# How do you write the equilibrium expression for hydrolysis of a Lewis acidic metal cation?

Jun 12, 2016

Well, let's take the hydrolysis of aluminum ion as an example, then.

1. ["Al"("H"_2"O")_6]^(3+)(aq) + stackrel("H"_2"O"(l)" ")(rightleftharpoons) ["Al"("H"_2"O")_5("OH")]^(2+)(aq) + "H"^(+)(aq)
2. ["Al"("H"_2"O")_5("OH")]^(2+)(aq) stackrel("H"_2"O"(l)" ")(rightleftharpoons) ["Al"("H"_2"O")_4("OH")_2]^(+)(aq) + "H"^(+)(aq)
3. ["Al"("H"_2"O")_4("OH")_2]^(+)(aq) stackrel("H"_2"O"(l)" ")(rightleftharpoons) "Al"("H"_2"O")_3("OH")_3(aq) + "H"^(+)(aq)
4. ${\text{Al"("H"_2"O")_3("OH")_3(aq) stackrel("H"_2"O"(l)" ")(rightleftharpoons) ["Al"("H"_2"O")_2("OH")_4]^(-)(aq) + "H}}^{+} \left(a q\right)$

The maximum coordination number of ${\text{Al}}^{3 +}$ is four, so it stops there.

This all adds up and cancels to give:

["Al"("H"_2"O")_6]^(3+)(aq) stackrel("H"_2"O"(l)" ")(rightleftharpoons) cancel(["Al"("H"_2"O")_5("OH")]^(2+)(aq)) + "H"^(+)(aq)
cancel(["Al"("H"_2"O")_5("OH")]^(2+)(aq)) stackrel("H"_2"O"(l)" ")(rightleftharpoons) cancel(["Al"("H"_2"O")_4("OH")_2]^(+)(aq)) + "H"^(+)(aq)
cancel(["Al"("H"_2"O")_4("OH")_2]^(+)(aq)) stackrel("H"_2"O"(l)" ")(rightleftharpoons) cancel("Al"("H"_2"O")_3("OH")_3(aq)) + "H"^(+)(aq)
cancel("Al"("H"_2"O")_3("OH")_3(aq)) stackrel("H"_2"O"(l)" ")(rightleftharpoons) ["Al"("H"_2"O")_2("OH")_4]^(-)(aq) + "H"^(+)(aq)
$\text{---------------------------------------------------------------------}$
color(blue)(["Al"("H"_2"O")_6]^(3+)(aq) stackrel("H"_2"O"(l)" ")(rightleftharpoons) ["Al"("H"_2"O")_(6-n)("OH")_n]^((3-n)+)(aq) + n"H"^(+)(aq))

You can then write up an equilibrium constant for the overall reaction as:

\mathbf(K_("hydrolysis") = (["Al"("H"_2"O")_(6-n)("OH")_n]^((3-n)+)["H"^(+)]^n)/(["Al"("H"_2"O")_6]^(3+))),

as you would with equilibrium reactions. (Products over reactants, omitting solids/liquids.)