Assuming chromium is present as ${\text{C}}{{\text{r}}^{3 + }}$, state an equation for its reaction with hydroxide ions, include state symbols.

State an expression for the solubility product constant, ${K_{{\text{sp}}}}$, for chromium(III) hydroxide.

The solubility product of chromium(III) hydroxide is $1.00 \times {10^{ - 33}}{\text{ mo}}{{\text{l}}^{\text{4}}}{\text{d}}{{\text{m}}^{ - 12}}$ at 298 K. Calculate the concentration, in ${\text{mol}}\,{\text{d}}{{\text{m}}^{ - 3}}$, of ${\text{C}}{{\text{r}}^{3 + }}$ in the solution, when chromium(III) hydroxide is precipitated.

${\text{C}}{{\text{r}}^{3 + }}{\text{(aq)}} + {\text{3O}}{{\text{H}}^ - }{\text{(aq)}} \rightleftharpoons {\text{Cr(OH}}{{\text{)}}_3}{\text{(s)}}$

correct equation;

correct state symbols;

Equilibrium or normal arrows can be used.

${K_{{\text{sp}}}} = {\text{[C}}{{\text{r}}^{3 + }}{\text{][O}}{{\text{H}}^ - }{{\text{]}}^3}$;

$$1.00 \times {10^{ - 33}} = {\text{[C}}{{\text{r}}^{3 + }}{\text{][O}}{{\text{H}}^ - }{{\text{]}}^{\text{3}}}$$ and ${\text{[C}}{{\text{r}}^{3 + }}{\text{]}} = 3 \times {\text{[O}}{{\text{H}}^ - }{\text{]}}$;

$1.00 \times {10^{ - 33}} = {\text{[C}}{{\text{r}}^{3 + }}{\text{](3[C}}{{\text{r}}^{3 + }}{\text{]}}{{\text{)}}^{\text{3}}} = {\text{ }}27 \times {{\text{[C}}{{\text{r}}^{3 + }}{\text{]}}^{\text{4}}}$

${\text{[C}}{{\text{r}}^{3 + }}{\text{]}} = 2.47 \times {10^{ - 9}}{\text{ (mol}}\,{\text{d}}{{\text{m}}^{ - 3}}{\text{)}}$;

Award [2] for correct final answer.

Award [1] for 5.62 $\times$ 10^–9.

Part (a) was generally done well and the majority of candidates scored 1 or 2 marks here. State symbols were sometimes omitted or incorrect, for example ${\text{C}}{{\text{r}}^{3 + }}{\text{(s)}}$ and ${\text{Cr(OH}}{{\text{)}}_{\text{3}}}{\text{(aq)}}$. Some did not know the formula ${\text{Cr(OH}}{{\text{)}}_{\text{3}}}$ and the coefficient of ${\text{O}}{{\text{H}}^ - }$ (3) was sometimes omitted.

In (b) about half the candidates gave the correct expression for ${K_{{\text{sp}}}}$. Some included the solid in the ${K_{{\text{sp}}}}$ expression and quite a few gave the inverted solubility constant expression; there was a lot of confusion with the ${K_{\text{c}}}$ expression.

In (c) only the best candidates did not have major difficulties calculating the solubility product. Most candidates failed to recognize the ${\text{[O}}{{\text{H}}^ - }{\text{]}}$ is three times that of ${\text{C}}{{\text{r}}^{3 + }}$ when calculating the ${\text{C}}{{\text{r}}^{3 + }}$ concentration.