Question

The solubility product for \(\mathrm{Zn}(\mathrm{OH})_{2}\) is \(3.0 \times 10^{-16}\). The formation constant for the hydroxo complex, \(\mathrm{Zn}(\mathrm{OH})_{4}{ }^{2-},\) is \(4.6 \times 10^{17}\). What concentration of \(\mathrm{OH}^{-}\) is required to dissolve 0.015 mol of \(\mathrm{Zn}(\mathrm{OH})_{2}\) in a liter of solution?

Short answer

The concentration of \(\mathrm{OH}^-\) required is approximately 0.0104 M.

Step-by-step solution

Understand the Equations

We have the compound \(\mathrm{Zn} ( \mathrm{OH} ) _2\), which dissociates according to its solubility product \(K_{sp}\): \(\mathrm{Zn} ( \mathrm{OH} ) _2 \rightleftharpoons \mathrm{Zn}^{2+} + 2 \mathrm{OH}^ -\) with \(K_{sp} = [\mathrm{Zn}^{2+}] [\mathrm{OH}^-]^2 = 3.0 \times 10^{-16}\). Additionally, the formation of the complex \(\mathrm{Zn} ( \mathrm{OH} ) _4^{2-}\) involves \(\mathrm{Zn}^{2+} + 4 \mathrm{OH}^- \rightleftharpoons \mathrm{Zn} ( \mathrm{OH} ) _4^{2-}\) with the formation constant \(K_f = \frac{[\mathrm{Zn} ( \mathrm{OH} ) _4^{2-}]}{[\mathrm{Zn}^{2+}][\mathrm{OH}^-]^4} = 4.6 \times 10^{17}\).

Setting Up the Equilibrium Expression

We can express that the dissolution of \(\mathrm{Zn} ( \mathrm{OH} ) _2\) in terms of the complex formation equilibrium: \(\mathrm{Zn} ( \mathrm{OH} ) _2 + 2 \mathrm{OH}^- \rightleftharpoons \mathrm{Zn} ( \mathrm{OH} ) _4^{2-}\). The effective equilibrium constant is \(K_{eff} = K_{sp} \times K_f\).

Calculate the Effective Equilibrium Constant

Substitute the values for \(K_{sp}\) and \(K_f\) to find \(K_{eff}\): \[K_{eff} = (3.0 \times 10^{-16})(4.6 \times 10^{17}) = 1.38 \times 10^2\]

Determine [OH^-] Concentration

Using the expression for \(K_{eff}\) and the dissolution process, set up the equation:\[K_{eff} = \frac{[\mathrm{Zn} ( \mathrm{OH} ) _4^{2-}]}{[\mathrm{OH}^-]^2}\]Assume that all \(0.015\, \mathrm{mol/L}\) of \(\mathrm{Zn} ( \mathrm{OH} ) _2\) is converted to \(\mathrm{Zn} ( \mathrm{OH} ) _4^{2-}\), yielding:\[1.38 \times 10^2 = \frac{0.015}{[\mathrm{OH}^-]^2}\]

Solve for [OH^-]

Rearrange the equation to find \([\mathrm{OH}^-]\):\[[\mathrm{OH}^-]^2 = \frac{0.015}{1.38 \times 10^2}\]\[[\mathrm{OH}^-]^2 = 1.087 \times 10^{-4}\]\[[\mathrm{OH}^-] = \sqrt{1.087 \times 10^{-4}}\]\[[\mathrm{OH}^-] \approx 0.0104 \text{ M}\].

Key concepts

Solubility Product Constant

The solubility product constant, often denoted as \( K_{sp} \), is a crucial concept in chemistry that helps us understand the extent to which a salt can dissolve in water. It emerges from the equilibrium condition of a slightly soluble salt when it dissolves. For example, when \( \mathrm{Zn(OH)}_2 \) dissolves, it dissociates into its ions following the equation:

• \( \mathrm{Zn(OH)}_2 \leftrightarrow \mathrm{Zn}^{2+} + 2 \mathrm{OH}^- \)

The equilibrium expression for this solubility process is given by:

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

This expression means that the product of the concentrations of the ions, each raised to the power of their stoichiometric coefficients, equals the \( K_{sp} \) value. For \( \mathrm{Zn(OH)}_2 \), this constant is \( 3.0 \times 10^{-16} \).
Understanding \( K_{sp} \) helps in predicting how much solute can dissolve in a given volume of solvent, allowing chemists to determine conditions needed for various applications.

Formation Constant

The formation constant, denoted by \( K_f \), is a measure of the stability of a complex ion in solution. It's derived from the equilibrium established when metal ions form a complex with other ions or molecules.
In the case of \( \mathrm{Zn(OH)}_4^{2-} \), it is formed by the combination of \( \mathrm{Zn}^{2+} \) ions with hydroxide ions:

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

The equilibrium constant for this reaction is given by:

• \( K_f = \frac{[\mathrm{Zn(OH)}_4^{2-}]}{[\mathrm{Zn}^{2+}][\mathrm{OH}^-]^4} = 4.6 \times 10^{17} \)

The high value of \( K_f \) indicates that the formation of \( \mathrm{Zn(OH)}_4^{2-} \) is highly favorable under the given conditions, meaning this complex ion is quite stable.
This constant is essential for understanding reactions involving complex ions and predicting the extent to which these complexes will form in a solution.

Equilibrium Constant Calculation

Calculating equilibrium constants is a fundamental skill in chemistry, allowing us to quantitatively describe chemical equilibria. In the case of solving how much \( \mathrm{OH}^- \) is required to dissolve \( \mathrm{Zn(OH)}_2 \), we utilize the concept of an effective equilibrium constant, \( K_{eff} \).
\( K_{eff} \) involves both the solubility product and the formation constant:

• \( K_{eff} = K_{sp} \times K_f \)

This compound constant helps us to tie together the dissolution of \( \mathrm{Zn(OH)}_2 \) and the formation of \( \mathrm{Zn(OH)}_4^{2-} \):

• \( \mathrm{Zn(OH)}_2 + 2 \mathrm{OH}^- \leftrightarrow \mathrm{Zn(OH)}_4^{2-} \)

Substituting the known values yields:

• \( K_{eff} = 1.38 \times 10^2 \)

This effective constant forms the basis of finding the required concentration of \( \mathrm{OH}^- \).
By rearranging the equilibrium expressions and substituting the appropriate concentrations, we can solve for the amount of hydroxide ions needed to achieve the specified equilibrium, hence allowing us to calculate the exact \( \mathrm{OH}^- \) concentration for dissolving 0.015 mol of \( \mathrm{Zn(OH)}_2 \).