CaF2(s) ⇄ Ca2+(aq) + 2 F−(aq) ΔH > 0 28. Dissolution of the slightly soluble salt CaF2 is shown by t — Equilibrium Chemistry Question

STEPS:

1. Identify the solubility equilibrium equation and the constant expression:
The dissolution of the slightly soluble salt calcium fluoride ($\text{CaF}_2$) is represented by the equilibrium equation:
$$\text{CaF}_2(s) \rightleftharpoons \text{Ca}^{2+}(aq) + 2\ \text{F}^-(aq) \quad \Delta H > 0 \quad$$
The solubility product constant expression for this reaction is:
$$K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2 \quad$$
Because solid $\text{CaF}_2(s)$ remains in contact with the solution, the system is at dynamic equilibrium.
2. Understand the common-ion effect:
When sodium fluoride ($\text{NaF}$), a highly soluble salt, is added to the saturated solution, it dissociates completely to release sodium ions ($\text{Na}^+$) and fluoride ions ($\text{F}^-$) into the mixture.
3. Apply Le Chatelier's principle to product addition:
* Because fluoride ($\text{F}^-$) is already present in the solution from the dissolved $\text{CaF}_2$, it acts as a common ion.
* Adding $\text{NaF}(s)$ dramatically increases the concentration of fluoride ions ($[\text{F}^-]$) in the system.
* To relieve this stress and consume the added product, Le Chatelier's principle dictates that the equilibrium position must shift to the left (towards the reactants).
4. Determine the effect of the shift on the calcium ion concentration:
* As the system shifts left, dissolved $\text{Ca}^{2+}$ ions are forced to react with the excess $\text{F}^-$ ions to precipitate out of the solution as solid $\text{CaF}_2(s)$.
* Consequently, the equilibrium concentration of calcium ions ($[\text{Ca}^{2+}]$) decreases. This confirms that Option D is the correct answer.

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WHY_OTHERS_WRONG:

• Option A is incorrect: While evaporating some water from a saturated solution forces more $\text{CaF}_2(s)$ to precipitate out of the solution, the remaining mixture still remains saturated. Because the temperature is kept constant, the value of the equilibrium constant $K_{sp}$ is unchanged, meaning the equilibrium concentrations of the dissolved ions—including $[\text{Ca}^{2+}]$—remain exactly the same.
• Option B is incorrect: Adding a strong acid like $\text{HNO}_3(aq)$ introduces $\text{H}^+(aq)$ ions into the beaker. Because fluoride ($\text{F}^-$) is the conjugate base of a weak acid ($\text{HF}$), the $\text{H}^+$ ions will react with the dissolved $\text{F}^-$ ions to form the weak electrolyte $\text{HF}(aq)$. This protonation reaction removes free $\text{F}^-$ product ions from the solubility equilibrium, causing the reaction to shift to the right to replace them, which increases $[\text{Ca}^{2+}]$, rather than decreasing it.
• Option C is incorrect: Sodium nitrate ($\text{NaNO}_3$) does not share any common ions with $\text{CaF}_2$. Although the addition of the solution initially dilutes the mixture, more solid $\text{CaF}_2(s)$ will dissolve from the bottom to replace the diluted ions until dynamic saturation equilibrium is re-established. Thus, there is no net decrease in the equilibrium concentration of $\text{Ca}^{2+}$.