Question

Calculate the equilibrium concentration of Pb$^{2+}$ ions in a solution of PbS containing 1 $\times$ 10$^{-11}$ moldm$^{-3}$ of sulphide ions. (Given $K_{sp}$ for PbS = 8.0 $\times$ 10$^{-28}$)}

• A. 4 $\times$ 10$^{-14}$
• B. 4 $\times$ 10$^{-17}$
• C. 8 $\times$ 10$^{-17}$
• D. 8 $\times$ 10$^{-11}$

Hint:

$K_{sp}$ is the product of the concentrations of the ions raised to the power of their stoichiometric coefficients.

Answer 1

The Correct Option is C

Step 1: Solubility Product Expression
For PbS: $K_{sp} = [Pb^{2+}][S^{2-}]$.
Step 2: Substitution
$8.0 \times 10^{-28} = [Pb^{2+}] \times (1 \times 10^{-11})$.
Step 3: Calculation
$[Pb^{2+}] = \frac{8.0 \times 10^{-28}}{1 \times 10^{-11}} = 8.0 \times 10^{-17}$ M.
Step 4: Conclusion
The equilibrium concentration is 8 $\times$ 10$^{-17}$ moldm$^{-3}$.
Final Answer:(C)