Problem 154 - A Complete Resource Book in Chemistry for JEE Main

Question

When \(60 \mathrm{ml}\) of \(0.1 \mathrm{M} \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) is mixed with \(40 \mathrm{ml}\) of \(0.125 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}, \mathrm{CaCO}_{3}\) precipitates. If \(\mathrm{K}_{\text {sp }}\) of \(\mathrm{CaCO}_{3}\) is \(5 \times 10^{-9} \mathrm{M}^{2}\), the \(\left[\mathrm{CO}_{3}^{2-}\right]\) in the resulting solution is (a) \(5 \times 10^{-8} \mathrm{M}\) (b) \(5 \times 10^{-9} \mathrm{M}\) (c) \(5 \times 10^{-6} \mathrm{M}\) (d) \(5 \times 10^{-7} \mathrm{M}\)

Step-by-Step Solution

Answer

The concentration of \( [\mathrm{CO_3^{2-}}] \) is \( 5 \times 10^{-7} \mathrm{M} \), option (d).

Step 1: Find Moles of Each Reactant

Calculate the moles of each reactant using the formula: \( ext{moles} = ext{volume} \times ext{molarity} \).- For \( \mathrm{Ca(NO_3)_2} \): \( 60 \, \mathrm{ml} = 0.060 \, \mathrm{L} \) and \( 0.1 \, \mathrm{M} \), thus \( ext{moles of } \mathrm{Ca^{2+}} = 0.060 \times 0.1 = 0.006 \).- For \( \mathrm{Na_2CO_3} \): \( 40 \, \mathrm{ml} = 0.040 \, \mathrm{L} \) and \( 0.125 \, \mathrm{M} \), thus \( ext{moles of } \mathrm{CO_3^{2-}} = 0.040 \times 0.125 = 0.005 \).

Step 2: Identify the Limiting Reactant

The reaction between \( \mathrm{Ca^{2+}} \) and \( \mathrm{CO_3^{2-}} \) forms \( \mathrm{CaCO_3} \) in a 1:1 molar ratio.As 0.006 moles of \( \mathrm{Ca^{2+}} \) and 0.005 moles of \( \mathrm{CO_3^{2-}} \) are available, \( \mathrm{CO_3^{2-}} \) is the limiting reactant. Thus, all 0.005 moles of \( \mathrm{CO_3^{2-}} \) will react.

Step 3: Calculate the Remaining Calcium Ions

Since \( \mathrm{CO_3^{2-}} \) is the limiting reactant, it fully reacts with 0.005 moles of \( \mathrm{Ca^{2+}} \), leaving \( 0.006 - 0.005 = 0.001 \) moles of \( \mathrm{Ca^{2+}} \) unreacted.

Step 4: Calculate Concentrations in Final Solution

The final volume is the sum of the volumes of the two solutions mixed: \( 60 + 40 = 100 \, \mathrm{ml} = 0.100 \, \mathrm{L} \).The concentration of unreacted \( \mathrm{Ca^{2+}} \) is \( \frac{0.001}{0.100} = 0.01 \, \mathrm{M} \).\( [\mathrm{CO_3^{2-}}] \) is determined by setting it equal to \( \frac{K_{\text{sp}}}{[\mathrm{Ca^{2+}}]} \), which is \( \frac{5 \times 10^{-9}}{0.01} = 5 \times 10^{-7} \mathrm{M} \).

Step 5: Choose the Correct Option

As calculated, the remaining \( [\mathrm{CO_3^{2-}}] = 5 \times 10^{-7} \mathrm{M} \), so option (d) is the correct choice.

Key Concepts

Solubility Product (Ksp)

The solubility product constant, abbreviated as \( K_{sp} \), is a special type of equilibrium constant that applies to the solubility of sparingly soluble ionic compounds. It is the product of the concentrations of the ions each raised to the power of their coefficients in the balanced equation. For calcium carbonate (\( \text{CaCO}_3 \)), \( K_{sp} \) expresses the product of the concentrations of calcium ions \( (\text{Ca}^{2+}) \) and carbonate ions \( (\text{CO}_3^{2-}) \), which must remain at or below a certain threshold for the precipitation equilibrium to remain. This value, \( 5 \times 10^{-9} \ \text{M}^2 \), informs us of the extent to which \( \text{CaCO}_3 \) can dissolve in water under a set of conditions.
The \( K_{sp} \) value is essential because it helps predict whether a precipitate will form in a solution. If the product of the ion concentrations exceeds this threshold, precipitation occurs. Otherwise, the ion concentrations are not sufficient to form a solid, and the compounds remain dissolved.

Precipitation Reaction

A precipitation reaction occurs when two aqueous solutions combine to form an insoluble solid, known as a precipitate. The formation of \( \text{CaCO}_3 \) from calcium nitrate \( (\text{Ca(NO}_3\text{)}_2) \) and sodium carbonate \( (\text{Na}_2\text{CO}_3) \) is an example of such a reaction. When we mix these two solutions, calcium ions \( (\text{Ca}^{2+}) \) and carbonate ions \( (\text{CO}_3^{2-}) \) react in a 1:1 molar ratio to form \( \text{CaCO}_3 \).
This type of chemical reaction is critical in various applications, from water treatment processes where contaminants are removed via precipitation, to understanding geological formations where minerals precipitate naturally. In laboratory contexts, knowing how to control precipitation reactions allows chemists to isolate and identify specific compounds from a mixture.

Limiting Reactant

Identifying the limiting reactant in a chemical reaction is crucial because it determines the maximum amount of product that can be generated. In our example of the precipitation of \( \text{CaCO}_3 \), the two ionic compounds \( \text{Ca}^{2+} \) and \( \text{CO}_3^{2-} \) have been mixed, and their respective moles calculated. Given the concentrations and volumes provided, 0.006 moles of \( \text{Ca}^{2+} \) and 0.005 moles of \( \text{CO}_3^{2-} \) were found.
Since \( \text{CO}_3^{2-} \) ions run out first, they are the limiting reactant, meaning the reaction ceases when all \( \text{CO}_3^{2-} \) ions have reacted. Knowing the limiting reactant helps predict not only the amount of precipitate formed but allows for efficient resource use in practical applications.

Molarity Calculations

Molarity is an essential concept when working with solutions in chemistry. It defines the number of moles of a solute per liter of solution. In our precipitation example, molarity calculations help us understand the concentrations of the compounds in the solutions before and after the reaction.
Initially, the moles of each ionic species were determined using the given molarity and volume, leading us to ascertain which ions remained after precipitation occurred. After mixing the solutions, the total volume was established as 0.1 L, allowing us to calculate the concentration of remaining \( \text{Ca}^{2+} \) ions. This calculation step was crucial to determine the leftover \( \text{CO}_3^{2-} \) concentration using the \( K_{sp} \) equation. By understanding molarity, students can measure and adjust concentrations in various chemical environments, facilitating precise experimentation and analysis.