CaF2(s) ⇌ Ca2+(aq) + 2 F−(aq) Ksp = 4.0 × 10−11 The concentration of F−(aq) in drinking water that i — Equilibrium Chemistry Question

STEPS:

1. Identify the solubility equilibrium equation and the $K_{sp}$ expression:
The dissolution of solid calcium fluoride ($\text{CaF}_2$) in water is represented by the equilibrium equation:
$$\text{CaF}_2(s) \rightleftharpoons \text{Ca}^{2+}(aq) + 2\ \text{F}^-(aq) \quad$$
The solubility product constant expression ($K_{sp}$) for this heterogeneous system excludes the solid reactant and is written as:
$$K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2 \quad$$

2. Understand the threshold for precipitation:
The maximum concentration of $\text{Ca}^{2+}$ that can coexist with a specified concentration of $\text{F}^-$ without causing precipitation of solid $\text{CaF}_2$ is reached when the system is exactly at saturation equilibrium. At this point, the ion product $Q_{sp}$ equals the solubility product constant $K_{sp}$. If the concentration of calcium ions exceeds this equilibrium limit, the excess ions will react and precipitate out of solution, pulling fluoride ions along with them and dropping the concentration of $\text{F}^-$ below the ideal level.

3. Substitute the given values into the $K_{sp}$ expression:
The problem provides the following values:
* $K_{sp} = 4.0 \times 10^{-11}$
* $[\text{F}^-] = 4.0 \times 10^{-5}\text{ M}$

Substitute these numbers into the $K_{sp}$ equation to solve for $[\text{Ca}^{2+}]$:
$$4.0 \times 10^{-11} = [\text{Ca}^{2+}](4.0 \times 10^{-5})^2 \quad$$

4. Calculate the squared fluoride concentration and isolate $[\text{Ca}^{2+}]$:
First, square the fluoride ion concentration:
$$(4.0 \times 10^{-5})^2 = (4.0)^2 \times (10^{-5})^2 = 16 \times 10^{-10}\text{ M}^2 = 1.6 \times 10^{-9}\text{ M}^2 \quad$$

Next, isolate and solve for the concentration of calcium ions:
$$[\text{Ca}^{2+}] = \frac{K_{sp}}{[\text{F}^-]^2} = \frac{4.0 \times 10^{-11}}{16 \times 10^{-10}} \quad$$
$$[\text{Ca}^{2+}] = \frac{4.0}{1.6} \times 10^{-11 - (-9)} \quad$$
$$[\text{Ca}^{2+}] = 2.5 \times 10^{-2}\text{ M} = \mathbf{0.025\text{ M}} \quad$$
This identifies Option B as the correct choice.

*

WHY_OTHERS_WRONG:

* Option A is incorrect (0.25 M): This value is ten times larger than the correct threshold, resulting from an exponent math error when dividing the powers of ten (e.g., calculating $\frac{10^{-11}}{10^{-10}}$ as $10^{-1}$ instead of $10^{-2}$).
* Option C is incorrect ($1.6 \times 10^{-6}\text{ M}$): This incorrect concentration arises if a student forgets to square the fluoride ion concentration in the solubility product expression, mistakenly solving $K_{sp} = [\text{Ca}^{2+}][\text{F}^-]$:
$$[\text{Ca}^{2+}] = \frac{4.0 \times 10^{-11}}{4.0 \times 10^{-5}} = 1.0 \times 10^{-6}\text{ M}$$
A student would then select this distractor by confusing $1.0 \times 10^{-6}\text{ M}$ with $1.6 \times 10^{-6}\text{ M}$.
* Option D is incorrect ($1.6 \times 10^{-15}\text{ M}$): This value is obtained if a student mistakenly multiplies the $K_{sp}$ value by the fluoride concentration (or its square) rather than dividing:
$$[\text{Ca}^{2+}] = K_{sp} \times [\text{F}^-] = (4.0 \times 10^{-11}) \times (4.0 \times 10^{-5}) = 1.6 \times 10^{-15}\text{ M}$$
This mathematically violates the basic algebraic rearrangement of the solubility product formula.