CAIE A-Level Chemistry 28.5.3 Calculations Using Kstab

Practise rearranging Kstab expressions and combining stability constants to solve complex-equilibrium problems.

Exam points

substitute equilibrium rather than initial concentrations into the correctly powered Kstab expression
rearrange the expression to calculate an unknown complex, metal-ion or free-ligand concentration
divide product and reactant stability constants in the correct order to obtain ligand-exchange Kc

[Maximum number: 1]

An aqueous solution of copper(II) sulfate is a blue colour due to the presence of $\left[\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$ complex ions.

The numerical value of the stability constant, $K_{\text {stab }}$ , of the $\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}\right]^{2+}$ complex ion is $1.40 \times 10^{13}$ .

In a particular solution the concentration of the $\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}\right]^{2+}$ complex ion is $0.0074 \mathrm{~mol} \mathrm{dm}{ }^{-3}$ and the concentration of $\mathrm{NH}_{3}$ is $0.57 \mathrm{~mol} \mathrm{dm}^{-3}$ .

Use your expression in (c)(iii) and the $K_{\text {stab }}$ value $1.40 \times 10^{13}$ to calculate the concentration of the $\left[\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$ complex ion in this solution.
concentration of $\left[\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}=$ $\mathrm{mol} \mathrm{dm}^{-3}$

$\begin{aligned} \left.\left[\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\right]=(0.0074) \div\left(1.4 \times 10^{13} \times 0.57^{4}\right) \left.\left[\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\right]=\mathbf{5 . 0 1} \times \mathbf{1 0}^{-\mathbf{1 5}} \text { min } 2 \text { sf ecf } \mathbf{6}(\mathbf{c})(\text { iii }) \end{aligned}$

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28.5.3—Kstab expressions to perform calculations question 1