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 case (a) where \( Q > K_{sp} \).

Step 1: Understand the Reaction and Ksp

The reaction for the dissolution of \( \mathrm{CaF}_2 \) in water is: \[ \mathrm{CaF}_2(s) \rightleftharpoons \mathrm{Ca}^{2+}(aq) + 2\mathrm{F}^{-}(aq) \] with a solubility product (\( K_{sp} \)) of \( 1.7 \times 10^{-10} \). This means \( [\mathrm{Ca}^{2+}] \times [\mathrm{F}^{-}]^2 \) must be less than \( K_{sp} \) for no precipitate to form.

Step 2: Calculate Ion Concentrations After Mixing

When equal volumes of two solutions are mixed, the concentration of each ion is halved. For example, if you mix 1 L of \(10^{-2} \mathrm{M} \) \( \mathrm{Ca}^{2+} \) with 1 L of \(10^{-3} \mathrm{M} \) \( \mathrm{F}^{-} \), the new concentration is \( \left[\mathrm{Ca}^{2+}\right] = 5 \times 10^{-3} \mathrm{M} \) and \( \left[\mathrm{F}^{-}\right] = 5 \times 10^{-4} \mathrm{M} \).

Step 3: Apply to Each Condition

(a) \( \left[\mathrm{Ca}^{2+}\right] = 5 \times 10^{-3} \), \( \left[\mathrm{F}^{-}\right] = 5 \times 10^{-4} \). \( Q = (5 \times 10^{-3})(5 \times 10^{-4})^2 = 1.25 \times 10^{-9} \).(b) \( \left[\mathrm{Ca}^{2+}\right] = 5 \times 10^{-4} \), \( \left[\mathrm{F}^{-}\right] = 5 \times 10^{-6} \). \( Q = (5 \times 10^{-4})(5 \times 10^{-6})^2 = 1.25 \times 10^{-13} \).(c) \( \left[\mathrm{Ca}^{2+}\right] = 5 \times 10^{-5} \), \( \left[\mathrm{F}^{-}\right] = 5 \times 10^{-5} \). \( Q = (5 \times 10^{-5})(5 \times 10^{-5})^2 = 1.25 \times 10^{-13} \).(d) \( \left[\mathrm{Ca}^{2+}\right] = 5 \times 10^{-6} \), \( \left[\mathrm{F}^{-}\right] = 5 \times 10^{-4} \). \( Q = (5 \times 10^{-6})(5 \times 10^{-4})^2 = 1.25 \times 10^{-13} \).

Step 4: Compare with Ksp

Only case (a) has \( Q > K_{sp} \) because \( Q = 1.25 \times 10^{-9} > 1.7 \times 10^{-10} \), meaning \( \mathrm{CaF}_2 \) will precipitate.

Key Concepts

Precipitation Reactions

Precipitation reactions involve the formation of a solid from a solution during a chemical reaction. When two solutions containing ions are mixed, a precipitate may form if the product of the ion concentrations exceeds the solubility product constant \( K_{sp} \). This scenario is crucial in understanding solubility equilibrium in chemistry.
Let's take the example of \( \mathrm{CaF}_2 \), which dissociates into calcium ions \( \mathrm{Ca}^{2+} \) and fluoride ions \( \mathrm{F}^- \) in water:
\[\mathrm{CaF}_2(s) \rightleftharpoons \mathrm{Ca}^{2+}(aq) + 2\mathrm{F}^-(aq)\]A precipitate of \( \mathrm{CaF}_2 \) will form if the product of \( [\mathrm{Ca}^{2+}] \) and \( [\mathrm{F}^-]^2 \) exceeds its \( K_{sp} \) of \( 1.7 \times 10^{-10} \). In simpler terms, if the concentration of ions produced is "too high," the solution cannot hold all the ions dissolved, leading to the excess forming a solid.

• A reaction reaches a point where some ions come out of the solution.
• A solid crash out occurs if \( Q > K_{sp} \), where \( Q \) is the reaction quotient.
• Understanding \( K_{sp} \) is crucial for predicting when and if precipitation will occur.

Ion Concentration Calculation

Calculating ion concentrations after mixing solutions is vital in determining whether a precipitate will form. When two solutions are combined in equal volumes, the concentrations of each ion are reduced to half of their original values.
For instance, if you start with a solution containing \( 10^{-2} \mathrm{M} \) calcium ions and another containing \( 10^{-3} \mathrm{M} \) fluoride ions, the new concentrations after mixing will be:

• \( \left[\mathrm{Ca}^{2+}\right] = 5 \times 10^{-3} \mathrm{M} \)
• \( \left[\mathrm{F}^{-}\right] = 5 \times 10^{-4} \mathrm{M} \)

This halving occurs because you are effectively diluting each solution by integrating them together.
Next, use these new concentrations to calculate \( Q \), the reaction quotient, by using the expression:\[Q = [\mathrm{Ca}^{2+}] \times [\mathrm{F}^-]^2\]By comparing \( Q \) to \( K_{sp} \), you can predict if a precipitate will form. If \( Q > K_{sp} \), precipitation is expected.

Chemical Equilibrium

Chemical equilibrium refers to a state where a reaction and its reverse occur at the same rate, leading to no net change in the concentration of reactants and products. In the context of precipitation reactions, chemical equilibrium helps in understanding how saturated solutions behave.
At equilibrium in the dissolution of \( \mathrm{CaF}_2 \):\[\mathrm{CaF}_2(s) \rightleftharpoons \mathrm{Ca}^{2+}(aq) + 2\mathrm{F}^{-}(aq)\]The forward reaction, where solid dissolves into ions, and the reverse reaction, where ions come together to form a solid, are balanced. The specific ratio of dissolved ions that results in no visible change is described by the solubility product \( K_{sp} \).

• Equilibrium ensures the system is stable.
• The concentrations of ions remain constant over time unless disturbed.
• The equilibrium state is where \( Q = K_{sp} \) for saturated solutions.

Understanding how chemical equilibrium works in these reactions is key to predicting ion behavior and the potential for precipitate formation.