chỉ phụ thuộc vào nhiệt độ