Problem 182 - A Complete Resource Book in Chemistry for JEE Main

Question

Silver acetate is a slightly soluble salt of a weak acid \(\left(\mathrm{K}_{\mathrm{a}}=1.75 \times 10^{-5}\right)\). At \(20^{\circ} \mathrm{C}, 100 \mathrm{~g}\) of water dissolves \(1.04 \mathrm{~g}\) of crystalline silver acetate. The density of saturated solution of silver acetate at \(20^{\circ} \mathrm{C}\) is \(1.01 \mathrm{~g} / \mathrm{cc}\). The solubility product constant for silver acetate at \(20^{\circ} \mathrm{C}\) (a) \(2.43 \times 10^{-3}\) (b) \(3.87 \times 10^{-3}\) (c) \(7.74 \times 10^{-5}\) (d) \(1.35 \times 10^{-5}\)

Step-by-Step Solution

Answer

The solubility product \(K_{sp}\) for silver acetate is \(3.87 \times 10^{-3}\), matching option (b).

Step 1: Calculate Molar Mass of Silver Acetate

The chemical formula of silver acetate is \( \text{AgC}_2\text{H}_3\text{O}_2 \). Calculate its molar mass using atomic masses: \( \text{Ag} = 107.87 \text{ g/mol} \), \( \text{C} = 12.01 \text{ g/mol} \), \( \text{H} = 1.01 \text{ g/mol} \), and \( \text{O} = 16.00 \text{ g/mol} \).\[ \text{Molar mass} = 107.87 + 2(12.01) + 3(1.01) + 2(16.00) = 166.91 \text{ g/mol} \]

Step 2: Calculate Moles of Silver Acetate Dissolved

Find the number of moles of silver acetate dissolved in 100 g of water using the mass given (1.04 g) and its molar mass (166.91 g/mol).\[ \text{Moles of silver acetate} = \frac{1.04}{166.91} \approx 0.00623 \text{ moles} \]

Step 3: Calculate Volume of Solution

Given the density of the solution is 1.01 g/cc, calculate the volume of the solution containing 100 g of water.\[ \text{Volume of solution} = \frac{100 \text{ g} + 1.04 \text{ g}}{1.01 \text{ g/cc}} \approx 100.396 \text{ cc} \]

Step 4: Calculate Concentration of Silver Acetate

Calculate the concentration in mol/L (molarity) by using the moles of silver acetate and the volume of the solution in liters.\[ \text{Concentration} = \frac{0.00623 \text{ moles}}{0.100396 \text{ L}} \approx 0.062 \text{ M} \]

Step 5: Determine Solubility Product \(K_{sp}\)

The dissolution of silver acetate can be expressed as:\[ \text{AgC}_2\text{H}_3\text{O}_2 (s) \rightleftharpoons \text{Ag}^+ (aq) + \text{C}_2\text{H}_3\text{O}_2^- (aq) \]The \(K_{sp}\) is defined as the product of the ion concentrations in a saturated solution:\[ K_{sp} = [\text{Ag}^+][\text{C}_2\text{H}_3\text{O}_2^-] \]Since the concentration of ions is equal in stoichiometry, \([\text{Ag}^+] = [\text{C}_2\text{H}_3\text{O}_2^-] = 0.062 \text{ M}\),\[ K_{sp} = (0.062)(0.062) = 0.003844 \approx 3.87 \times 10^{-3} \]

Step 6: Choose the Correct Option

Pause to review the possible answers and match the calculated \(K_{sp}\) value with the given options. The result is \(3.87 \times 10^{-3}\).Therefore, the correct answer is option (b).

Key Concepts

Silver Acetate

Silver acetate, known by its chemical formula \( \text{AgC}_2\text{H}_3\text{O}_2 \), is a compound of silver and acetic acid. It is characterized as a slightly soluble salt in water, meaning that only a small amount dissolves in water to form a solution. This characteristic makes silver acetate an interesting subject when studying solubility products.
When silver acetate dissolves in water, it dissociates into silver ions \( \text{Ag}^+ \) and acetate ions \( \text{C}_2\text{H}_3\text{O}_2^- \). This process is important to understand because it helps to comprehend how the dissolution equilibrium in a solution is achieved.

• **Chemical Behavior:** Despite its rarity in everyday applications, silver acetate is often used in laboratories to explore the solubility of salts and investigate equilibrium processes.
• **Importance in Chemistry:** The study of silver acetate aids in understanding how substances dissolve and reach equilibrium, essential for fields like pharmacology and biochemistry.

Knowledge of compounds like silver acetate gives us the foundation to predict and calculate solubility products, helping to inform decisions in both industrial and clinical settings.

Molar Mass Calculation

Understanding the molar mass of a compound like silver acetate is crucial for several chemical calculations. The molar mass is the weight of one mole of a substance, and for \( \text{AgC}_2\text{H}_3\text{O}_2 \), it can be calculated by summing the atomic masses of its constituent elements.
Using standard atomic masses:

• Silver (Ag): 107.87 g/mol
• Carbon (C): 12.01 g/mol (2 atoms)
• Hydrogen (H): 1.01 g/mol (3 atoms)
• Oxygen (O): 16.00 g/mol (2 atoms)

Thus, the molar mass of silver acetate is calculated as:\[107.87 + 2(12.01) + 3(1.01) + 2(16.00) = 166.91 \, \text{g/mol} \]This calculation is a straightforward example of molar mass determination, a foundational skill in chemistry crucial for converting mass into moles, which is critical for many reaction stoichiometry problems.

Dissolution Equilibrium

Dissolution equilibrium refers to the state reached when the rate of dissolving of a solute equals the rate of precipitation from the solution. In the context of silver acetate, this involves learning how the compound dissociates into ions and how these ions achieve equilibrium in solution.
The equation representing the dissolution of silver acetate is:\[\text{AgC}_2\text{H}_3\text{O}_2 (s) \rightleftharpoons \text{Ag}^+ (aq) + \text{C}_2\text{H}_3\text{O}_2^- (aq) \]This equation indicates that in a saturated solution, the concentration of silver ions and acetate ions are equal, and the product of these concentrations defines the solubility product \( K_{sp} \).
For silver acetate, since \([\text{Ag}^+] = [\text{C}_2\text{H}_3\text{O}_2^-] = 0.062 \, \text{M}\), the solubility product is calculated as:\[K_{sp} = (0.062)^2 = 0.003844 \]Achieving dissolution equilibrium is key in numerous practical applications, from drug delivery systems to environmental science, providing insight into how substances interact within solutions.