Question

The solubility product of PbCl2 at 298 K is $3.2 \times 10^{-5}$. What is its solubility in moldm$^{-3}$?

• A. $8 \times 10^{-6}$
• B. $2 \times 10^{-2}$
• C. $5.6 \times 10^{-3}$
• D. $5.0 \times 10^{-2}$

Hint:

The solubility product ($K_{sp}$) is a measure of the extent to which a sparingly soluble ionic compound dissolves in a solvent. For a salt $M_{x}A_{y}$, $K_{sp} = [M^{y+}]^{x}[A^{x-}]^{y}$. Remember to define solubility ($S$) in terms of ion concentrations and correctly power the concentration of each ion.

Answer 1

The Correct Option is B

Step 1: Write the dissolution equilibrium and $K_{sp$ expression}\nThe dissolution of lead(II) chloride ($PbCl_{2}$) in water is represented by the following equilibrium: \[PbCl_{2}(s) \rightleftharpoons Pb^{2+}(aq) + 2Cl^{-}(aq)\] If $S$ represents the molar solubility of $PbCl_{2}$ in moldm$^{-3}$, then at equilibrium: \[[Pb^{2+}] = S\] \[[Cl^{-}] = 2S\] The solubility product constant ($K_{sp}$) expression is: \[K_{sp} = [Pb^{2+}][Cl^{-}]^{2}\] \[K_{sp} = (S)(2S)^{2}\] \[K_{sp} = 4S^{3}\]
Step 2: Substitute the given $K_{sp$ value and solve for $S$}\nWe are given $K_{sp} = 3.2 \times 10^{-5}$. Substitute this value into the $K_{sp}$ expression: \[4S^{3} = 3.2 \times 10^{-5}\] Divide by 4 to find $S^{3}$: \[S^{3} = \frac{3.2 \times 10^{-5{4}\] \[S^{3} = 0.8 \times 10^{-5}\] To make it easier to take the cube root, rewrite $0.8 \times 10^{-5}$ as $8 \times 10^{-6}$: \[S^{3} = 8 \times 10^{-6}\] Now, take the cube root of both sides to find $S$: \[S = \sqrt[3]{8 \times 10^{-6\] \[S = (8)^{1/3} \times (10^{-6})^{1/3}\] \[S = 2 \times 10^{-2} \text{ moldm}^{-3}\]
Step 3: Conclude the solubility\nThe solubility of $PbCl_{2}$ is $2 \times 10^{-2}$ moldm$^{-3}$.