Problem 262 - A Complete Resource Book in Chemistry for JEE Main

Question

pH of \(0.005 \mathrm{M}\) calcium acetate (pKa of \(\mathrm{CH}_{3} \mathrm{COOH}\) \(=4.74\) ) is \(\quad\) [2002] (a) \(7.37\) (b) \(9.37\) (c) \(9.26\) (d) \(8.37\)

Step-by-Step Solution

Answer

The pH of 0.005 M calcium acetate is approximately 8.13; none of the given options.

Step 1: Understanding the Compound

Calcium acetate dissociates in water to produce acetate ions, ext{CH}_3 ext{COO}^-. As acetate ( ext{CH}_3 ext{COO}^-) is the conjugate base of acetic acid ( ext{CH}_3 ext{COOH}), it will affect the pH by creating a basic solution when it reacts with water.

Step 2: Determine the Concentration of Acetate Ions

Since calcium acetate ext{(Ca(CH}_3 ext{COO)}_2 ext{)} dissociates completely in water, each molecule produces two acetate ions. Thus, for a solution of 0.005 M calcium acetate, the concentration of acetate ions is \( 2 \times 0.005 \ \mathrm{M} = 0.01 \ \mathrm{M} \).

Step 3: Calculate the Base Dissociation Constant (Kb)

Use the relationship \( K_w = K_a \times K_b \), where \( K_w = 1.0 \times 10^{-14} \) at 25°C. The value of \( K_a = 10^{-pK_a} \) for acetic acid is \( 10^{-4.74} \). Solve for \( K_b \): \[ K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{10^{-4.74}} = 1.8 \times 10^{-10} \].

Step 4: Set Up the Base Dissociation Equation

Acetate ion will react with water: \( \text{CH}_3\text{COO}^- + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COOH} + \text{OH}^- \). The expression for the equilibrium constant is \( K_b = \frac{[\text{CH}_3\text{COOH}][\text{OH}^-]}{[\text{CH}_3\text{COO}^-]} \).

Step 5: Solve for OH- Concentration

Assume \([\text{OH}^-] = x\). The initial concentration of \( \text{CH}_3\text{COO}^- \) is \(0.01\) M. At equilibrium, \([\text{OH}^-] = x\) and \([\text{CH}_3\text{COO}^-] = 0.01 - x \approx 0.01\, \text{M}\) since \(x\) will be small. Therefore:\[ K_b = \frac{x^2}{0.01} \].Solving for \(x\): \[ x^2 = 1.8 \times 10^{-10} \times 0.01 \],\[ x^2 \approx 1.8 \times 10^{-12} \],\[ x \approx 1.34 \times 10^{-6} \text{ M} \, (\text{OH})^- \].

Step 6: Calculate pOH and then pH

\( \text{pOH} = -\log[\text{OH}^-] \), \[ \text{pOH} = -\log(1.34 \times 10^{-6}) \approx 5.87 \]. Since \(\text{pH} + \text{pOH} = 14\): \[ \text{pH} = 14 - 5.87 = 8.13 \].

Key Concepts

Calcium Acetate

Calcium acetate, represented by the chemical formula \( \text{Ca(CH}_3\text{COO})_2 \), is a water-soluble calcium salt. It has interesting properties when it interacts with water due to its ability to dissociate into ions. When calcium acetate dissolves in water, it splits into one calcium ion \( \text{Ca}^{2+} \) and two acetate ions \( \text{CH}_3\text{COO}^- \). This complete dissociation of calcium acetate is instrumental in determining the solution's pH as it directly increases the concentration of acetate ions in the solution.

• Calcium acetate readily releases acetate ions in solution.
• The acetate ion is crucial for influencing the basic nature of the solution.

Understanding this dissociation process is key to comprehending how calcium acetate acts as a source of its conjugate base, the acetate ion, which affects the pH of the solution.

Base Dissociation Constant

The base dissociation constant, \( K_b \), is a crucial factor when calculating the pH of a solution containing a conjugate base like acetate. It represents the equilibrium constant for the reaction where a base accepts a proton to form its conjugate acid and hydroxide ions \( \text{OH}^- \). For the acetate ion, the reaction can be represented as:
\[ \text{CH}_3\text{COO}^- + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COOH} + \text{OH}^- \]
To calculate \( K_b \), we use the relationship between the ionization constant of water \( K_w \), the acidic dissociation constant \( K_a \) of acetic acid, and \( K_b \):

• \( K_w = 1.0 \times 10^{-14} \)
• For acetic acid, \( K_a = 10^{-4.74} \).
• So, \( K_b = \frac{K_w}{K_a} \).

By solving the equation, we find \( K_b = 1.8 \times 10^{-10} \), indicating how readily the acetate ion will produce hydroxide ions in solution.

OH- Concentration

The concentration of hydroxide ions, \( \text{OH}^- \), reflects the basicity of a solution. When calcium acetate dissolves, acetate ions react with water to produce hydroxide ions, as shown in the dissociation equation. This OH- concentration is vital for calculating the pH of the solution.
Here’s how to calculate it:

• Start by setting \( [\text{OH}^-] = x \).
• The initial acetate concentration is the sum of its contributions from complete dissociation, which is \( 0.01 \text{ M} \).
• Use the expression for \( K_b \): \( K_b = \frac{x^2}{0.01} \).
• Substitute \( 1.8 \times 10^{-10} \) for \( K_b \) and solve for \( x \).
• After solving, \( x \approx 1.34 \times 10^{-6} \text{ M} \).

This value represents the OH- concentration, which determines the solution's degree of basicity and is essential for the subsequent calculation of pOH and pH.

Conjugate Base

A conjugate base is what remains after an acid loses a proton. In the case of acetic acid \( \text{CH}_3\text{COOH} \), when it donates a proton, it turns into its conjugate base, the acetate ion \( \text{CH}_3\text{COO}^- \).
The role of the conjugate base is crucial in influencing the pH of a solution, especially when substances like calcium acetate are involved. This is because:

• The acetate ion can accept a proton from water, reflecting its ability to act as a base.
• This reaction forms acetic acid and hydroxide ions, which contribute to the solution's basic nature.

Understanding the concept of conjugate bases helps explain why certain solutions are basic or acidic, helping you predict changes in pH in response to the presence of specific ions.