To calculate the molar solubility of lead(II) carbonate (PbCO₃) when its solubility product constant ([tex]\( K_{sp} \)[/tex]) is given as [tex]\( 1.5 \times 10^{-15} \)[/tex] at [tex]\( 25^{\circ} \, C \)[/tex], we need to follow these steps:
1. Understand the Dissociation Equation:
Lead(II) carbonate (PbCO₃) dissociates in water according to the following equilibrium equation:
[tex]\[
\text{PbCO}_3 (s) \leftrightharpoons \text{Pb}^{2+} (aq) + \text{CO}_3^{2-} (aq)
\][/tex]
2. Define the Molar Solubility:
Let [tex]\( S \)[/tex] represent the molar solubility of PbCO₃. This means at equilibrium, the concentrations of [tex]\( \text{Pb}^{2+} \)[/tex] and [tex]\( \text{CO}_3^{2-} \)[/tex] will both be [tex]\( S \)[/tex].
3. Write the Solubility Product Expression:
The expression for the solubility product [tex]\( K_{sp} \)[/tex] in terms of [tex]\( S \)[/tex] is:
[tex]\[
K_{sp} = [\text{Pb}^{2+}][\text{CO}_3^{2-}]
\][/tex]
Substituting [tex]\( S \)[/tex] for the concentrations of [tex]\( \text{Pb}^{2+} \)[/tex] and [tex]\( \text{CO}_3^{2-} \)[/tex]:
[tex]\[
K_{sp} = S \cdot S = S^2
\][/tex]
4. Solve for [tex]\( S \)[/tex]:
Given [tex]\( K_{sp} = 1.5 \times 10^{-15} \)[/tex],
[tex]\[
S^2 = 1.5 \times 10^{-15}
\][/tex]
To find [tex]\( S \)[/tex], take the square root of both sides:
[tex]\[
S = \sqrt{1.5 \times 10^{-15}}
\][/tex]
5. Calculate the Square Root:
[tex]\[
S \approx 3.872983346207417 \times 10^{-8} \, \text{M}
\][/tex]
Therefore, the molar solubility of [tex]\( \text{PbCO}_3 \)[/tex] is approximately [tex]\( 3.9 \times 10^{-8} \, \text{M} \)[/tex].
Out of the given options, the closest and correct value is:
[tex]\[ \boxed{3.9 \times 10^{-8} \, \text{M}} \][/tex]
