[VISUAL] 29. Equimolar samples of Pb(OH)2(s), PbI2(s), and PbF2(s) are placed in three separate beak — Equilibrium Chemistry Question

STEPS:

1. Understand the physical system and equilibrium equations:
Each of the three beakers contains water and a solid lead salt in contact with its saturated solution. These represent heterogeneous solubility equilibria, which can be written as:
* Beaker 1: $\text{Pb(OH)}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\ \text{OH}^-(aq)$
* Beaker 2: $\text{PbI}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\ \text{I}^-(aq)$
* Beaker 3: $\text{PbF}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\ \text{F}^-(aq)$

2. Formulate the solubility product constant ($K_{sp}$) expressions:
* All three salts share the exact same stoichiometry, having a formula of the type $\text{XY}_2$ where 1 mole of solid dissolves to yield 1 mole of lead cations ($\text{Pb}^{2+}$) and 2 moles of anions.
* Because they have identical stoichiometry, they share the same mathematical form for their $K_{sp}$ expressions:
$$K_{sp} = [\text{Pb}^{2+}][\text{Anion}^-]^2$$
* If we define the molar solubility of each salt as $s$, then at equilibrium:
$$[\text{Pb}^{2+}] = s$$
$$[\text{Anion}^-] = 2s$$
* Substituting these into the equilibrium expression gives:
$$K_{sp} = (s)(2s)^2 = 4s^3 \implies s = \sqrt{\frac{K_{sp}}{4}}$$

3. Compare molar solubilities ($s$) using the given $K_{sp}$ values:
* Because all three salts share the same stoichiometry, their molar solubilities (and consequently, their equilibrium $[\text{Pb}^{2+}]$ concentrations) are directly proportional to their $K_{sp}$ values. A smaller $K_{sp}$ value mathematically guarantees a smaller molar solubility ($s$) and a lower $[\text{Pb}^{2+}]$ concentration.
* Looking at the provided data table:
* $K_{sp}$ of $\text{Pb(OH)}_2 = \mathbf{1.2 \times 10^{-15}}$
* $K_{sp}$ of $\text{PbI}_2 = 1.4 \times 10^{-8}$
* $K_{sp}$ of $\text{PbF}_2 = 4.0 \times 10^{-8}$
* Comparing these values, the $K_{sp}$ of $\text{Pb(OH)}_2$ is seven orders of magnitude smaller than the other two constants ($1.2 \times 10^{-15} \ll 1.4 \times 10^{-8} < 4.0 \times 10^{-8}$).

4. Conclude:
* Because $\text{Pb(OH)}_2$ has by far the smallest solubility product constant, it dissolves to the least extent.
* It will therefore yield the lowest equilibrium concentration of lead ions ($[\text{Pb}^{2+}]$) among the three systems, making Option A the correct answer.

*

WHY_OTHERS_WRONG:

• Option B is incorrect ($\text{PbI}_2(s)$ only): While $\text{PbI}_2$ is a slightly soluble salt, its $K_{sp}$ of $1.4 \times 10^{-8}$ is much larger than that of $\text{Pb(OH)}_2$. This means $\text{PbI}_2$ will dissolve to a significantly greater extent, resulting in a much higher $[\text{Pb}^{2+}]$ concentration at equilibrium (approximately $1.5 \times 10^{-3}\text{ M}$, compared to only $6.7 \times 10^{-6}\text{ M}$ for $\text{Pb(OH)}_2$).
• Option C is incorrect ($\text{PbF}_2(s)$ only): $\text{PbF}_2$ has the largest $K_{sp}$ value ($4.0 \times 10^{-8}$) among the three listed compounds. It will dissolve to the greatest extent and produce the *highest* concentration of $[\text{Pb}^{2+}]$ in this group, rather than the lowest.
• Option D is incorrect: This option falsely claims that both $\text{PbI}_2(s)$ and $\text{PbF}_2(s)$ yield the same, lowest concentration. They have different $K_{sp}$ values and therefore will not have the same concentration at equilibrium. Additionally, both of their dissolved solutions contain a far higher concentration of $[\text{Pb}^{2+}]$ than a saturated solution of $\text{Pb(OH)}_2$.