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 \(\mathrm{K}_{s p}\) 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 \( \mathrm{CO}_3^{2-} \) in the resulting solution is \( 5 \times 10^{-7} \text{ M} \) (option d).

Step 1: Calculate Moles of Reactants

First, calculate the moles of \( \mathrm{Ca}^{2+} \) and \( \mathrm{CO}_3^{2-} \) provided in the solution. For \( \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_2 \), \[ \text{Moles} = \text{Volume (L)} \times \text{Molarity} = 0.060 \times 0.1 = 0.006 \text{ moles} \] For \( \mathrm{Na}_2 \mathrm{CO}_3 \), \[ \text{Moles} = \text{Volume (L)} \times \text{Molarity} = 0.040 \times 0.125 = 0.005 \text{ moles} \]

Step 2: Determine the Limiting Reactant

The reaction in which \( \text{Ca}^{2+} \) and \( \text{CO}_3^{2-} \) ions form \( \text{CaCO}_3 \) as a precipitate is a 1:1 mole reaction. Thus, \( \text{CaCO}_3 \) formation is limited by the lower number of moles, which is \( 0.005 \text{ moles} \) for \( \mathrm{CO}_3^{2-} \).

Step 3: Calculate Excess \( \text{Ca}^{2+} \) Ions

Subtract the moles of consumed \( \text{Ca}^{2+} \) from the initial moles to find excess. \[ \text{Excess} = 0.006 - 0.005 = 0.001 \text{ moles} \]

Step 4: Calculate Total Volume of the Solution

The total volume after mixing is the sum of the volumes of both solutions: \[ 60 \text{ ml} + 40 \text{ ml} = 100 \text{ ml} = 0.1 \text{ L} \]

Step 5: Calculate \( \text{[CO}_3^{2-}\text{]} \) at Equilibrium

At equilibrium, the moles of \( \text{CO}_3^{2-} \) is determined by the solubility product \( K_{sp} \). The concentration of \( \text{Ca}^{2+} \) ions at equilibrium must be calculated using \( K_{sp} \) expression: \[ K_{sp} = [\text{Ca}^{2+}][\text{CO}_3^{2-}] = 5 \times 10^{-9} \] Given \([\text{Ca}^{2+}] = \frac{0.001}{0.1} = 0.01 \text{ M} \), solve for \([\text{CO}_3^{2-}]\): \[ [\text{CO}_3^{2-}] = \frac{K_{sp}}{[\text{Ca}^{2+}]} = \frac{5 \times 10^{-9}}{0.01} = 5 \times 10^{-7} \text{ M} \]

Key Concepts

Ksp Calculation

The solubility product constant, often represented as \( K_{sp} \), is a measure used in chemistry to describe the concentration equilibrium of a solid substance dissolving in a liquid. In simple terms, it tells us how much of that solid can dissolve in a solution.
For calcium carbonate \( \text{CaCO}_3 \), the dissociation can be represented as:

• \( \text{CaCO}_3 \ (s) \rightleftharpoons \text{Ca}^{2+} \ (aq) + \text{CO}_3^{2-} \ (aq) \)

The \( K_{sp} \) is defined by the equation:

• \[ K_{sp} = [\text{Ca}^{2+}][\text{CO}_3^{2-}] \]

Where \([\text{Ca}^{2+}]\) and \([\text{CO}_3^{2-}]\) are the molar concentrations of the ions in solution at equilibrium.
In our exercise, the given \( K_{sp} \) for calcium carbonate is \( 5 \times 10^{-9} \text{ M}^2 \). By knowing this, we can calculate how much carbonate ion \( \text{CO}_3^{2-} \) remains in solution after precipitation.

Limiting Reactant

In any chemical reaction, a limiting reactant is the substance that is completely consumed when the reaction occurs and thus determines the amount of product that can be formed. In other words, it limits the reaction.
In a 1:1 stoichiometric reaction like the formation of \( \text{CaCO}_3 \) from \( \text{Ca}^{2+} \) and \( \text{CO}_3^{2-} \), you need equal moles of both reactants to form the product without any leftovers.
From the calculations:

• \( 0.006 \text{ mol} \) is the amount of \( \text{Ca}^{2+} \)
• \( 0.005 \text{ mol} \) is the amount of \( \text{CO}_3^{2-} \)

The \( \text{CO}_3^{2-} \) is the limiting reactant as it has a lower initial mole quantity, regulating the formation of the \( \text{CaCO}_3 \) precipitate.

Precipitation Reaction

A precipitation reaction occurs when two soluble salts in aqueous solutions react to form an insoluble salt, or precipitate. The important aspect of identifying a precipitation reaction is recognizing when the resultant compound exceeds its solubility product \( K_{sp} \), leading to a solid formation.
When \( \text{Ca}^{2+} \) ions from \( \text{Ca(NO}_3)_2 \) and \( \text{CO}_3^{2-} \) ions from \( \text{Na}_2\text{CO}_3 \) come together, they form \( \text{CaCO}_3 \):

• \( \text{Ca}^{2+}_{(aq)} + \text{CO}_3^{2-}_{(aq)} \to \text{CaCO}_3_{(s)} \)

This solid \( \text{CaCO}_3 \) drops out of solution as a precipitate when its concentration product exceeds \( K_{sp} \). Knowing the threshold for precipitation helps predict when a reaction will produce a solid.

Molar Concentration Calculation

Calculating the molar concentration is key to understanding the behavior of solutions in chemistry. Molarity, or molar concentration, indicates the number of moles of a solute present in one liter of solution, often expressed as \( \text{mol/L} \).
To determine the leftover or equilibrium concentration of ions in a solution, you must account for any that participated in reactions and calculate based on the total volume of the solution.
After the precipitation of \( \text{CaCO}_3 \), there are leftover \( \text{Ca}^{2+} \) ions. The remaining concentration can be calculated as:

• Excess \( \text{Ca}^{2+} \) moles: \( 0.001 \text{ mol} \)
• New volume: \( 0.1 \text{ L} \)
• Concentration: \( [\text{Ca}^{2+}] = \frac{0.001}{0.1} = 0.01 \text{ M} \)

This concentration allows us to find the equilibrium \( [\text{CO}_3^{2-}] \) using the \( K_{sp} \) expression, which reflects the dynamic balance reached after precipitation.