Problem 65 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The solubility product \(\left(K_{s p}\right)\) of \(\mathrm{Ca}(\mathrm{OH})_{2}\) at \(25^{\circ} \mathrm{C}\) is \(4.42 \times 10^{-5}\). A 500 \(\mathrm{mL}\). of saturated solution of \(\mathrm{Ca}(\mathrm{OH})_{2}\) is mixed with equal volume of \(0.4 \mathrm{M} \mathrm{NaOH}\). How much \(\mathrm{Ca}(\mathrm{OH})_{2}\) in milligrams is precipitated?

Step-by-Step Solution

Answer

788 mg of Ca(OH)2 is precipitated.

Step 1: Determine Initial Concentrations

First, we need to find the concentration of \( \text{Ca(OH)}_2 \) in the saturated solution before mixing. In a saturated solution, \( \text{Ca(OH)}_2 \) is in equilibrium with its ions, so: \[ \text{Ca(OH)}_2 (s) \rightleftharpoons \text{Ca}^{2+} (aq) + 2 \text{OH}^- (aq) \]The solubility product expression is:\[ K_{sp} = [\text{Ca}^{2+}][\text{OH}^-]^2 \]Let \( s \) be the solubility of \( \text{Ca(OH)}_2 \) in mol/L. Therefore, \([\text{Ca}^{2+}] = s\) and \([\text{OH}^-] = 2s\). Then:\[ K_{sp} = s(2s)^2 = 4s^3 \]Given \( K_{sp} = 4.42 \times 10^{-5} \), we solve for \( s \):\[ 4s^3 = 4.42 \times 10^{-5} \]\[ s^3 = 1.105 \times 10^{-5} \]\[ s = \sqrt[3]{1.105 \times 10^{-5}} \approx 0.022 \text{ M} \]

Step 2: Calculate Hydroxide Ion Contribution

Before adding \( \text{NaOH} \), the \([\text{OH}^-]\) concentration in the \( \text{Ca(OH)}_2 \) solution is \( 2s = 2 \times 0.022 = 0.044 \text{ M}\).Now, calculate the \([\text{OH}^-]\) concentration contributed by the \( 0.4 \text{ M NaOH} \). Since the solution is mixed in equal volumes (500 mL each), the concentration of \( \text{NaOH} \) after mixing becomes:\[ \frac{0.4 \text{ M}}{2} = 0.2 \text{ M} \]

Step 3: Combine OH- Contributions and Set Equilibrium

After mixing, the total \([\text{OH}^-]\) is the sum of the contributions from \( \text{Ca(OH)}_2 \) and \( \text{NaOH} \):\[ 0.044 \text{ M} + 0.2 \text{ M} = 0.244 \text{ M} \]For precipitation to occur, the \( \text{Q} \), the reaction quotient, must be greater than \( K_{sp} \):\[ Q = s \times (0.244)^2 \approx 0.022 \times 0.059536 \approx 0.00131 \]Since \( Q > K_{sp} \), precipitation occurs.

Step 4: Amount of Ca(OH)2 Precipitated

Since precipitation occurs until \( Q = K_{sp} \), the required \( [\text{OH}^-] \) at equilibrium is obtained by solving:\[ K_{sp} = 4.42 \times 10^{-5} = s (0.244)^2 \approx s \times 0.059536 \]Therefore:\[ s = \frac{4.42 \times 10^{-5}}{0.059536} \approx 7.42 \times 10^{-4} \text{ M} \]The precipitation, in moles, is the difference in \([\text{Ca}^{2+}]\): initial moles before mixing minus the equilibrium moles:Initial: \( 0.022 \text{ M} \) in 0.5 L \( = 0.011 \text{ mol} \)Equilibrium: \( 7.42 \times 10^{-4} \text{ M} \) in 0.5 L \( = 3.71 \times 10^{-4} \text{ mol} \)Precipitated: \( 0.011 - 3.71 \times 10^{-4} = 0.01063 \text{ mol} \)

Step 5: Convert Precipitated Amount to Mass

Convert moles of the precipitated \( \text{Ca(OH)}_2 \) to grams (molecular weight of \( \text{Ca(OH)}_2 \) is 74.1 g/mol):\[ 0.01063 \text{ mol} \times 74.1 \text{ g/mol} = 0.788 \text{ g} \]Convert to milligrams:\[ 0.788 \text{ g} = 788 \text{ mg} \]

Key Concepts

Ca(OH)2

Calcium hydroxide, represented as \( \text{Ca(OH)}_2 \), is an inorganic compound also known as slaked lime. It is a slightly soluble compound in water, forming a saturated solution that is in equilibrium with its ions. When \( \text{Ca(OH)}_2 \) dissolves in water, it dissociates into calcium ions \( \text{Ca}^{2+} \) and hydroxide ions \( \text{OH}^- \):

• \( \text{Ca(OH)}_2 (s) \rightleftharpoons \text{Ca}^{2+} (aq) + 2 \text{OH}^- (aq) \)
• The solubility is represented by \( s \) (mol/L), indicating the concentration of \( \text{Ca}^{2+} \) ions.
• \( \text{OH}^- \) concentration would be \( 2s \) because of the stoichiometry.

This equilibrium concept is essential in calculating how much \( \text{Ca(OH)}_2 \) can dissolve under given conditions, which is important in the context of reactions involving precipitation.

Precipitation

Precipitation in chemistry refers to the process where dissolved ions in a solution form an insoluble solid. In the context of \( \text{Ca(OH)}_2 \), precipitation occurs when the solution contains more \( \text{Ca}^{2+} \) and \( \text{OH}^- \) ions than can be dissolved, leading to the formation of solid \( \text{Ca(OH)}_2 \) precipitate.

• Precipitation occurs when the product of ionic concentrations (reaction quotient \( Q \)) exceeds the solubility product constant \( K_{sp} \).
• In our case, when mixing a saturated solution with \( \text{NaOH} \), the overall \( \text{OH}^- \) concentration increases.
• If the reaction quotient \( Q \) is greater than \( K_{sp} \), \( \text{Ca(OH)}_2 \) begins to precipitate.

This concept is crucial for understanding how concentrations affect the solubility equilibrium and when solids will form.

Reaction Quotient

The reaction quotient \( Q \) is a measure used to determine the direction a reaction will proceed to reach equilibrium. Similar to the equilibrium constant, it is calculated using concentrations or pressures at any point during the reaction, not just at equilibrium.

• The formula for the reaction quotient in the dissolution of \( \text{Ca(OH)}_2 \) is similar to \( K_{sp} \): \( Q = [\text{Ca}^{2+}][\text{OH}^-]^2 \).
• Comparing \( Q \) to \( K_{sp} \) helps predict if a precipitate will form.
• If \( Q > K_{sp} \), more solid will precipitate as the system shifts to decrease the ion concentration.

Understanding \( Q \) allows us to make predictions and understand non-equilibrium states within a chemical reaction.

Equilibrium

Equilibrium in a chemical reaction occurs when the rates of the forward and reverse reactions are equal, so the concentrations of reactants and products remain constant over time. In the case of \( \text{Ca(OH)}_2 \), this equilibrium involves the dissolved ions and the solid compound.

• The solubility product constant \( K_{sp} \) quantifies the equilibrium between the solid and its dissolved ions.
• When additional ions are added (such as from \( \text{NaOH} \)), the system will adjust to compensate until a new equilibrium is established, possibly resulting in precipitation.
• This dynamic response helps explain why precipitation can occur when solutions are mixed, depending on the concentrations of the involved ions.

Equilibrium concepts are foundational for predicting and understanding reactions in biochemical and industrial contexts.