Problem 51 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The \(K_{s p}\) of \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\) is \(1.1 \times 10^{-12}\) at \(298 \mathrm{~K}\). The solubility (in \(\mathrm{mol} / \mathrm{L}\) ) of \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\) in a \(0.1 \mathrm{M} \mathrm{AgNO}_{3}\) solution is (a) \(1.1 \times 10^{-11}\) (b) \(1.1 \times 10^{-10}\) (c) \(1.1 \times 10^{-12}\) (d) \(1.1 \times 10^{-9}\)

Step-by-Step Solution

Answer

The solubility is \(1.1 \times 10^{-10}\, \text{mol/L}\). Answer: (b).

Step 1: Write the Dissociation Equation

Silver chromate, \(\text{Ag}_2\text{CrO}_4\), dissociates in water according to the following equation: \[\text{Ag}_2\text{CrO}_4 (s) \rightleftharpoons 2\, \text{Ag}^+ (aq) + \text{CrO}_4^{2-} (aq).\]

Step 2: Expression for the Solubility Product, \(K_{sp}\)

The solubility product expression for silver chromate is given by: \[K_{sp} = [\text{Ag}^+]^2[\text{CrO}_4^{2-}].\] Given that \(K_{sp} = 1.1 \times 10^{-12}\).

Step 3: Consider Initial Concentration

The \(0.1\, \text{M} \text{AgNO}_3\) solution already contains \(0.1\, \text{M}\) of \(\text{Ag}^+\). Let the solubility of \(\text{Ag}_2\text{CrO}_4\) be \(s\) mol/L.

Step 4: Setup New Equilibrium Concentrations

Upon dissolving, the concentration of \(\text{Ag}^+\) becomes \((0.1 + 2s)\) and the concentration of \(\text{CrO}_4^{2-}\) becomes \(s\).

Step 5: Solve the \(K_{sp}\) Expression

Substitute the expressions into the \(K_{sp}\) equation: \[1.1 \times 10^{-12} = (0.1 + 2s)^2 \cdot s.\] Assuming \(s\) is very small compared to \(0.1\), approximate: \[(0.1)^2 \cdot s \approx 1.1 \times 10^{-12}.\]

Step 6: Calculate \(s\)

Solve for \(s\) in the equation \[(0.1)^2 \cdot s = 1.1 \times 10^{-12}:\] \[0.01 \cdot s = 1.1 \times 10^{-12},\] \[s = \frac{1.1 \times 10^{-12}}{0.01} = 1.1 \times 10^{-10}.\]

Step 7: Select the Correct Answer

The calculated solubility \(s\) is \(1.1 \times 10^{-10} \text{ M/L}\). Therefore, the correct answer is (b) \(1.1 \times 10^{-10}\).

Key Concepts

Solubility Product

The solubility product constant, often abbreviated as \(K_{sp}\), is a critical concept in chemistry that helps us understand the solubility of sparingly soluble salts in aqueous solutions. It provides a quantitative measure of the extent to which a compound will dissolve to form a saturated solution. The solubility product is unique for every slightly soluble ionic compound at a given temperature.

The \(K_{sp}\) value is the product of the molar concentrations of the ions, each raised to the power of their stoichiometric coefficients in the dissociation equation. For example, if a salt dissolves into two cations \([A^+]\) and one anion \([B^-]\), then \(K_{sp} = [A^+]^2 [B^-]\).

Factors such as temperature and the presence of common ions can influence the dissolution and therefore affect the value of \(K_{sp}\). When a common ion is already present in the solution, it can decrease the solubility of the salt due to the common ion effect.

Silver Chromate

Silver chromate, denoted as \(\text{Ag}_2\text{CrO}_4\), is a chemical compound composed of silver and chromate ions. It is well known for its characteristic reddish-brown color. In water, silver chromate is considered a sparingly soluble salt. This means it does not dissolve extensively in water, leading to its low solubility product, which is significant for calculating its solubility.

When introduced into an aqueous solution, silver chromate dissociates into silver ions \(\text{Ag}^+\) and chromate ions \(\text{CrO}_4^{2-}\). Understanding the solubility of silver chromate is important in various chemical processes, including those in analytical chemistry where it might be used in precipitation reactions or in determining the concentration of other ions in solution through its distinct color changes.

Dissolution Equation

The dissolution equation shows the process by which a solid substance dissolves in a solvent, splitting into its constituent ions. For silver chromate, the dissolution process in water is represented by the equation:

• \(\text{Ag}_2\text{CrO}_4 (s) \rightleftharpoons 2\, \text{Ag}^+ (aq) + \text{CrO}_4^{2-} (aq)\).

This equation tells us that one mole of silver chromate produces two moles of silver ions and one mole of chromate ions. It's crucial to accurately balance these equations to understand the stoichiometry of the reaction fully. The dissolution equation helps in forming the basis for calculating the \(K_{sp}\) and further determines the solubility of the compound in a given solution.

Moreover, in solutions where a common ion is present, such as a solution of \(\text{AgNO}_3\) in this example, the equation must be adjusted by adding the ion concentration from the external source. This involves modifying the equilibrium concentrations in the context of the solubility product equation to calculate true solubility accurately. This new equilibrium is solved using the \(K_{sp}\) expression to find the correct solubility, considering the external \(\text{Ag}^+\) ion concentration contributed by \(\text{AgNO}_3\).