Question

The pH of 0.01M BOH solution is 10. What is its degree of dissociation? (Given $ K_b $ of BOH is $ 1 \times 10^{-6} $)

• A. 10%
• B. 5%
• C. 2%
• D. 1%

Hint:

Use the relationship between pH and pOH, and the definition of the degree of dissociation for a weak base.

Answer 1

The Correct Option is D

To solve this problem, we need to find the degree of dissociation of BOH in a 0.01M solution given that the pH is 10. Here's the step-by-step solution:

Step 1: Calculate the pOH. The solution's pH is given as 10. Since the relationship between pH and pOH is: pH + pOH = 14, we find:

pOH = 14 - 10 = 4

Step 2: Calculate the OH⁻ concentration. The pOH is related to the hydroxide ion concentration by the formula: [OH⁻] = 10^-pOH.

[OH⁻] = 10^-4 = 1 × 10⁻⁴ M

Step 3: Use the K_b expression for BOH. The base dissociation constant K_b can be expressed as:

K_b = [BH⁺][OH⁻] / [BOH]

Assuming the degree of dissociation is α, then:[BOH] = 0.01M(1 - α) ≈ 0.01M since α is small.

[BH⁺] = [OH⁻] = 1 × 10⁻⁴ M

Substitute these into the K_b expression:

1 × 10⁻⁶ = (1 × 10⁻⁴)² / (0.01(1 - α)) ≈ (1 × 10⁻⁴)² / 0.01

1 × 10⁻⁶ = 1 × 10⁻⁸ / 0.01 = 1 × 10⁻⁸ × 100

1 × 10⁻⁶ = 1 × 10⁻⁶

Step 4: Solve for α (degree of dissociation). Since [OH⁻] = α[BOH] = α × 0.01 M, and we calculated [OH⁻] as 1 × 10⁻⁴ M:

α × 0.01 = 1 × 10⁻⁴

α = (1 × 10⁻⁴) / 0.01 = 0.01 or 1%

Thus, the degree of dissociation of the BOH solution is 1%.