Question

Most hydroxides are not very soluble in water. For example, \(K_{\mathrm{sp}}\) for nickel(II) hydroxide, \(\mathrm{Ni}(\mathrm{OH})_{2},\) is \(2.0 \times 10^{-15}\) at 25 . \(\mathrm{C}\). How many grams of nickel(II) hydroxide dissolve per liter at 25 ' \(\mathrm{C}\) ?

Short answer

The solubility of nickel(II) hydroxide in water at 25°C is \(0.000735 \, \mathrm{g/L}\).

Step-by-step solution

Write the solubility equilibrium expression

For the dissolution of nickel(II) hydroxide in water, the balanced chemical equation is: \( \mathrm{Ni(OH)_2 (s) \rightleftharpoons Ni^{2+} (aq) + 2OH^{-} (aq)} \) The solubility equilibrium expression for nickel(II) hydroxide is the solubility product constant, Ksp: \(K_{sp}= [\mathrm{Ni^{2+}}][\mathrm{OH^{-}}]^2\)

Set up an ICE table

Let x be the molar solubility of nickel(II) hydroxide. An ICE (Initial, Change, Equilibrium) table will help us keep track of the concentrations: \( \begin{array}{c|ccc} & [\mathrm{Ni^{2+}}] & [\mathrm{OH^{-}}] \\ \hline \text {Initial} & 0 & 0 \\ \hline \text {Change} & +x & +2x \\ \hline \text {Equilibrium} & x & 2x \end{array}\)

Substitute the concentrations from the ICE table into the Ksp expression

Substitute the equilibrium concentrations from the ICE table into the equilibrium expression and solve for x: \(K_{sp} = [\mathrm{Ni^{2+}}][\mathrm{OH^{-}}]^2\) \(2.0 \times 10^{-15} = (x)(2x)^2\)

Solve for x

Solve the equation for x, the molar solubility of nickel(II) hydroxide: \(2.0 \times 10^{-15} = 4x^3\) \(x^3 = \frac{2.0 \times 10^{-15}}{4}\) \(x^3 = 5.0 \times 10^{-16}\) \(x = \sqrt[3]{5.0 \times 10^{-16}}\) \(x = 7.93 \times 10^{-6} \, \mathrm{mol/L}\)

Convert molar solubility to grams per liter

The molar solubility of nickel(II) hydroxide is 7.93 x 10^{-6} mol/L. Now we need to convert this to grams per liter. To do this, we will multiply the molar solubility by the molar mass of nickel(II) hydroxide (58.71 g/mol for Ni + 34.02 g/mol for 2 OH = 92.73 g/mol): \(7.93 \times 10^{-6} \, \mathrm{mol/L} \times 92.73 \, \mathrm{g/mol} = 0.000735 \, \mathrm{g/L}\)

Write the final answer

The solubility of nickel(II) hydroxide in water at 25°C is 0.000735 grams per liter.

Key concepts

Solubility Product Constant

The solubility product constant, denoted as \(K_{sp}\), is a vital concept in understanding solubility equilibrium. It represents the product of the concentrations of the ions produced by dissolving a solid ionic compound in water, each raised to the power corresponding to their coefficients in the balanced chemical equation.

For nickel(II) hydroxide, \(K_{sp}\) is given for the equilibrium reaction:

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

The \(K_{sp}\) expression is:

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

The value \(2.0 \times 10^{-15}\) shows that only a small amount of nickel(II) hydroxide dissolves in water, indicating its low solubility.

Understanding \(K_{sp}\) helps predict whether a precipitate will form when solutions are mixed, based on the product of ion concentrations compared to \(K_{sp}\).

Molar Solubility

Molar solubility is the number of moles of a solute that can dissolve in a liter of solution before the solution becomes saturated. For nickel(II) hydroxide, it is represented by \(x\) in the solution process.

The concept is deeply related to \(K_{sp}\) because the molar solubility will determine the concentration of ions in a saturated solution. The step-by-step solution calculates \(x = 7.93 \times 10^{-6} \) mol/L for nickel(II) hydroxide.

This value represents the equilibrium concentration of \([\mathrm{Ni^{2+}}]\) in the solution, with \([\mathrm{OH^{-}}]\) being twice as much, due to two hydroxide ions per nickel ion. Converting molar solubility to grams per liter involves multiplying by the molar mass, showing the practical mass amount that dissolves.

ICE Table

The ICE Table (Initial, Change, Equilibrium) is a systematic way to determine the concentrations of species involved in a chemical equilibrium. It organizes information about initial concentrations and the changes they undergo to reach equilibrium.

For nickel(II) hydroxide:

• Initial: Both \([\mathrm{Ni^{2+}}]\) and \([\mathrm{OH^{-}}]\) are initially zero, since we start with solid \(\mathrm{Ni(OH)_2}\).
• Change: Upon dissolution, \([\mathrm{Ni^{2+}}]\) increases by \(x\) and \([\mathrm{OH^{-}}]\) increases by \(2x\). This reflects stoichiometry in the reaction.
• Equilibrium: Using these changes, the concentrations at equilibrium are \(x\) for \([\mathrm{Ni^{2+}}]\) and \(2x\) for \([\mathrm{OH^{-}}]\).

The ICE table clearly lays out the calculations needed to substitute into the \(K_{sp}\) expression, simplifying the problem-solving process.

Equilibrium Concentrations

Equilibrium concentrations are the concentrations of reactants and products in a reaction system at equilibrium. In this context, it refers to \([\mathrm{Ni^{2+}}]\) and \([\mathrm{OH^{-}}]\) when nickel(II) hydroxide is at saturation.

From the ICE table, we determine these concentrations:

• \([\mathrm{Ni^{2+}}]\) is \(x = 7.93 \times 10^{-6} \) mol/L.
• \([\mathrm{OH^{-}}]\) is \(2x = 1.586 \times 10^{-5} \) mol/L, considering the stoichiometry of the dissolution equation.

These concentrations, when plugged back into the \(K_{sp}\) expression, confirm that the system is at equilibrium by reproducing the known \(K_{sp}\) value. Understanding equilibrium concentrations is crucial for predicting the extent of reactions, especially in ionic solutions.