Step 1: Understanding the Question:
The question asks for the equilibrium concentration of lead ions (Pb\(^{2+}\)) in a solution containing a known concentration of sulfide ions (S\(^{2-}\)). This is a problem involving the solubility product constant (K\(_{sp}\)).
Step 2: Key Formula or Approach:
For a sparingly soluble salt like lead(II) sulfide (PbS), the dissolution equilibrium is:
\[
\text{PbS}_{(s)} \rightleftharpoons \text{Pb}^{2+}_{(aq)} + \text{S}^{2-}_{(aq)}
\]
The solubility product expression is:
\[
K_{sp} = [\text{Pb}^{2+}] [\text{S}^{2-}]
\]
We are given K\(_{sp}\) and [S\(^{2-}\)], and we need to find [Pb\(^{2+}\)].
Step 3: Detailed Explanation:
Given values:
- K\(_{sp}\) for PbS = 8.0 \( \times \) 10\(^{-28}\)
- Concentration of sulphide ions, [S\(^{2-}\)] = 1 \( \times \) 10\(^{-11}\) mol/dm\(^3\) (or M)
Let the equilibrium concentration of Pb\(^{2+}\) be [Pb\(^{2+}\)].
Rearrange the K\(_{sp}\) formula to solve for [Pb\(^{2+}\)]:
\[
[\text{Pb}^{2+}] = \frac{K_{sp}}{[\text{S}^{2-}]}
\]
Substitute the given values into the equation:
\[
[\text{Pb}^{2+}] = \frac{8.0 \times 10^{-28}}{1 \times 10^{-11}}
\]
\[
[\text{Pb}^{2+}] = 8.0 \times 10^{(-28 - (-11))}
\]
\[
[\text{Pb}^{2+}] = 8.0 \times 10^{-17}
\]
The equilibrium concentration of Pb\(^{2+}\) ions is 8.0 \( \times \) 10\(^{-17}\) mol/dm\(^3\). Note that Pb\(^{++}\) is another way to write Pb\(^{2+}\).
Step 4: Final Answer:
The calculated concentration of Pb\(^{2+}\) is 8.0 \( \times \) 10\(^{-17}\) M, which matches option (C).