Question
Explain the following:
$\mathrm{BF}_3$ does not hydrolyse.
$\mathrm{BF}_3$ does not hydrolyse.
✓
Answer
Unlike other boron halides, $\mathrm{BF}_3$ does not hydrolyse completely. Instead, it hydrolyses incompletely to form boric acid and fluoroboric acid. This is because the HF first formed reacts with $\mathrm{H}_3 \mathrm{BO}_3$.
$\text{BF}_3+3\text{H}_2\text{O}\rightarrow\text{H}_3\text{BO}_3+3\text{HF}\ ]\times4$
$\frac{\text{H}_3\text{BO}_3+4\text{HF}\rightarrow\text{H}^++[\text{BF}_4]^-+3\text{H}_2\text{O]}\times3}{4\text{BF}_3+3\text{H}_2\text{O}\rightarrow\text{H}_3\text{BO}_3+3[\text{BF}_4]^-+3\text{H}^+}$
$\text{BF}_3+3\text{H}_2\text{O}\rightarrow\text{H}_3\text{BO}_3+3\text{HF}\ ]\times4$
$\frac{\text{H}_3\text{BO}_3+4\text{HF}\rightarrow\text{H}^++[\text{BF}_4]^-+3\text{H}_2\text{O]}\times3}{4\text{BF}_3+3\text{H}_2\text{O}\rightarrow\text{H}_3\text{BO}_3+3[\text{BF}_4]^-+3\text{H}^+}$
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