To 100 ml of a solution, which contains lead ions, of is added. How much lead remains in the solutio — Ionic Equilibrium Chemistry Question
Question
To 100 ml of a solution, which contains $8.32 \times 10^{-3}\text{ g}$ lead ions, $10^{-4}\text{ moles}$ of $\text{H}_2\text{SO}_4$ is added. How much lead remains in the solution unprecipitated? $K_{sp} \text{ of PbSO}_4 = 1.6 \times 10^{-7}$. ($\text{Pb} = 208$)
💡 Solution & Explanation
Moles of $\text{Pb}^{2+} = \frac{8.32 \times 10^{-3}}{208} = 4 \times 10^{-5}\text{ moles}$. In 100 ml, $[\text{Pb}^{2+}] = 4 \times 10^{-4}\text{ M}$. Moles of added $\text{SO}_4^{2-} = 10^{-4}\text{ moles}$, so $[\text{SO}_4^{2-}] = 10^{-3}\text{ M}$. Let $x$ be the concentration of $\text{Pb}^{2+}$ that precipitates. $(4 \times 10^{-4} - x)(10^{-3} - x) = 1.6 \times 10^{-7}$. Solving the quadratic equation gives $x = 2 \times 10^{-4}\text{ M}$. The remaining $[\text{Pb}^{2+}] = 4 \times 10^{-4} - 2 \times 10^{-4} = 2 \times 10^{-4}\text{ M}$. In 100 ml, moles remaining $= 2 \times 10^{-5}\text{ moles}$. Mass of Pb remaining $= 2 \times 10^{-5} \times 208 = 4.16 \times 10^{-3}\text{ g}$.