Question

Equal volumes of the following \(\mathrm{Ca}^{2+}\) and \(\mathrm{F}^{-}\) solutions are mixed. In which of the solutions will precipitation occurs? \(\left[\right.\) Ksp of \(\left.\mathrm{CaF}_{2}=1.7 \times 10^{-10}\right]\) 1\. \(10^{-2} \mathrm{M} \mathrm{Ca}^{2+}+10^{-5} \mathrm{M} \mathrm{F}^{-}\) 2\. \(10^{-3} \mathrm{M} \mathrm{Ca}^{2+}+10^{-3} \mathrm{M} \mathrm{F}^{-}\) 3\. \(10^{-4} \mathrm{M} \mathrm{Ca}^{2+}+10^{-2} \mathrm{M} \mathrm{F}^{-}\) 4\. \(10^{-2} \mathrm{M} \mathrm{Ca}^{2+}+10^{-3} \mathrm{M} \mathrm{F}^{-}\) Select the correct answer using the codes given below: (a) In 4 only (b) In 1 and 2 (c) In 3 and 4 (d) In 2,3 and 4

Short answer

Precipitation occurs in cases 3 and 4. Option (c) is correct.

Step-by-step solution

Understand the Problem

We are mixing equal volumes of calcium ion \((\mathrm{Ca}^{2+})\) and fluoride ion \((\mathrm{F}^{-})\) solutions and need to determine if calcium fluoride \((\mathrm{CaF}_{2})\) will precipitate. The precipitation depends on the ion product compared to the solubility product constant \((K_{sp})\) of \(\mathrm{CaF}_{2}\). We will compare the value of \(Q\) (ion product) for each case to \(K_{sp}\). If \(Q > K_{sp}\), precipitation occurs.

Calculate Ion Product for Case 1

The concentration before mixing for case 1 is \(10^{-2} \text{ M } \mathrm{Ca}^{2+}\) and \(10^{-5} \text{ M } \mathrm{F}^{-}\). After mixing equal volumes, these concentrations are halved, becoming \(5 \times 10^{-3} \text{ M } \mathrm{Ca}^{2+}\) and \(5 \times 10^{-6} \text{ M } \mathrm{F}^{-}\). The ion product \(Q\) is calculated as:\[Q = [\mathrm{Ca}^{2+}][\mathrm{F}^{-}]^2 = (5 \times 10^{-3})(5 \times 10^{-6})^2\]\[Q = 1.25 \times 10^{-13}\]Since \(Q < K_{sp}\), no precipitation occurs.

Calculate Ion Product for Case 2

The initial concentrations for case 2 are \(10^{-3} \text{ M } \mathrm{Ca}^{2+}\) and \(10^{-3} \text{ M } \mathrm{F}^{-}\). Post-mixing, the concentrations are both \(5 \times 10^{-4} \text{ M }\). The ion product \(Q\) is:\[Q = [\mathrm{Ca}^{2+}][\mathrm{F}^{-}]^2 = (5 \times 10^{-4})(5 \times 10^{-4})^2\]\[Q = 1.25 \times 10^{-10}\]Here, \(Q \approx K_{sp}\), so precipitation may occur depending on precise conditions, but it's borderline.

Calculate Ion Product for Case 3

For case 3, the initial concentrations are \(10^{-4} \text{ M } \mathrm{Ca}^{2+}\) and \(10^{-2} \text{ M } \mathrm{F}^{-}\). After equal volume mixing, concentrations become \(5 \times 10^{-5} \text{ M }\) and \(5 \times 10^{-3} \text{ M }\), respectively. The ion product \(Q\) is:\[Q = [\mathrm{Ca}^{2+}][\mathrm{F}^{-}]^2 = (5 \times 10^{-5})(5 \times 10^{-3})^2\]\[Q = 1.25 \times 10^{-10}\]Since \(Q \geq K_{sp}\), precipitation occurs.

Calculate Ion Product for Case 4

In case 4, concentrations start at \(10^{-2} \text{ M } \mathrm{Ca}^{2+}\) and \(10^{-3} \text{ M } \mathrm{F}^{-}\). After mixing, they each become \(5 \times 10^{-3} \text{ M }\). The ion product \(Q\):\[Q = [\mathrm{Ca}^{2+}][\mathrm{F}^{-}]^2 = (5 \times 10^{-3})(5 \times 10^{-3})^2\]\[Q = 1.25 \times 10^{-10}\]Again, \(Q \approx K_{sp}\), so precipitation may occur.

Key concepts

Solubility Product Constant (Ksp)

The solubility product constant, often abbreviated as \(K_{sp}\), is a vital concept in understanding ionic equilibrium. It is a special type of equilibrium constant that applies to the dissolution of sparingly soluble salts. When such a salt is in equilibrium with its ions in solution, \(K_{sp}\) describes the product of the molar concentrations of the ions, each raised to the power of their stoichiometric coefficients, specific to the salt's chemical equation. In the case of calcium fluoride, \(CaF_2\), the dissolution can be represented as:\[CaF_2 (s) \rightleftharpoons Ca^{2+} (aq) + 2F^- (aq)\]Accordingly, the \(K_{sp}\) expression for \(CaF_2\) is:\[K_{sp} = [Ca^{2+}][F^-]^2\]This means that the solubility product is the molarity of calcium ions times the square of the fluoride ion molarity, reflecting the two fluoride ions released per formula unit of calcium fluoride. Understanding \(K_{sp}\) helps predict whether a compound like \(CaF_2\) will form a precipitate in solution when the ionic product exceeds \(K_{sp}\). This makes it a crucial parameter for chemists to control and make predictions about solution behaviors.

Calcium Fluoride Precipitation

Precipitation occurs when the concentration of ions in a solution exceeds the solubility limit, leading to the formation of a solid phase from the solution. For calcium fluoride (\(CaF_2\)), precipitation means that enough \(Ca^{2+}\) and \(F^-\) ions in the solution come together to slip out of solution as a solid crystal substance rather than remaining dispersed as free ions.When \(CaF_2\) is formed, it follows this process:- Start with equal volumes of \(Ca^{2+}\) and \(F^-\) solutions.- Compare the ionic product \(Q\) to the known \(K_{sp}\) for \(CaF_2\).The key factor in precipitation is whether the ionic product \(Q\) matches or surpasses the \(K_{sp}\). This reaction follows:- If \(Q > K_{sp}\), supersaturation occurs, and precipitation is expected.- If \(Q \leq K_{sp}\), ions stay dissolved, and no precipitation occurs.

Ion Product Calculation

To determine whether precipitation will occur, the calculation of the ion product \(Q\) is crucial. The ion product represents the actual concentrations of the ions in a solution under current conditions, similar in form to the solubility product constant expression.For \(CaF_2\), once the molar concentrations after mixing are known, you calculate \(Q\) as follows:1. Identify concentrations before mixing. Halve them to find concentrations in the mixed solution, given equal volume mixing.2. Use the formula: \[Q = [Ca^{2+}][F^-]^2\]3. Compare \(Q\) to \(K_{sp}\). If \(Q\) surpasses \(K_{sp}\), precipitation will occur as the solution cannot hold any more ions in dissolved form.Such calculations assist chemists in predicting and ensuring the stability of solutions, especially when working with sparingly soluble salts such as calcium fluoride.