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 \(\mathrm{K}_{\mathrm{sp}}\) 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: Calculate Initial Moles of Reactants

First, calculate the initial moles of each reactant. For \( \mathrm{Ca}(\mathrm{NO}_3)_2 \):\[\text{Moles of } \mathrm{Ca}^{2+} = 0.060 \, \text{L} \times 0.1 \, \text{M} = 0.006 \, \text{moles}\]For \( \mathrm{Na}_2 \mathrm{CO}_3 \):\[\text{Moles of } \mathrm{CO}_3^{2-} = 0.040 \, \text{L} \times 0.125 \, \text{M} = 0.005 \, \text{moles}\]

Step 2: Determine Limiting Reactant

From the reaction \( \mathrm{Ca}^{2+} + \mathrm{CO}_3^{2-} \rightarrow \mathrm{CaCO}_3 \), we need equal moles of \( \mathrm{Ca}^{2+} \) and \( \mathrm{CO}_3^{2-} \). We have more moles of \( \mathrm{Ca}^{2+} \) (0.006) than \( \mathrm{CO}_3^{2-} \) (0.005). Thus, \( \mathrm{CO}_3^{2-} \) is the limiting reactant and will completely react.

Step 3: Calculate Remaining \( \mathrm{Ca}^{2+} \) After Reaction

Since \( \mathrm{CO}_3^{2-} \) is the limiting reactant, all 0.005 moles of it will react with an equivalent amount of \( \mathrm{Ca}^{2+} \). This leaves:\[0.006 \, \text{moles of } \mathrm{Ca}^{2+} - 0.005 \, \text{moles} = 0.001 \, \text{moles} \text{of } \mathrm{Ca}^{2+}\text{ remaining}.\]

Step 4: Determine Total Volume of Solution

The total volume of the solution after mixing is:\[60 \, \mathrm{ml} + 40 \, \mathrm{ml} = 100 \, \mathrm{ml} = 0.1 \, \text{L}\]

Step 5: Calculate Concentration of \( \mathrm{Ca}^{2+} \) and \( \mathrm{CO}_3^{2-} \)

After the reaction, the remaining \( \mathrm{Ca}^{2+} \) is:\[\text{Concentration of } \mathrm{Ca}^{2+} = \frac{0.001 \, \text{moles}}{0.1 \, \text{L}} = 0.01 \, \text{M}\]The concentration of \( \mathrm{CO}_3^{2-} \) in equilibrium needs to satisfy the solubility product \( K_{sp} \):\[K_{sp} = [\mathrm{Ca}^{2+}][\mathrm{CO}_3^{2-}] = 5 \times 10^{-9} \]

Step 6: Solve for Equilibrium \( \mathrm{CO}_3^{2-} \) Concentration

Using \( K_{sp} \) and the concentration of \( \mathrm{Ca}^{2+} \):\[(0.01)[\mathrm{CO}_3^{2-}] = 5 \times 10^{-9}\]Solve for \( [\mathrm{CO}_3^{2-}] \):\[[\mathrm{CO}_3^{2-}] = \frac{5 \times 10^{-9}}{0.01} = 5 \times 10^{-7} \, \text{M}\]

Step 7: Conclusion: Identify the Correct Option

The final concentration of \( \left[\mathrm{CO}_3^{2-}\right] \) in the solution is \( 5 \times 10^{-7} \, \text{M} \). This corresponds to option (d).

Key Concepts

Stoichiometry

Stoichiometry is the study of the quantitative relationships between the amounts of reactants and products in a chemical reaction. It's based on the laws of conservation of mass and constant proportions. In our exercise, we are dealing with the reaction between calcium nitrate, Ca(NO₃)₂, and sodium carbonate, Na₂CO₃, to form calcium carbonate, CaCO₃, which precipitates. Key steps include:

• Determining the moles of each reactant initially present.
• Using these amounts to identify the limiting reactant, which determines the extent of the reaction.
• Calculating the remaining concentrations after the reaction based on stoichiometry.

Giving attention to the details of stoichiometric calculations ensures the correct determination of reactant and product quantities. This foundational knowledge is crucial for mastering equilibrium and solubility concepts in chemistry.

Solubility Product Constants (Ksp)

The solubility product constant (Ksp) is a key concept that helps predict the solubility of a sparingly soluble compound. It represents the maximum product of the ion concentrations that can be achieved before the solution becomes supersaturated and a precipitate forms. In this exercise, Ksp applies to the precipitation of calcium carbonate, CaCO₃.For our example:

• The equilibrium of dissolution is governed by Ksp at the point when no more solid can dissolve.
• If \( \text{CaCO}_3 \) forms and dissolves back into its ions, the concentrations of \([\text{Ca}^{2+}]\) and \([\text{CO}_3^{2-}]\) must satisfy the equation \([\text{Ca}^{2+}][\text{CO}_3^{2-}] = 5 \times 10^{-9} \mathrm{M}^2\).

Essentially, calculating with Ksp lets us determine the potential concentrations of ions in solution at equilibrium, helping predict if a precipitation will occur or stay dissolved.

Limiting Reactant Analysis

In a chemical reaction, the limiting reactant is the substance that is entirely consumed when the reaction completes. It limits the amount of product formed because there is insufficient quantity to react with the other reactants that may be in excess. Identifying the limiting reactant is crucial for accurate reaction analysis.Steps include:

• Calculating initial moles of each reactant available.
• Determining which reactant will be consumed first using the stoichiometric ratio from the balanced equation.
• After identifying the limiting reactant, calculate how much of the other reactants remain unreacted.

In our exercise, carbonate \( (\text{CO}_3^{2-}) \) acts as the limiting reactant, reacting completely with calcium ions to form calcium carbonate precipitate. Recognizing the limiting reactant helps understand which material governs the reaction yield and subsequent equilibrium conditions.