Question

If the ionic product of \(\text{Ni(OH)}_2\) is \(1.9 \times 10^{-15}\), then the molar solubility of \(\text{Ni(OH)}_2\) in \(1.0\text{ M NaOH}\) is

• A. \( 2.9 \times 10^{-18}\text{ M} \)
• B. \( 1.9 \times 10^{-13}\text{ M} \)
• C. \( 1.9 \times 10^{-15}\text{ M} \)
• D. \( 2.9 \times 10^{-14}\text{ M} \)

Hint:

For any sparingly soluble metal hydroxide of the form \(\text{M(OH)}_x\) dissolved in a strong base of molarity \(C\), the common ion effect simplifies the solubility calculation to a direct shortcut formula: \( s = \frac{K_{sp}}{C^x} \). In this scenario, substituting the parameters yields \( s = \frac{1.9 \times 10^{-15}}{(1.0)^2} = 1.9 \times 10^{-15}\text{ M}\).

Answer 1

The Correct Option is B

Concept: This problem involves calculating solubility in the presence of a common ion:
• Sodium hydroxide (\(\text{NaOH}\)) is a strong electrolyte that dissociates completely in water, producing a high concentration of hydroxyl ions (\(\text{OH}^-\)).
• Nickel hydroxide (\(\text{Ni(OH)}_2\)) is a sparingly soluble salt that sets up a dynamic equilibrium with its constituent ions in solution.
• The ionic product of a saturated solution is equal to its solubility product constant (\(K_{sp}\)). Due to Le Chatelier's principle and the common ion effect, the presence of \(\text{OH}^-\) ions from \(\text{NaOH}\) shifts the equilibrium backward, drastically suppressing the molar solubility (\(s\)) of \(\text{Ni(OH)}_2\).

Step 1: Determining total ion concentrations at equilibrium.
First, look at the complete dissociation of the strong base: \[ \text{NaOH}(aq) \rightarrow \text{Na}^+(aq) + \text{OH}^-(aq) \] Since \([\text{NaOH}] = 1.0\text{ M}\), it provides \([\text{OH}^-] = 1.0\text{ M}\). Next, let the molar solubility of \(\text{Ni(OH)}_2\) in this solution be \(s\). The equilibrium dissociation is: \[ \begin{array}{lcccc} \text{Reaction:} & \text{Ni(OH)}_2(s) & \rightleftharpoons & \text{Ni}^{2+}(aq) & + & 2\text{OH}^-(aq) \text{Equilibrium (M):} & & & s & & 1.0 + 2s \end{array} \] Because the ionic product (\(K_{sp}\)) is exceptionally small (\(1.9 \times 10^{-15}\)), the value of \(s\) will be negligible. Therefore, we can safely apply the algebraic approximation: \[ 1.0 + 2s \approx 1.0\text{ M} \]

Step 2: Solving for the molar solubility ($s$) using the solubility product expression.
Write the expression for the solubility product constant (\(K_{sp}\)): \[ K_{sp} = [\text{Ni}^{2+}][\text{OH}^-]^2 \] Substitute the equilibrium concentrations and the given value of the ionic product into the formula: \[ 1.9 \times 10^{-15} = (s)(1.0)^2 \] \[ 1.9 \times 10^{-15} = s \times 1.0 \] \[ s = 1.9 \times 10^{-15}\text{ M} \]