Question

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.

Short answer

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

Step-by-step solution

Write the balanced chemical equation for the dissolving process.

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

Write the expression for Ksp.

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: \[ K_{sp} = [Ni^{2+}][OH^-]^2 \]

Calculate the molar concentration of dissolved ions.

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 \(\mathrm{Ni}(\mathrm{OH})_{2}\): \[ [OH^-] = 2 \times \frac{0.14}{92.71} \text{mol/L} \]

Calculate Ksp using the equilibrium concentrations of ions.

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

Final Answer:

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

Key concepts

Nickel(II) Hydroxide

Nickel(II) hydroxide, with the chemical formula \(\mathrm{Ni(OH)}_2\), is a green solid often used in industries, particularly in rechargeable battery technology. It is a rather sparingly soluble compound, which means it dissolves very slightly in water. This can often be a point of confusion, as students try to calculate the solubility or reaction products of such compounds.
The dissolution process can be represented by the equation:

• \(\mathrm{Ni(OH)}_2 (s) \rightleftharpoons \mathrm{Ni^{2+}} (aq) + 2\mathrm{OH^-} (aq)\)

Here, the compound breaks down into its ionic components. This reaction is an essential part of understanding how ionic equilibrium is established in solutions. The relative insolubility of nickel(II) hydroxide governs how it reacts in other chemical processes.

Chemical Equilibrium

Chemical equilibrium occurs when a chemical reaction ceases to change the concentration of reactants and products over time. For sparingly soluble salts like nickel(II) hydroxide, this happens when the rate of dissolving equals the rate of precipitation. Even though nickel(II) hydroxide barely dissolves, the small amount that does leads to a saturated solution. In this scenario, the equilibrium state can be described by the solubility product constant, \(K_{sp}\).
The \(K_{sp}\) expresses the specific concentrations of ions at equilibrium:

• \(K_{sp} = [\mathrm{Ni^{2+}}][\mathrm{OH^-}]^2\)

This equation implies that if you know one concentration, you can calculate the others. Given its low solubility, the \(K_{sp}\) for nickel(II) hydroxide is very small, signifying the equilibrium significantly favors the solid form over the dissociated ions. Understanding equilibrium is critical for predicting how much of a solid will dissolve under certain conditions.

Concentration Calculations

Calculating concentrations involves converting the given mass of a compound into moles and then determining how much dissolves into a liter of water. This calculation is crucial for understanding how much of a compound can actually achieve equilibrium in solution. For nickel(II) hydroxide:
Start with the mass and convert it using the molar mass:

• Moles of \(\mathrm{Ni(OH)}_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{0.14\, \text{g}}{92.71\, \text{g/mol}}\)

This leads to the concentration of \(\mathrm{Ni^{2+}}\) ions. As each mole of nickel(II) hydroxide generates one mole of \(\mathrm{Ni^{2+}}\) and two moles of \(\mathrm{OH^-}\), the hydroxide ion concentration is twice that of the nickel ions.

• \([\mathrm{OH^-}] = 2 \times [\mathrm{Ni^{2+}}]\)

These concentrations are then used in the \(K_{sp}\) equation to find the precise solubility product constant, which reflects the maximum amount of \(\mathrm{Ni(OH)}_2\) that can dissolve at equilibrium.