Google
Câu hỏi: Hãy cho biết các phân tử và ion sau là axit, bazơ hay lưỡng tính theo thuyết Bron – stêt: \(HI, C{H_3}CO{O^ - },{H_2}PO_4^ - , PO_4^{3 - }, N{H_3},{S^{2 - }}, HPO_4^{2 - }\) Giải thích.
Lời giải chi tiết
Axit: \(HI + {H_2}O \to{H_3}O^++ {I^ - }\)
Bazơ: \(C{H_3}CO{O^ - },{S^{2 - }}, PO_4^{3 - }, N{H_3}\)
$\begin{array}{l}
C{H_3}CO{O^ - } + {H_2}O \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over
{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} C{H_3}COOH + O{H^ - }\\
PO_4^{3 - } + {H_2}O \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over
{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} HPO_4^{2 - } + O{H^ - }\\
{S^{2 - }} + {H_2}O \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over
{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} H{S^ - } + O{H^ - }\\
N{H_3} + {H_2}O \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over
{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} NH_4^ + + O{H^ - }
\end{array}$
Lưỡng tính: \(HPO_4^{2 - },{H_2}PO_4^ - \)
$\begin{array}{l}
HPO_4^{2 - } + {H_2}O \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over
{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} PO_4^{3 - } + {H_3}{O^ + }\\
HPO_4^{2 - } + {H_2}O \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over
{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} {H_2}P{O_4}^ - + O{H^ - }\\
{H_2}P{O_4}^ - + {H_2}O \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over
{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} HPO_4^{2 - } + {H_3}{O^ + }\\
{H_2}P{O_4}^ - + {H_2}O \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over
{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} {H_3}P{O_4} + O{H^ - }
\end{array}$
Axit: \(HI + {H_2}O \to{H_3}O^++ {I^ - }\)
Bazơ: \(C{H_3}CO{O^ - },{S^{2 - }}, PO_4^{3 - }, N{H_3}\)
$\begin{array}{l}
C{H_3}CO{O^ - } + {H_2}O \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over
{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} C{H_3}COOH + O{H^ - }\\
PO_4^{3 - } + {H_2}O \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over
{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} HPO_4^{2 - } + O{H^ - }\\
{S^{2 - }} + {H_2}O \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over
{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} H{S^ - } + O{H^ - }\\
N{H_3} + {H_2}O \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over
{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} NH_4^ + + O{H^ - }
\end{array}$
Lưỡng tính: \(HPO_4^{2 - },{H_2}PO_4^ - \)
$\begin{array}{l}
HPO_4^{2 - } + {H_2}O \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over
{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} PO_4^{3 - } + {H_3}{O^ + }\\
HPO_4^{2 - } + {H_2}O \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over
{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} {H_2}P{O_4}^ - + O{H^ - }\\
{H_2}P{O_4}^ - + {H_2}O \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over
{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} HPO_4^{2 - } + {H_3}{O^ + }\\
{H_2}P{O_4}^ - + {H_2}O \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over
{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} {H_3}P{O_4} + O{H^ - }
\end{array}$