NESA Chemistry Solution Equilibria

$\text{Cd}_3(\text{PO}_4)_2(s) \rightleftharpoons 3\text{Cd}^{2+}(aq) + 2\text{PO}_4^{3-}(aq)$

Let $s \text{ mol } \text{Cd}_3(\text{PO}_4)_2(s)$ dissolve per litre

$K_{sp} = [\text{Cd}^{2+}]^3[\text{PO}_4^{3-}]^2 = (3s)^3(2s)^2 = 108s^5$

$s = \sqrt[5]{\frac{2.53 \times 10^{-33}}{108}} = 1.19 \times 10^{-7}$

$[\text{Cd}^{2+}] = 3s = 3.56 \times 10^{-7} \text{ mol L}^{-1}$

If only barium ions are present, they will give a precipitate in step 1 of this procedure ($\text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq) \rightarrow \text{BaSO}_4(s)$), so the conclusion that only barium ions are present will be correct.

If lead(II) ions are present, they will also precipitate at step 1 of this procedure. This will leave insufficient lead(II) ions in solution to give a precipitate in step 3. The conclusion reached will be that only barium ions are present, which is incorrect.

[Mg$^{2+}$] = $6 \times 10^{-3}$
[OH$^-$] = $1 \times 10^{-2}$
[CO$_3^{2-}$] = $2 \times 10^{-3}$

[Mg$^{2+}$][OH$^-$]$^2$ = $6 \times 10^{-7}$
$K_{sp} = 5.61 \times 10^{-12}$

[Mg$^{2+}$][CO$_3^{2-}$] = $1.2 \times 10^{-5}$
$K_{sp} = 6.82 \times 10^{-6}$

Mg(OH)$_2$ is a lot less soluble than MgCO$_3$.

Mg(OH)$_2$ will precipitate. The Mg$^{2+}$ is in excess. Its concentration will drop to approximately $6 \times 10^{-3} - \frac{1 \times 10^{-2}}{2} = 1 \times 10^{-3}$.

[Mg$^{2+}$][CO$_3^{2-}$] becomes $2 \times 10^{-6}$ which is less than $K_{sp}$.

MgCO$_3$ won't precipitate.