Given \(K_{s p}\) and the equilibrium concentration of one ion, calculate the equilibrium concentration of the other ion. (a) cadmium(II) hydroxide: \(K_{\text {sp }}=2.5 \times 10^{-14} ;\left[\mathrm{Cd}^{2+}\right]=1.5 \times 10^{-6} M\) (b) copper(II) arsenate \(\left(\mathrm{Cu}_{3}\left(\mathrm{AsO}_{4}\right)_{2}\right): K_{\mathrm{sp}}=7.6 \times 10^{-36} ;\left[\mathrm{AsO}_{4}^{3-}\right]=\) \(2.4 \times 10^{-4} M\) (c) zinc oxalate: \(K_{s p}=2.7 \times 10^{-8} ;\left[\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}\right]=8.8 \times 10^{-3} M\)

Question: Calculate the equilibrium concentration of ions for (a) hydroxide ions when the equilibrium concentration of cadmium(II) ions is \(1.5 \times 10^{-6} M\) and the \(K_{sp}\) for cadmium(II) hydroxide is \(2.5 \times 10^{-14}\), (b) copper(II) ions when the equilibrium concentration of arsenate ions is \(2.4 \times 10^{-4} M\) and the \(K_{sp}\) for copper(II) arsenate is \(7.6 \times 10^{-36}\), and (c) zinc(II) ions when the equilibrium concentration of oxalate ions is \(8.8 \times 10^{-3} M\) and the \(K_{sp}\) for zinc oxalate is \(2.7 \times 10^{-8}\). Answer: (a) \([\mathrm{OH^-}] = 4.09 \times 10^{-5} M\), (b) \([\mathrm{Cu^{2+}}] = 1.1 \times 10^{-9} M\), (c) \([\mathrm{Zn^{2+}}] = 3.07\times10^{-6} M\)