Question #44174

A hypothetical metal ion \ce{M^{3+}} forms a sparingly soluble hydroxide \ce{M(OH)3 (s)} with $K_{sp} = 1.0 \times 10^{-33}$ at 298 K. In the presence of excess aqueous \ce{OH^-}, \ce{M^{3+}} forms two soluble complex ions: a tetrahedral $sp^3$ hybridized complex \ce{[M(OH)4]^-} and an octahedral $sp^3d^2$ hybridized complex \ce{[M(OH)6]^{3-}}.

The overall formation constant ($\beta_4$) for \ce{[M(OH)4]^-} is $1.0 \times 10^{32}$. The equilibrium constant ($K$) for the reaction \ce{[M(OH)4]^- (aq) + 2OH^- (aq) \rightleftharpoons [M(OH)6]^{3-} (aq)} is $1.0 \times 10^3$.

A suspension is prepared by adding 0.11 moles of \ce{M(OH)3 (s)} to 1.0 L of water. Solid \ce{NaOH} is then added slowly to the suspension. Assuming no change in volume, calculate the total moles of \ce{NaOH} that must be added to just completely dissolve the \ce{M(OH)3 (s)} precipitate.