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

(a) \(5.1 \times 10^{-5} \) M

Step 1: Understand the Reaction and Precipitation Condition

The reaction in question is the dissolution of barium nitrate into a sodium carbonate solution. This can lead to the precipitation of barium carbonate, represented by the equilibrium: \(\text{Ba}^{2+} + \text{CO}_{3}^{2-} \rightleftharpoons \text{BaCO}_3\). The precipitate (i.e., \(\text{BaCO}_3\)) forms when the ion product \([\text{Ba}^{2+}] [\text{CO}_3^{2-}]\) exceeds the solubility product \(K_{sp}\) of barium carbonate: \(K_{sp} = 5.1 \times 10^{-9}\).

Step 2: Set Up the Inequality for Precipitation

For precipitation to begin, we require that the ion product \([\text{Ba}^{2+}][\text{CO}_3^{2-}]\) is at least equal to \(K_{sp}\). Therefore, \([\text{Ba}^{2+}][\text{CO}_3^{2-}] \geq 5.1 \times 10^{-9}\). Since the concentration of \([\text{CO}_3^{2-}]\) in the solution is given as \(1.0 \times 10^{-4} \text{ M}\), we substitute into the inequality: \([\text{Ba}^{2+}] \times 1.0 \times 10^{-4} \geq 5.1 \times 10^{-9}\).

Step 3: Solve for \([\text{Ba}^{2+}]\)

We solve for the concentration of barium ions \([\text{Ba}^{2+}]\) where precipitation begins by manipulating the inequality: \([\text{Ba}^{2+}] \geq \frac{5.1 \times 10^{-9}}{1.0 \times 10^{-4}}\). This simplifies to \([\text{Ba}^{2+}] \geq 5.1 \times 10^{-5} \text{ M}\).

Step 4: Compare with Given Options

Now that we have the critical concentration \([\text{Ba}^{2+}]\) for precipitation as \(5.1 \times 10^{-5} \text{ M}\), we compare it with the options given. Option (a) \(5.1 \times 10^{-5} \text{ M}\) matches this concentration.

Key Concepts

Solubility Product

The term "solubility product" often written as \( K_{sp} \), refers to the equilibrium constant for the dissolution of a solid substance into ions in a solution. It is a vital concept in understanding when a compound will dissolve or precipitate in a given solvent.

The solubility product is unique for each ionic compound and varies with temperature. For an equilibrium process where a slightly soluble ionic compound dissolves in water, forming a saturated solution, the equilibrium can be expressed as:

• \( aA_{(s)} \rightleftharpoons bB^{+}_{(aq)} + cC^{-}_{(aq)} \)

At equilibrium, the solubility product \( K_{sp} \) is defined as follows:

• \( K_{sp} = [B^{+}]^b[C^{-}]^c \)

This means \( K_{sp} \) is the product of the concentrations of the dissolved ions raised to the power of their stoichiometric coefficients. In the case of barium carbonate \( \text{BaCO}_3 \), the ion balance is: \( \text{Ba}^{2+} + \text{CO}_3^{2-} \rightleftharpoons \text{BaCO}_3 \). This tells us that \( K_{sp} \) for \( \text{BaCO}_3 \) is the product of the concentration of \( \text{Ba}^{2+} \) and \( \text{CO}_3^{2-} \).

If the ion product (concentrations of ions multiplied) exceeds \( K_{sp} \), the system has surpassed its solubility limit, leading to the formation of a precipitate.

Barium Carbonate

Barium carbonate, \( \text{BaCO}_3 \), is a white, insoluble salt that forms when barium ions \( \text{Ba}^{2+} \) combine with carbonate ions \( \text{CO}_3^{2-} \). It often arises in reactions taking place in aqueous solutions, especially when barium salts are present along with a source of carbonate ions.

This compound is sparingly soluble in water, which means it doesn’t dissolve completely in typical conditions. Its low solubility is quantified by its \( K_{sp} \) value, which is \( 5.1 \times 10^{-9} \) for \( \text{BaCO}_3 \). Due to this low \( K_{sp} \), starting with just tiny concentrations of \( \text{Ba}^{2+} \) and \( \text{CO}_3^{2-} \) in a solution could lead to the precipitation of barium carbonate.

Understanding the behavior of \( \text{BaCO}_3 \) is important in laboratory settings and in industries dealing with barium chemicals. Precipitation of \( \text{BaCO}_3 \) can be useful, for example, in removing barium from solutions industrially.

Ion Product

The ion product is a key factor in determining whether a precipitate will form in a solution. It is calculated just like the solubility product (\( K_{sp} \)), but it represents the current state of the solution, not the equilibrium state.

For a given reaction, such as the formation of \( \text{BaCO}_3 \) from \( \text{Ba}^{2+} \) and \( \text{CO}_3^{2-} \), it's defined as:

• Ion Product = \([\text{Ba}^{2+}][\text{CO}_3^{2-}]\)

In a saturated solution, the ion product equals the \( K_{sp} \).

If the ion product exceeds \( K_{sp} \), the solution is supersaturated, and a precipitate, like barium carbonate, will start to form. Conversely, if the ion product is less than \( K_{sp} \), the solution is unsaturated, and no precipitate will form. This simple comparison provides an effective way of predicting and understanding precipitation reactions in chemical processes.