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Tính nồng độ mol \({H^ + }\) (mol/lít) trong các dung dịch sau:
a) \(C{H_3}COONa\,\,0,1M\,\,\)(\({K_b}\) của \(C{H_3}CO{O^ – }\) là \(5,{71.10^{ – 10}}\))
b) \(N{H_4}Cl\,\,0,10M\) (\({K_a}\)của \(NH_4^ + \) là \(5,{56.10^{ – 10}}\))
a)
\(C{H_3}CO{O^ – } + {H_2}O \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over {\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} C{H_3}COOH + O{H^ – }\)
Trước thủy phân: 0,1
Thủy phân: x \( \to \) x \( \to \) x
Sau thủy phân: (0,1 – x) x x
Ta có \({K_b} = {{\left[ {C{H_3}COOH} \right]\left[ {O{H^ – }} \right]} \over {\left[ {C{H_3}CO{O^ – }} \right]}} = 5,{71.10^{ – 10}}\)
\(\Rightarrow {{xx} \over {\left( {0,1 – x} \right)}} = 5,{71.10^{ – 10}}\)
Vì x<<0,1 \( \Rightarrow \left( {0,1 – x} \right) \approx 0,1 \)
\(\Rightarrow xx = 0,1.5,{71.10^{ – 10}} = 0,{571.10^{ – 10}} \)
\(\Rightarrow x = 0,{76.10^{ – 5}}\)
\(\eqalign{ & \left[ {O{H^ – }} \right] = 0,{76.10^{ – 5}}\,\,mol/l \cr & \Rightarrow \left[ {O{H^ – }} \right].\left[ {{H^ + }} \right] = {10^{ – 14}}\cr& \Rightarrow \left[ {{H^ + }} \right] = {{{{10}^{ – 14}}} \over {{{0.76.10}^{ – 5}}}} = 1,{3.10^{ – 9}}\,\,mol/l \cr} \)
\(NH_4^ + + {H_2}O \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over {\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} N{H_3} + {H_3}{O^ + }\)
Trước thủy phân: 0,1
Thủy phân: x \( \to \) x \( \to \) x
Sau thủy phân: (0,1 – x) x x
Ta có \({K_a} = {{\left[ {N{H_3}} \right]\left[ {{H_3}{O^ + }} \right]} \over {\left[ {N{H_4}^ + } \right]}} = 5,{56.10^{ – 10}}\)
\(\Rightarrow {{xx} \over {\left( {0,1 – x} \right)}} = 5,{56.10^{ – 10}}\)
Vì x<<0,1 \( \Rightarrow \left( {0,1 – x} \right) \approx 0,1 \)
\(\Rightarrow xx = 0,1.5,{56.10^{ – 10}} = 0,{556.10^{ – 10}}\)
\(\Rightarrow x = 0,{75.10^{ – 5}}\)
\(\left[ {{H_3}{O^ + }} \right] = 0,{75.10^{ – 5\,\,}}\,\,mol/l\)