Approximately 0.14 g nickel(II) hydroxide, \(\mathrm{Ni}(\mathrm{OH})_{2}(s),\) dissolves per liter of water at \(20^{\circ} \mathrm{C}\). Calculate \(K_{\mathrm{sp}}\) for \(\mathrm{Ni}(\mathrm{OH})_{2}(s)\) at this temperature.

The balanced equation for the dissolution of nickel(II) hydroxide in water is: \[ Ni(OH)_2 (s) \rightleftharpoons Ni^{2+} (aq) + 2OH^- (aq) \]

Ksp is the solubility product constant that represents the equilibrium between the solid and its dissolved ions in a saturated solution. The Ksp expression is given by the product of the equilibrium concentrations of the ions raised to their respective stoichiometric coefficients. For the dissolving process of nickel(II) hydroxide, the Ksp expression is: \[ K_{sp} = [Ni^{2+}][OH^-]^2 \]

We know that 0.14 g of \(\mathrm{Ni}(\mathrm{OH})_{2}\) dissolves per liter of water. To find the molar concentration of the ions, we need to first calculate the moles of nickel(II) hydroxide dissolved: Moles of $\mathrm{Ni}(\mathrm{OH})_{2} = \frac{\text{mass}}{\text{molar mass}} = \frac{0.14\text{ g}}{1\text{ L} \cdot (58.69 + 2\times 17.01)\text{ g/mol}} = \frac{0.14}{92.71} \text{mol/L}\] Since there is a 1:1 ratio between the moles of \(\mathrm{Ni}(\mathrm{OH})_{2}\) dissolved and the moles of \(Ni^{2+}\) ions formed, the molar concentration of \(Ni^{2+}\) is equal to that of \(\mathrm{Ni}(\mathrm{OH})_{2}\): \[ [Ni^{2+}] = \frac{0.14}{92.71} \text{mol/L} \] Since the dissolution of \(\mathrm{Ni}(\mathrm{OH})_{2}\) produces 2 moles of \(OH^-\) ions per mole of \(\mathrm{Ni}(\mathrm{OH})_{2}\), the concentration of \(OH^-\) ions is twice the concentration of \(\mathrm{Ni}(\mathrm{OH})_{2}\): \[ [OH^-] = 2 \times \frac{0.14}{92.71} \text{mol/L} \]

Now that we have the equilibrium concentrations of the ions, we can plug them into the Ksp expression: \[ K_{sp} = [Ni^{2+}][OH^-]^2 = \left(\frac{0.14}{92.71}\right) \left(2 \times \frac{0.14}{92.71}\right)^2\] \[ K_{sp} = \frac{0.14}{92.71} \times \frac{0.14^2}{8537.6141} \] \[ K_{sp} \approx 4.52 \times 10^{-16} \]

The solubility product constant, \(K_{sp}\), for \(\mathrm{Ni}(\mathrm{OH})_{2}(s)\) at \(20^{\circ} \mathrm{C}\) is approximately \(4.52 \times 10^{-16}\).