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Câu hỏi: Viết các biểu thức tính hằng số phân li axit \({K_a}\) hoặc hằng số phân li bazơ\({K_b}\) của các axit và bazơ sau: \(HClO, Br{O^ - }, HN{O_2}, NO_2^ - \)
Lời giải chi tiết
$HClO \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over
{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} {H^ + } + Cl{O^ - }$
\({K_a} = {{\left[ {{H^ + }} \right]\left[ {Cl{O^ - }} \right]} \over {\left[ {HClO} \right]}}\)
$Br{O^ - } + {H_2}O \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over
{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} HBrO + O{H^ - }$
\({K_b} = {{\left[ {HBrO} \right]\left[ {O{H^ - }} \right]} \over {\left[ {Br{O^ - }} \right]}}\)
$HN{O_2} \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over
{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} {H^ + } + N{O_2}^ - $
\({K_a} = {{\left[ {{H^ + }} \right]\left[ {NO_2^ - } \right]} \over {\left[ {HN{O_2}} \right]}}\)
$N{O_2}^ - + {H_2}O \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over
{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} HN{O_2} + O{H^ - }$
\({K_b} = {{\left[ {HN{O_2}} \right]\left[ {O{H^ - }} \right]} \over {\left[ {NO_2^ - } \right]}}\)
$HClO \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over
{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} {H^ + } + Cl{O^ - }$
\({K_a} = {{\left[ {{H^ + }} \right]\left[ {Cl{O^ - }} \right]} \over {\left[ {HClO} \right]}}\)
$Br{O^ - } + {H_2}O \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over
{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} HBrO + O{H^ - }$
\({K_b} = {{\left[ {HBrO} \right]\left[ {O{H^ - }} \right]} \over {\left[ {Br{O^ - }} \right]}}\)
$HN{O_2} \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over
{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} {H^ + } + N{O_2}^ - $
\({K_a} = {{\left[ {{H^ + }} \right]\left[ {NO_2^ - } \right]} \over {\left[ {HN{O_2}} \right]}}\)
$N{O_2}^ - + {H_2}O \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over
{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} HN{O_2} + O{H^ - }$
\({K_b} = {{\left[ {HN{O_2}} \right]\left[ {O{H^ - }} \right]} \over {\left[ {NO_2^ - } \right]}}\)