Approximately \(0.14 \mathrm{~g}\) nickel(II) hydrox?de, \(\mathrm{N}_{1}(\mathrm{OH})_{2}(s), \mathrm{dis}-\) solves per liter of water at \(20^{\circ} \mathrm{C}\). Calculate \(K_{\mathrm{sp}}\) for \(\mathrm{Ni}(\mathrm{OH})_{2}(s)\) at this temperature.

First, we need to write the dissociation equation for nickel(II) hydroxide in water: \(Ni(OH)_2 (s) \rightleftharpoons Ni^{2+} (aq) + 2OH^{-} (aq)\)

Now, let's use the solubility information given in the problem to determine the equilibrium concentrations of the ions. Solubility of \(Ni(OH)_2 = 0.14 \mathrm{g\ per\ liter}\) First, we need to convert this value into moles per liter (M) using the molar mass of \(Ni(OH)_2\), which is approximately 92.71 g/mol. Molar solubility = \(\frac{0.14 \ \mathrm{g}}{92.71 \ \mathrm{g/mol}} \approx 1.51 \times 10^{-3} \ \mathrm{M}\) Since \(Ni(OH)_2\) dissociates into 1 mole of \(Ni^{2+}\) ions and 2 moles of \(OH^-\) ions, at equilibrium, the concentrations of these ions would be: \[ [Ni^{2+}] = 1.51 \times 10^{-3} \ \mathrm{M} \] \[ [OH^-] = 2 \times 1.51 \times 10^{-3} \ \mathrm{M} = 3.02 \times 10^{-3} \ \mathrm{M} \]

Now that we have the equilibrium concentrations of the ions, we can calculate the solubility product constant (Ksp) using the Ksp expression for \(Ni(OH)_2\): \[K_{sp} = [Ni^{2+}] [OH^-]^2\] Substituting the equilibrium concentrations, we get: \[ K_{sp} = (1.51 \times 10^{-3})(3.02 \times 10^{-3})^2 \] \[ K_{sp} \approx 4.36 \times 10^{-11} \] So, the Ksp of \(Ni(OH)_2\) at 20°C is approximately \(4.36 \times 10^{-11}\).