Question

The pH of a saturated solution of a metal hydroxide of formula X(OH)\textsubscript{2 is 12.0 at 298 K. What is the solubility product of the metal hydroxide at 298 K (in mol\textsuperscript{3} L\textsuperscript{--3})?}

• A. $2 \times 10^{-6}$
• B. $1 \times 10^{-7}$
• C. $5 \times 10^{-5}$
• D. $2 \times 10^{-5}$
• E. $5 \times 10^{-7}$

Hint:

For a salt of type $AB_2$, $K_{sp}$ in terms of molar solubility ($s$) is $4s^3$. Here, $s$ is equal to the concentration of $X^{2+}$, which we found to be $5 \times 10^{-3}$.

Answer 1

Answer: (E)

Concept: The solubility product ($K_{sp}$) of a sparingly soluble salt is the equilibrium constant for the dissolution of the solid in water.
• Dissociation Equation: For $X(OH)_2 \rightleftharpoons X^{2+} + 2OH^-$.
• Relation to pH: The concentration of $OH^-$ ions can be derived from the pH of the solution.
• Ksp Formula: $K_{sp} = [X^{2+}][OH^-]^2$.

Step 1: Calculate [OH\textsuperscript{--}] from pH. Given $pH = 12.0$. Since $pH + pOH = 14$ at 298 K: \[ pOH = 14 - 12 = 2 \] The concentration of hydroxide ions is: \[ [OH^-] = 10^{-pOH} = 10^{-2} \text{ M} = 0.01 \text{ M} \]

Step 2: Determine [X\textsuperscript{2+}] and calculate K\textsubscript{sp}. From the stoichiometry of $X(OH)_2$, for every 2 moles of $OH^-$ produced, 1 mole of $X^{2+}$ is produced. \[ [X^{2+}] = \frac{[OH^-]}{2} = \frac{10^{-2}}{2} = 0.5 \times 10^{-2} = 5 \times 10^{-3} \text{ M} \] Now, calculate the solubility product: \[ K_{sp} = [X^{2+}][OH^-]^2 = (5 \times 10^{-3}) \times (10^{-2})^2 \] \[ K_{sp} = 5 \times 10^{-3} \times 10^{-4} = 5 \times 10^{-7} \]