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 precipitate being to form? \(\left(\mathrm{K}_{s p}\right.\) 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

The concentration of \(\mathrm{Ba^{2+}}\) at which precipitate begins to form is \(5.1 \times 10^{-5} \mathrm{M}\). (Option a)

Step 1: Write the dissolution equation

The dissolution of barium carbonate in water can be described by its equilibrium reaction: \( \mathrm{BaCO_3(s)} \rightleftharpoons \mathrm{Ba^{2+}}(aq) + \mathrm{CO_3^{2-}}(aq) \). This represents the dissociation of solid barium carbonate into barium ions and carbonate ions in the solution.

Step 2: Use of Solubility Product

The solubility product constant, \(K_{sp}\), is given for \(\mathrm{BaCO_3}\) as \(5.1 \times 10^{-9}\). This can be expressed as: \(K_{sp} = [\mathrm{Ba^{2+}}][\mathrm{CO_3^{2-}}]\).

Step 3: Substitute Known Values

We are given that the concentration of \(\mathrm{Na}_2\mathrm{CO}_3\) is \(1.0 \times 10^{-4} \mathrm{M}\). In solution, \(\mathrm{Na}_2\mathrm{CO}_3\) dissociates completely to give \([\mathrm{CO_3^{2-}}] = 1.0 \times 10^{-4} \mathrm{M}\).

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

Using the relation from the solubility product, set \([\mathrm{Ba^{2+}}] [\mathrm{CO_3^{2-}}] = K_{sp}\). Substitute the known concentration of \([\mathrm{CO_3^{2-}}]\): \([\mathrm{Ba^{2+}}] \times (1.0 \times 10^{-4}) = 5.1 \times 10^{-9}\).

Step 5: Calculate \([\mathrm{Ba^{2+}}]\) Concentration

Rearranging gives \([\mathrm{Ba^{2+}}] = \frac{5.1 \times 10^{-9}}{1.0 \times 10^{-4}}\). Calculate this: \([\mathrm{Ba^{2+}}] = 5.1 \times 10^{-5} \mathrm{M}\).

Key Concepts

Solubility Product

The solubility product, \(K_{sp}\), is a vital concept when studying chemical equilibrium, particularly in precipitation reactions. It represents the level at which a solute dissolves in solution to maintain an equilibrium with its undissolved form. For an ionic compound \(AB\) that dissolves according to \(AB \rightleftharpoons A^{+} + B^{-}\), the \(K_{sp}\) is expressed as: \[K_{sp} = [A^{+}][B^{-}]\]This equation shows how the concentration of dissolved ions relate to the solubility product constant.

• A lower \(K_{sp}\) means the compound is less soluble.
• The ion concentrations at equilibrium define the product outflow in solutions.

For \(\text{BaCO}_3\), \(K_{sp} = 5.1 \times 10^{-9}\). This low value indicates that \(\text{BaCO}_3\) is only sparingly soluble in water. When such systems are disturbed by adding a common ion, the equilibrium shifts to reduce the disturbance, according to Le Chatelier's Principle. The concentration of ions influences the point at which the solution becomes saturated and starts forming a precipitate.

Precipitation Reaction

A precipitation reaction occurs when two solutions combine to form an insoluble solid, known as the precipitate. This happens because the product of the ion concentrations in solution exceeds the solubility product.When \(\text{Ba(NO}_3\text{)}_2\) is dissolved in \(\text{Na}_2\text{CO}_3\) solution, \(\text{Ba}^{2+}\) ions react with \(\text{CO}_3^{2-}\) ions to form the insoluble precipitate \(\text{BaCO}_3\):\[\text{Ba}^{2+} + \text{CO}_3^{2-} \rightarrow \text{BaCO}_3(s)\]Precipitation is indicative of the solution reaching saturation levels of a compound.

• If \(\left[\text{Ba}^{2+}\right] \times \left[\text{CO}_3^{2-}\right] > K_{sp}\), precipitation occurs.
• The threshold known as the solubility product helps predict if a compound will precipitate under specific concentrations.

By carefully calculating ionic concentrations, chemists can predict the precipitation point, ensuring solutions are mixed in ways that avoid unwanted precipitate formation in reactions or industrial processes.

Ionic Equilibria

Ionic equilibria involve understanding how ions behave in a solution where various equilibria can coexist. Specific to solubility and precipitation, it’s vital to grasp how different ions in solution interact and reach a dynamic balance.In solutions containing salts, equilibrium is dynamic:\[AB (s) \rightleftharpoons A^{+} (aq) + B^{-} (aq)\]Here, the designing and predicting of reactions depend on understanding the relation:

• Ionic strength impacts how precisely ions meet \(K_{sp}\) requirements.
• The concept guides the addition of ions that lead to a shift in balance, either way.

For example, using \(\text{Na}_2\text{CO}_3\) introduces \(\text{CO}_3^{2-}\) into the solution, fostering an environment for equilibrium to offset until the ion product matches \(K_{sp}\). Understanding these principles helps manage chemical equilibria which is crucial in separation processes and influencing reaction kinetics.