Problem 115 - A Complete Resource Book in Chemistry for JEE Main

Question

Equal volumes of the following \(\mathrm{Ca}^{2+}\) and \(\mathrm{F}^{-}\)solutions are mixed. In which of the solutions will precipitation occurs? \(\left[\mathrm{Ksp}\right.\) 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

Step-by-Step Solution

Answer

In solution 4 only (option a).

Step 1: Calculate Initial Concentrations After Mixing

When equal volumes of two solutions are mixed, their concentrations are halved. For example, if we mix equal volumes of \(10^{-2} \text{ M} \ Ca^{2+}\) and \(10^{-5} \text{ M} \ F^{-}\), their concentrations become \(5 \times 10^{-3} \text{ M} \ Ca^{2+}\) and \(5 \times 10^{-6} \text{ M} \ F^{-}\) respectively after mixing.

Step 2: Determine Ion Product (Q) for Each Mixture

For the dissolution of calcium fluoride \(CaF_2\), the relation is \([Ca^{2+}][F^{-}]^2\). Calculate \(Q\) for each mixture:1. \(Q_1 = (5 \times 10^{-3})(5 \times 10^{-6})^2 = 1.25 \times 10^{-13}\)2. \(Q_2 = (5 \times 10^{-4})(5 \times 10^{-4})^2 = 1.25 \times 10^{-10}\)3. \(Q_3 = (5 \times 10^{-5})(5 \times 10^{-3})^2 = 1.25 \times 10^{-10}\)4. \(Q_4 = (5 \times 10^{-3})(5 \times 10^{-4})^2 = 1.25 \times 10^{-9}\)

Step 3: Compare Ion Product (Q) with \(K_{sp}\) of \(CaF_2\)

The \(K_{sp}\) of \(CaF_2\) is \(1.7 \times 10^{-10}\). Compare \(Q\) with \(K_{sp}\) to determine if precipitation occurs:1. \(Q_1 = 1.25 \times 10^{-13} < 1.7 \times 10^{-10}\) (No precipitation)2. \(Q_2 = 1.25 \times 10^{-10} < 1.7 \times 10^{-10}\) (No precipitation)3. \(Q_3 = 1.25 \times 10^{-10} < 1.7 \times 10^{-10}\) (No precipitation)4. \(Q_4 = 1.25 \times 10^{-9} > 1.7 \times 10^{-10}\) (Precipitation occurs)

Key Concepts

Chemical Equilibrium

Chemical equilibrium is a fundamental concept in chemistry where the rates of the forward and reverse reactions are equal, resulting in a stable concentration of reactants and products. When solutions are mixed, as seen in the original exercise with calcium fluoride \(CaF_2\), the system may reach equilibrium where the solid \(CaF_2\) will either remain dissolved or start to precipitate, depending on the ion concentrations. The point at which this balance occurs is dictated by the solubility product constant, \(K_{sp}\).

• In our scenario, the \(Ca^{2+}\) and \(F^{-}\) ions come together to check if they can sustain an equilibrium with solid \(CaF_2\).
• If the solid forms, it means equilibrium hasn't been established purely in the dissolved state.

As reactions proceed to equilibrium, no net change in concentration is observed, illustrating the dynamic but stable nature of equilibrium.

Precipitation Reactions

Precipitation reactions occur when the concentration of ions exceeds their solubility, causing a solid to form from a solution. In the exercise, equal volumes of various concentrations of \(Ca^{2+}\) and \(F^{-}\) solutions are mixed to potentially form a precipitate of \(CaF_2\).

• The conditions under which precipitation happens depend on comparing the ion product, denoted as \(Q\), against the known solubility product \(K_{sp}\).
• When \(Q\) exceeds \(K_{sp}\), it indicates that the solution has exceeded its capacity to hold ions in a dissolved state, thus a precipitate will form.

In our problem, precipitation successfully occurs when the ion product exceeds the \(K_{sp}\) for calcium fluoride, as evidenced in the mixture with the highest concentration of ions.

Ion Product

The ion product, often represented by \(Q\), is similar to the expression of the equilibrium constant for solutions that aren't at equilibrium.

• The ion product formula for calcium fluoride is given by \([Ca^{2+}][F^{-}]^2\).
• By calculating \(Q\) for each ion combination, we assess whether the current concentrations allow for continued dissolution or if a precipitate forms.

In the example, the \(Q\) values allowed us to compare different mixtures of \(Ca^{2+}\) and \(F^{-}\) to the solubility product \(K_{sp} = 1.7 \times 10^{-10}\).In mixture 4, where \(Q = 1.25 \times 10^{-9}\), the ion product surpasses the solubility product, leading to precipitation. Understanding \(Q\) is vital for predicting solution behavior before equilibrium is achieved.