Dissolving a precipitate

The principle

• A precipitate dissolves further if you remove one of its ions from solution.
• Removing a product shifts the dissolving equilibrium to the right, so more solid dissolves.

• This is the exact opposite of the common ion effect, which adds an ion and shifts the equilibrium left.

Method 1: add acid to remove a basic anion

• If the anion of the solid is a base, adding a strong acid protonates it and removes it from solution.

For magnesium hydroxide:

$$\text{Mg(OH)}_2(s) \rightleftharpoons \text{Mg}^{2+} + 2\text{OH}^-$$

$$\text{OH}^- + \text{H}_3\text{O}^+ \rightarrow 2\text{H}_2\text{O}$$

• The acid consumes the hydroxide, so $[\text{OH}^-]$ falls, the first equilibrium shifts right, and the solid dissolves.

• This only works if the anion is basic. Hydroxides, carbonates and fluorides dissolve in acid; chlorides, nitrates and sulfates essentially do not, because $\text{Cl}^-$, $\text{NO}_3^-$ and $\text{SO}_4^{2-}$ are very weak bases.

Method 2: form a complex ion

• If the cation forms a complex ion with an added reagent, that reagent removes the cation.

For silver chloride with ammonia:

$$\text{AgCl}(s) \rightleftharpoons \text{Ag}^+ + \text{Cl}^-$$

$$\text{Ag}^+ + 2\text{NH}_3 \rightleftharpoons [\text{Ag}(\text{NH}_3)_2]^+$$

• The ammonia ties up the silver ion, so $[\text{Ag}^+]$ falls, the first equilibrium shifts right, and the precipitate dissolves.

The nine complex ions

The specification says knowledge of these will be assumed, and the list is given in the resource booklet:

$$[\text{Ag}(\text{CN})_2]^- \quad [\text{Ag}(\text{NH}_3)_2]^+ \quad [\text{Al}(\text{OH})_4]^-$$

$$[\text{Cu}(\text{NH}_3)_4]^{2+} \quad [\text{Pb}(\text{OH})_4]^{2-} \quad [\text{Zn}(\text{OH})_4]^{2-}$$

$$[\text{Zn}(\text{NH}_3)_4]^{2+} \quad [\text{Ni}(\text{NH}_3)_6]^{2+} \quad [\text{Ni}(\text{CN})_4]^{2-}$$

• Note the hydroxide complexes: $\text{Al}(\text{OH})_3$, $\text{Pb}(\text{OH})_2$ and $\text{Zn}(\text{OH})_2$ dissolve in excess sodium hydroxide by forming $[\text{Al}(\text{OH})_4]^-$, $[\text{Pb}(\text{OH})_4]^{2-}$ and $[\text{Zn}(\text{OH})_4]^{2-}$.
• Less familiar complex ions may appear, but the question will supply the information.

The three-way summary

Addition	Effect on the ion	Equilibrium shift	Solubility
common ion	adds one of the ions	left	decreases
acid (basic anion)	removes the anion	right	increases
complexing agent	removes the cation	right	increases
ion with nothing in common	no change	none	unchanged

A hydroxide that goes both ways

• Adding a little sodium hydroxide to a solution of $\text{Zn}^{2+}$ precipitates $\text{Zn(OH)}_2$ — the hydroxide is a common ion and also a reactant.
• Adding excess sodium hydroxide redissolves it, by forming $[\text{Zn}(\text{OH})_4]^{2-}$.
• So the same reagent first precipitates and then dissolves, depending on how much is added.