Problem 72 - A Complete Resource Book in Chemistry for JEE Main

Question

The precipitate of \(\mathrm{CaF}_{2}\left(\mathrm{Ksp}=1.7 \times 10^{-10}\right)\) is obtained when equal volumes of the following are mixed (a) \(10^{-2} \mathrm{M} \mathrm{Ca}^{2+}+10^{-3} \mathrm{MF}^{-}\) (b) \(10^{-3} \mathrm{M} \mathrm{Ca}^{2+}+10^{-5} \mathrm{MF}^{-}\) (c) \(10^{-4} \mathrm{M} \mathrm{Ca}^{2+}+10^{-4} \mathrm{MF}^{-}\) (d) \(10^{-5} \mathrm{M} \mathrm{Ca}^{2+}+10^{-3} \mathrm{MF}^{-}\)

Step-by-Step Solution

Answer

The precipitate forms in option (a).

Step 1: Understanding the Concept

To determine if a precipitate forms, compare the reaction quotient (Q) to the solubility product constant (Ksp). If Q > Ksp, a precipitate will form. Here, the compound is \( \text{CaF}_2 \) and Ksp is given as \( 1.7 \times 10^{-10} \).

Step 2: Writing the Solubility Product Expression

The dissolution of \( \text{CaF}_2 \) can be represented as: \[ \text{CaF}_2 (s) \rightleftharpoons \text{Ca}^{2+} (aq) + 2\text{F}^- (aq) \]. The Ksp expression is \( K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2 \).

Step 3: Calculate the Reaction Quotient \( Q \) for (a)

For options a: equal volumes mean concentrations are halved. \([\text{Ca}^{2+}] = 5 \times 10^{-3} \text{ M}\) and \([\text{F}^-] = 5 \times 10^{-4} \text{ M}\). Compute \( Q \): \[ Q = (5 \times 10^{-3})(5 \times 10^{-4})^2 = 1.25 \times 10^{-9} \].

Step 4: Calculate the Reaction Quotient \( Q \) for (b)

For option b: \([\text{Ca}^{2+}] = 5 \times 10^{-4} \text{ M}\) and \([\text{F}^-] = 5 \times 10^{-6} \text{ M}\). Compute \( Q \): \[ Q = (5 \times 10^{-4})(5 \times 10^{-6})^2 = 1.25 \times 10^{-14} \].

Step 5: Calculate the Reaction Quotient \( Q \) for (c)

For option c: \([\text{Ca}^{2+}] = 5 \times 10^{-5} \text{ M}\) and \([\text{F}^-] = 5 \times 10^{-5} \text{ M}\). Compute \( Q \): \[ Q = (5 \times 10^{-5})(5 \times 10^{-5})^2 = 1.25 \times 10^{-13} \].

Step 6: Calculate the Reaction Quotient \( Q \) for (d)

For option d: \([\text{Ca}^{2+}] = 5 \times 10^{-6} \text{ M}\) and \([\text{F}^-] = 5 \times 10^{-4} \text{ M}\). Compute \( Q \): \[ Q = (5 \times 10^{-6})(5 \times 10^{-4})^2 = 1.25 \times 10^{-13} \].

Step 7: Compare \( Q \) with \( K_{sp} \)

\( Q \) in (a) is \(1.25 \times 10^{-9}\), which is greater than \( K_{sp} = 1.7 \times 10^{-10} \). Thus, a precipitate forms. In all other options, \( Q < K_{sp} \), so no precipitate forms.

Key Concepts

Solubility Product (Ksp)

The solubility product, often represented as \(K_{sp}\), is an important concept in understanding chemical equilibrium, particularly in the context of sparingly soluble salts. It is the product of the concentrations of the ions of a compound in a saturated solution raised to the power of their respective stoichiometric coefficients.
For the compound calcium fluoride \(\text{CaF}_2\), the dissolution can be described by the equation: \[\text{CaF}_2 (s) \rightleftharpoons \text{Ca}^{2+} (aq) + 2\text{F}^- (aq)\]The \(K_{sp}\) expression is written as: \[K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2\] In this expression, \(K_{sp}\) provides a measure of the extent to which \(\text{CaF}_2\) dissociates into its ions in solution.
Since \(K_{sp}\) is a constant at a given temperature, it helps in predicting whether a precipitate will form under specific conditions when various solutions are combined. By comparing \(K_{sp}\) with a calculated value from a mixture, decisions can be made about reactions and solubility.

Reaction Quotient (Q)

The reaction quotient, symbolized by \(Q\), is a measure that helps in predicting the direction in which a reaction needs to shift to reach equilibrium. It is calculated similarly to the equilibrium constant, using the concentrations of the reactants and products.
In scenarios involving precipitation, \(Q\) is compared to the solubility product constant, \(K_{sp}\). The expression for \(Q\) when mixing two solutions is: \[Q = [\text{Ca}^{2+}][\text{F}^-]^2\] where each concentration is adjusted based on the volume change upon mixing.

• If \(Q > K_{sp}\), the solution is supersaturated, and a precipitate will form because the product of the ion concentrations exceeds the solubility product.
• If \(Q = K_{sp}\), the solution is at equilibrium and is exactly saturated.
• If \(Q < K_{sp}\), no precipitation occurs as the solution is unsaturated and more solute can still dissolve.

Hence, \(Q\) is a practical tool for determining the immediate state and future progress of a reaction.

Precipitation Reaction

A precipitation reaction occurs when two soluble salts in aqueous solution react to form an insoluble solid, known as a precipitate. This typically happens when the product of the ionic concentrations in a solution exceeds their \(K_{sp}\).
Such reactions are crucial in various applications, such as in analytical chemistry for detecting the presence of specific ions through selective precipitation. For \(\text{CaF}_2\), as given by the exercise, precipitation occurs when the combined concentrations of calcium and fluoride ions exceed the solubility threshold represented by \(K_{sp}=1.7 \times 10^{-10}\).
Identifying whether a precipitation reaction will occur involves calculating the reaction quotient \(Q\) from the mixed ionic concentrations and comparing it to the solubility product. Only when \(Q\) surpasses \(K_{sp}\) does precipitation happen, signifying a shift toward solid formation and thus signaling a change from ionic dispersion to a solid form.

Calcium Fluoride Precipitation

When solutions containing calcium ions \((\text{Ca}^{2+})\) and fluoride ions \((\text{F}^-)\) are mixed, precipitate formation depends on the concentrations of these ions compared to \(K_{sp}\) for \(\text{CaF}_2\).

• A typical scenario occurs when the combined concentration achieves a supersaturated state, where \(Q\) surpasses \(1.7 \times 10^{-10}\), triggering precipitation.
• In the exercise, solution (a) presents a situation where initial higher concentrations result in \(Q > K_{sp}\) with the calculation showing \(1.25 \times 10^{-9}\), leading to precipitation.
• Other mixtures produced \(Q\) values smaller than \(K_{sp}\), thus no precipitation occurred, demonstrating the specificity needed in ion concentration to form \(\text{CaF}_2\) as a solid.

Understanding this process is essential for applications in water treatment where controlled precipitation removes unwanted ions, ensuring safe drinking water and effective waste management.