The value of Ksp for PbCl2 is 1.6 × 10^-5. What is the lowest concentration of Cl−(aq) that would be — Equilibrium Chemistry Question

STEPS:

1. Identify the Dissolution Equation and $K_{sp}$ Expression: The compound $PbCl_2$ dissociates in water according to the following equilibrium: $PbCl_2(s) \rightleftharpoons Pb^{2+}(aq) + 2Cl^-(aq)$. The corresponding solubility product expression is $K_{sp} = [Pb^{2+}][Cl^-]^2$.
2. Determine the Concentration of Lead(II) Ions: Because $Pb(NO_3)_2$ is a highly soluble strong electrolyte, it dissociates completely in aqueous solution. A $0.010\text{ M}$ solution of $Pb(NO_3)_2$ results in a $[Pb^{2+}]$ of $0.010\text{ M}$ (or $1.0 \times 10^{-2}\text{ M}$).
3. Identify the Condition for Precipitation: Precipitation begins the moment the ion product ($Q$) reaches the value of the solubility product constant ($K_{sp}$). To find the "lowest concentration" needed to begin the process, we set the $K_{sp}$ expression equal to the given constant.
4. Set Up the Calculation: Substitute the known values into the $K_{sp}$ equation:
$$1.6 \times 10^{-5} = (0.010) \times [Cl^-]^2$$
5. Solve for $[Cl^-]^2$: Divide both sides by $0.010$ (which is $10^{-2}$):
$$[Cl^-]^2 = \frac{1.6 \times 10^{-5}}{10^{-2}} = 1.6 \times 10^{-3}$$
6. Calculate the Final Chloride Concentration: Take the square root of the result. To perform this mentally, it is helpful to rewrite $1.6 \times 10^{-3}$ as $16 \times 10^{-4}$:
$$[Cl^-] = \sqrt{16 \times 10^{-4}} = \mathbf{4.0 \times 10^{-2}\text{ M}}$$
7. Conclusion: The lowest concentration of $Cl^-$ needed to begin precipitation is $4.0 \times 10^{-2}\text{ M}$, which matches option E.

WHY_OTHERS_WRONG:

• A) $1.6 \times 10^{-7}\text{ M}$: This value is far too small to reach the $K_{sp}$ threshold. A student might arrive here through a significant miscalculation of the exponents.
• B) $4.0 \times 10^{-4}\text{ M}$: This is likely the result of a decimal error when taking the square root of $1.6 \times 10^{-3}$.
• C) $1.6 \times 10^{-3}\text{ M}$: This value is equal to $[Cl^-]^2$. A student would choose this if they successfully completed the division but forgot to take the final square root required by the stoichiometry of the $PbCl_2$ formula.
• D) $2.6 \times 10^{-3}\text{ M}$: This value does not result from any standard application of the given variables and is likely a distractor.