Question

Zinc hydroxide, \(\mathrm{Zn}(\mathrm{OH})_{2}\left(K_{\mathrm{sp}}=4.1 \times 10^{-17}\right)\), is nearly insoluble in water but is more soluble in strong base because \(\mathrm{Zn}^{2+}\) forms the soluble complex ion \(\mathrm{Zn}(\mathrm{OH})_{4}^{2-}\left(K_{\mathrm{f}}=3 \times 10^{15}\right)\) (a) What is the molar solubility of \(\mathrm{Zn}(\mathrm{OH})_{2}\) in pure water? (You may ignore \(\mathrm{OH}^{-}\) from the self-dissociation of water.) (b) What is the \(\mathrm{pH}\) of the solution in part (a)? (c) What is the molar solubility of \(\mathrm{Zn}(\mathrm{OH})_{2}\) in \(0.10 \mathrm{M} \mathrm{NaOH}\) ?

Short answer

(a) Molar solubility in water: \(2.15 \times 10^{-6}\) M. (b) pH: 8.63. (c) Molar solubility in 0.10 M NaOH: \(6.5 \times 10^{-3}\) M.

Step-by-step solution

Determine the solubility expression

For the dissolution of \( \mathrm{Zn(OH)}_2 \) in water: \( \mathrm{Zn(OH)}_2 (s) \rightleftharpoons \mathrm{Zn}^{2+} (aq) + 2\mathrm{OH}^- (aq) \), the solubility product \( K_{\mathrm{sp}} \) is given by \( K_{\mathrm{sp}} = [\mathrm{Zn}^{2+}][\mathrm{OH}^-]^2 \). Since \( K_{\mathrm{sp}} = 4.1 \times 10^{-17} \), let the solubility of \( \mathrm{Zn(OH)}_2 \) be \( s \), then \( [\mathrm{Zn}^{2+}] = s \) and \( [\mathrm{OH}^-] = 2s \). Substitute to get the expression: \( K_{\mathrm{sp}} = s(2s)^2 = 4s^3 \).

Calculate molar solubility in pure water

Solve the expression from Step 1: \( 4s^3 = 4.1 \times 10^{-17} \). Dividing both sides by 4 gives: \( s^3 = 1.025 \times 10^{-17} \). Taking the cube root of both sides, \( s = \sqrt[3]{1.025 \times 10^{-17}} \approx 2.15 \times 10^{-6} \). This is the molar solubility of \( \mathrm{Zn(OH)}_2 \) in pure water.

Calculate the pH of the solution

From the molar solubility, \( [\mathrm{OH}^-] = 2s = 2(2.15 \times 10^{-6}) = 4.30 \times 10^{-6} \). The pOH of the solution is \( -\log[\mathrm{OH}^-] \approx -\log(4.30 \times 10^{-6}) \approx 5.37 \). Therefore, the pH is \( 14 - \text{pOH} = 8.63 \).

Calculate molar solubility in 0.10 M NaOH

In a 0.10 M NaOH solution, \( [\mathrm{OH}^-] = 0.10 \), which shifts the equilibrium: \( \mathrm{Zn(OH)}_2 + 2\mathrm{OH}^- \rightleftharpoons \mathrm{Zn(OH)}_4^{2-} \). The reaction for forming the complex ion is \( \mathrm{Zn}^{2+} + 4\mathrm{OH}^- \rightleftharpoons \mathrm{Zn(OH)}_4^{2-} \). With \( K_f = 3 \times 10^{15} \), and approximating that the increase in \( \mathrm{OH}^- \) due to solubility is negligible, the solubility \( s' \) is given by: \( K_f \cdot K_{sp} / [\mathrm{OH}^-]^2 = [\mathrm{Zn(OH)}_4^{2-}] \). Substituting the values gives: \( s' = \sqrt{\frac{3 \times 10^{15} \times 4.1 \times 10^{-17}}{(0.10)^4}} \). Calculating gives \( s' \approx 6.5 \times 10^{-3} \).

Key concepts

Solubility Product (Ksp)

The solubility product, shortened as \( K_{sp} \), is a constant that helps us understand the solubility of sparingly soluble compounds like zinc hydroxide. It refers to the product of the concentrations of the ions in a saturated solution, each raised to the power of their stoichiometric coefficients. For \( \mathrm{Zn(OH)}_2 \), the solubility equilibrium in water is written as:

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

Thus, the expression for \( K_{sp} \) is given by:

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

When calculating solubility, we assign \( s \) as the solubility: the concentration of \( \mathrm{Zn}^{2+} \) becomes \( s \) and \( \mathrm{OH}^- \) becomes \( 2s \), making the equation \( 4s^3 = K_{sp} \). Solving for \( s \) gives the molar solubility in pure water, crucial for predicting how much zinc hydroxide can dissolve. Understanding \( K_{sp} \) is fundamental to determining the saturation point of any ionic compound in a solution.

Complex Ion Formation

Zinc ions can form a complex ion under certain conditions, which increases the solubility of zinc hydroxide. When \( \mathrm{Zn^{2+}} \) ions encounter a large excess of hydroxide ions \(( \mathrm{OH}^- )\), they can form a stable complex, \( \mathrm{Zn(OH)}_4^{2-} \). The formation is facilitated by the complexation equilibrium:

• \( \mathrm{Zn^{2+} + 4OH^- \rightleftharpoons Zn(OH)_4^{2-}} \)

The stability of this complex ion is measured by the formation constant \( K_f \), which for \( \mathrm{Zn(OH)}_4^{2-} \), is very large, \( 3 \times 10^{15} \). A high \( K_f \) value means that once formed, the complex ion is very stable. In the presence of a strong base, like 0.10 M NaOH, the formation of this complex makes additional zinc ions dissolve, enhancing the solubility significantly. This effect underlines the critical role of complex ions in solubility dynamics, especially in metal hydroxides.

Effect of pH on Solubility

pH levels have a direct impact on the solubility of compounds like zinc hydroxide. The dissolution equilibrium of \( \mathrm{Zn(OH)_2} \) releases \( \mathrm{OH^-} \) ions, and altering the pH affects how much zinc hydroxide can dissolve in water. In pure water, the molar solubility is directly related to the concentration of hydroxide ions, influencing the pH:

• More \( \mathrm{OH^-} \) raises the pH, making the solution more basic.

In part (a) of the exercise, the solubility of \( \mathrm{Zn(OH)}_2 \) was modest, resulting in a pH of 8.63, indicating a basic solution. However, in an alkaline environment, like 0.10 M NaOH, the excess \( \mathrm{OH^-} \) shifts the equilibrium towards dissolution, facilitated by complex ion formation as explained earlier. This results in significant enhancement of zinc hydroxide solubility beyond what is possible in neutral water. Therefore, by adjusting the pH, often through adding acids or bases, we can control the solubility behavior of metal hydroxides, an essential principle in chemical equilibrium and practical applications.