Question #62993

You're dealing with the hydrolisis of boron trichloride, `"BCl"_3`.

This reaction will produce boric acid, which you'll see written both as `"H"_3"BO"_3` and as `"B"("OH")_3`, and hydrochloric acid, `"HCl"`.

The balanced chemical equation would indeed look like this

`"BCl"_text(3(aq]) + 3"H"_2"O"_text((l]) -> "B"("OH")_text(3(aq]) + 3"HCl"_text((aq])`

I'll try to explain what's going on here without going into too much detail. Mind you, this is a simplified version of what actually goes on, but it will give you an idea of how things work here.

The boron atom in boron trichloride, which is `"sp"^2` hybridized, has an empty 2p-orbital.


This means that it can act as a Lewis acid, i.e .accept a pair of electrons from the oxygen atom of a water molecule.

This will lead to the formation of an adduct, which is simply the result of the addition of two molecules.

Once this adduct, which can be represented as `["H"_2"O"-"BCl"_3]^(-)`, is formed, a hydrogen atom located on the water molecule will jump off, leaving an `"OH"-` group attached to the boron trichloride.

This will trigger the replacement of the `"B"-"Cl"` bonds with stronger `"B"-"O"` bonds, resulting in the formation of boric acid, `"B"("OH")_3`.

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As an interest follow-up, boric acid acts as an acid not because it donates protons, `"H"^(+)`, but because it pulls another `-"OH"` group from water to form the tetrahydroxyborate anion, `["B"("OH")_4]^(-)`

`"B"("OH")_text(3(aq]) + "H"_2"O"_text((l]) rightleftharpoons ["B"("OH")_text(4(aq])]^(-) + "H"_text((aq])^(+)`