Problem 274 - A Complete Resource Book in Chemistry for JEE Main

Question

Solid \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) is gradually dissolved in a \(1.0 \times 10^{-4}\) \(\mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}\) solution. At what concentration of \(\mathrm{Ba}^{2+}\) will a precipitate being to form? \(\left(\mathrm{K}_{\mathrm{sp}}\right.\) for \(\left.\mathrm{Ba} \mathrm{CO}_{3}=5.1 \times 10^{-9}\right)\) (a) \(5.1 \times 10^{-5} \mathrm{M}\) (b) \(8.1 \times 10^{-8} \mathrm{M}\) (c) \(8.1 \times 10^{-7} \mathrm{M}\) (d) \(4.1 \times 10^{-5} \mathrm{M}\)

Step-by-Step Solution

Answer

At \(5.1 \times 10^{-5} \text{ M} \) \( \text{Ba}^{2+} \), a precipitate will form; answer is (a).

Step 1: Understanding the Problem

We need to determine the concentration of \( \text{Ba}^{2+} \) ions at which \( \text{BaCO}_3 \) begins to precipitate out from a \( 1.0 \times 10^{-4} \text{ M} \) \( \text{Na}_2\text{CO}_3 \) solution. The solubility product constant \( K_{sp} \) for \( \text{BaCO}_3 \) is given as \( 5.1 \times 10^{-9} \).

Step 2: Write the Solubility Product Expression

The solubility product expression for \( \text{BaCO}_3 \) is given by: \[ K_{sp} = [\text{Ba}^{2+}][\text{CO}_3^{2-}] \]We know the value of \( K_{sp} \) and the initial concentration of \( \text{CO}_3^{2-} \).

Step 3: Substitute the Known Values

Substitute the known values into the expression:\[ 5.1 \times 10^{-9} = [\text{Ba}^{2+}](1.0 \times 10^{-4}) \]

Step 4: Solve for \([\text{Ba}^{2+}]\)

Rearrange the equation to solve for \([\text{Ba}^{2+}]\): \[ [\text{Ba}^{2+}] = \frac{5.1 \times 10^{-9}}{1.0 \times 10^{-4}} \]Calculate the result:\[ [\text{Ba}^{2+}] = 5.1 \times 10^{-5} \text{ M} \]

Step 5: Match with Options

The calculated concentration matches option (a), which is \( 5.1 \times 10^{-5} \text{ M} \).

Key Concepts

Solubility Product Constant (Ksp)

The concept of the solubility product constant, noted as \( K_{sp} \), plays a crucial role in understanding when a compound will precipitate. It represents the equilibrium state between a solid and its dissolved ions in a saturated solution.

This constant is unique for each compound at a specific temperature and is expressed in terms of the concentration of its ions. For example, for a compound \( ext{BaCO}_3 \), the solubility product expression is \( K_{sp} = [ ext{Ba}^{2+}][ ext{CO}_3^{2-}] \). This equation shows that the product of the concentrations of the barium ions \( ext{Ba}^{2+} \) and carbonate ions \( ext{CO}_3^{2-} \) must equal \( K_{sp} \) for the solution to be at equilibrium.

When solving problems involving \( K_{sp} \), it is critical to:

• Identify the compound's dissociation into its respective ions.
• Write the solubility product expression.
• Substitute known values to find unknown concentrations.

Knowing \( K_{sp} \) allows us to determine the concentration at which a compound will start to precipitate when it exceeds the product of the ionic concentrations.

Precipitation

Precipitation occurs when the product of the concentrations of ions in a solution exceeds the solubility product constant, \( K_{sp} \). At this point, the solution can no longer hold all the ions in a dissolved state, resulting in the formation of a solid precipitate.

This process is vital in various chemical applications, from separating substances in a mixture to purifying compounds. In our specific case, precipitation begins once enough \( ext{Ba}^{2+} \) ions are added to a \( ext{Na}_2 ext{CO}_3 \) solution, leading the solution to reach saturation and exceed the \( K_{sp} \).

To predict when precipitation will occur:

• Use the \( K_{sp} \) expression to calculate the threshold concentration for ions.
• Compare the concentrations in the solution to the calculated threshold.
• If the product of the ion concentrations surpasses \( K_{sp} \), precipitation will start.

Understanding these steps helps in predicting precipitation, essential in laboratory practices and industrial processes alike.

Ionic Equilibrium

Ionic equilibrium is the state achieved in a solution when the rates of dissolution and precipitation of a compound are equal. At this point, the concentration of ions remains constant over time.

Equilibrium is an essential concept in chemistry, helping us explore how dissolved ions in a solution interact. For mixtures like \( ext{Na}_2 ext{CO}_3 \) with \( ext{Ba}^{2+} \) ions, ionic equilibrium dictates the concentrations of each ion that can coexist without further dissolution or precipitation.

Key aspects of ionic equilibrium involve:

• Recognizing that at equilibrium, the rate of ion formation from the solid equals the rate of the reverse process.
• Relating equilibrium concentrations to \( K_{sp} \) to determine theoretical behaviors of solutions.
• Using equilibrium conditions to adjust concentrations, predict outcomes, and control chemical reactions efficiently.

Understanding ionic equilibrium helps in manipulating and predicting the behaviors of solutions in both natural and experimental contexts.