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 is approximately \(3.87 \times 10^{-3}\), corresponding to option (b).

Step 1: Determine the molar mass of silver acetate

Silver acetate is \( \text{AgC}_2\text{H}_3\text{O}_2 \). Calculate its molar mass: \( \text{Ag} = 107.87 \, \text{g/mol}, \text{C} = 12.01 \, \text{g/mol}, \text{H} = 1.01 \, \text{g/mol}, \text{O} = 16.00 \, \text{g/mol} \). Thus, \( \text{Molar Mass} = 107.87 + 2(12.01) + 3(1.01) + 2(16.00) = 166.92 \, \text{g/mol} \).

Step 2: Convert grams of silver acetate to moles

Given 1.04 g of silver acetate dissolves in 100 g of water, calculate moles: \[ \text{Moles of AgC}_2\text{H}_3\text{O}_2 = \frac{1.04 \, \text{g}}{166.92 \, \text{g/mol}} = 0.00623 \, \text{mol} \]

Step 3: Calculate concentration of ions in solution

Since each molecule of silver acetate dissociates into one silver ion \( \text{Ag}^+ \) and one acetate ion \( \text{C}_2\text{H}_3\text{O}_2^- \), the concentration of each ion is \( 0.00623 \, \text{mol} \) in 100 g water with a solution density of 1.01 g/cc. Convert 100 g to volume: \( \text{Volume} = \frac{100}{1.01} = 99.01 \, \text{cc} \), which equals 99.01 mL or 0.09901 L. Thus, \( \text{Concentration} = \frac{0.00623}{0.09901} = 0.0629 \, \text{M} \).

Step 4: Calculate the solubility product \( K_{sp} \)

The solubility product \( K_{sp} \) for the reaction is \( K_{sp} = [\text{Ag}^+][\text{C}_2\text{H}_3\text{O}_2^-] \). Substitute the concentration of ions: \( K_{sp} = (0.0629)(0.0629) = 0.00396 \). Thus, \( K_{sp} = 3.96 \times 10^{-3} \).

Step 5: Choose the correct answer

Compare the calculated \( K_{sp} \) to the given options: \( K_{sp} = 3.96 \times 10^{-3} \) is closest to option (b) \( 3.87 \times 10^{-3} \). Thus, option (b) is the correct answer.

Key Concepts

Silver Acetate

Silver acetate is a chemical compound with the formula \( \text{AgC}_2\text{H}_3\text{O}_2 \). It is known for its slightly soluble nature in water. Structurally, silver acetate consists of a silver ion \( \text{Ag}^+ \) and an acetate ion \( \text{C}_2\text{H}_3\text{O}_2^- \). Given its limited solubility, calculating the precise solubility product or \( K_{sp} \) is crucial, especially in predicting how much of the compound will dissolve in water.

Silver acetate often appears as a white crystalline salt and is utilized in several chemical processes. Despite being slightly soluble, it can decompose under certain conditions and release silver ions and acetate ions into the solution. Understanding its solubility behavior is essential for various applications, such as in chemical synthesis and material science.

Molar Mass Calculation

The molar mass calculation of a compound like silver acetate is essential for converting mass into moles, a necessary step in various chemical calculations. To determine the molar mass of silver acetate, add the atomic masses of each element within its formula:

• Silver \( (\text{Ag}) = 107.87 \ \text{g/mol} \)
• Carbon \( (\text{C}) = 12.01 \ \text{g/mol} \), having two carbon atoms in the formula
• Hydrogen \( (\text{H}) = 1.01 \ \text{g/mol} \), with three hydrogen atoms
• Oxygen \( (\text{O}) = 16.00 \ \text{g/mol} \), with two oxygen atoms

Adding these together gives Silver Acetate its molar mass of 166.92 g/mol:
\[ 107.87 + 2(12.01) + 3(1.01) + 2(16.00) = 166.92 \, \text{g/mol} \]

This foundational step translates the mass of dissolved silver acetate into moles, which are then used in subsequent concentration calculations.

Ion Concentration Calculation

Calculating the ion concentration in a solution is key to understanding its reactivity and behavior. When silver acetate dissolves slightly in water, each molecule dissociates into one \( \text{Ag}^+ \) and one \( \text{C}_2\text{H}_3\text{O}_2^- \) ion. The number of moles of dissolved silver acetate corresponds to the total number of each ion produced in the solution:

Given 1.04 g of silver acetate in 100 g of water, we previously calculated 0.00623 moles of silver acetate. Since it dissociates equally into ions, the concentration of each ion is therefore also 0.00623 mol. To find the concentration per liter, consider the solution volume, derived via:

\[ \text{Volume} = \frac{100 \, \text{g}}{1.01 \, \text{g/cc}} = 0.09901 \, \text{L} \]

Thus, the concentration for each ion is:
\[ \text{Concentration} = \frac{0.00623}{0.09901} = 0.0629 \, \text{M} \]

This concentration is pivotal for determining solubility products and other equilibrium considerations.

Acid Dissociation Constant

The acid dissociation constant, \( K_a \), is an essential parameter in understanding the strength of a weak acid, such as the acetate ion in silver acetate. An acid dissociation constant indicates how well an acid can donate a proton in water, determining the equilibrium position of its dissociation.

For acetate ions \( \text{C}_2\text{H}_3\text{O}_2^- \), the provided dissociation constant is \( K_a = 1.75 \times 10^{-5} \). This relatively small value suggests that acetate is a weak acid, meaning it has a minor tendency to dissociate fully in water, remaining mostly intact in its ionized form.

Understanding the \( K_a \) value provides insight into the equilibrium state of the acetate ions in solution, and consequently, their behavior in reactions that may alter pH or involve buffering effects.