Question

The pH of monoacidic base is 10. Calculate its percentage dissociation in 0.01 M solution at 298 K ?

• A. 10%
• B. 5%
• C. 2%
• D. 1%

Hint:

$\text{pOH} = 14 - \text{pH}$ and $[OH^-] = c\alpha$ for weak monoacidic bases.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given the pH of a monoacidic base solution and its concentration. We need to find the percentage dissociation (\( \alpha \times 100 \)).
Step 2: Key Formula or Approach:
1. \( pOH = 14 - pH \)
2. \( [OH^{-}] = 10^{-pOH} \)
3. For a weak base, \( [OH^{-}] = C \alpha \), where \( C \) is concentration and \( \alpha \) is degree of dissociation.
4. Percentage dissociation = \( \alpha \times 100 \)
Step 3: Detailed Explanation:
Given: \( pH = 10 \), \( C = 0.01\text{ M} = 10^{-2}\text{ M} \).
At 298 K:
\[ pOH = 14 - 10 = 4 \]
\[ [OH^{-}] = 10^{-pOH} = 10^{-4}\text{ M} \]
Using the relation for degree of dissociation:
\[ \alpha = \frac{[OH^{-}]}{C} = \frac{10^{-4}}{10^{-2}} = 10^{-2} = 0.01 \]
Calculating percentage dissociation:
\[ \text{Percentage dissociation} = \alpha \times 100 = 0.01 \times 100 = 1% \]
Step 4: Final Answer:
The percentage dissociation of the monoacidic base is 1%.