CAIE A-Level Chemistry 25.1.9 Ksp and Solubility Calculations

Practise converting between molar solubility, equilibrium ion concentrations and Ksp with correct stoichiometry.

Exam points

let molar solubility be s and express each ion concentration as its coefficient times s
substitute the concentrations with powers into Ksp before solving for s or the constant
distinguish supplied ion concentration from solubility when a common ion is already present

[Maximum number: 3]

Iodine is found naturally in compounds in many different oxidation states.

An orange precipitate of $\mathrm{HgI}_{2}$ forms when $\mathrm{Hg}^{2+}$ ions are added to KI(aq). The solubility of $\mathrm{HgI}_{2}$ at $25^{\circ} \mathrm{C}$ is $1.00 \times 10^{-7} \mathrm{gdm}^{-3}$ .

Calculate the solubility product, $K_{\mathrm{sp}}$ , of $\mathrm{HgI}_{2}$ . Include units in your answer.
$\left[M_{\mathrm{r}}: \mathrm{HgI}_{2}, 454.4\right]$
value of $K_{s p}=$
units =

\begin{array}{ll} \text { M1: }\left(\left[\mathrm{Hg}^{2+}\right]\right)=1.00 \times 10^{-7} \div 454.4 =2.20 \times 10^{-10}\left(\mathrm{~mol} \mathrm{dm}^{-3}\right) \text { M2: } K_{\mathrm{sp}}=\left[\mathrm{Hg}^{2+}\right]\left[\mathrm{I}^{-}\right]^{2}=4\left[\mathrm{Hg}^{2+}\right]^{3} =\mathbf{4 . 2 6} \times \mathbf{1 0}^{-\mathbf{2 9}} \mathrm{ecf} \text { M3: units }=\mathrm{mol}^{3} \mathrm{dm}^{-9} \mathrm{ecf} \end{array}

All question bank results loaded

25.1.9—Ksp from concentrations and vice versa question 1