Question

What is the pH of \(2 \times 10^{-3}\ \text{M}\) solution of a monacidic weak base if it ionises to the extent of \(5\%\)?

• A. \(14\)
• B. \(10\)
• C. \(4\)
• D. \(2\)

Hint:

For weak bases: \[ [OH^-] = C\alpha \] where:
• \(C\) = concentration
• \(\alpha\) = degree of ionization Then use: \[ pOH = -\log[OH^-] \] and finally: \[ pH = 14 - pOH \] Always convert percentage ionization into decimal form before substitution.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Weak bases do not ionize completely in solution. The concentration of hydroxide ions (\( [OH^-] \)) produced depends on the molarity (\( C \)) and the degree of ionization (\( \alpha \)).
For a monacidic weak base:
\[ BOH \rightleftharpoons B^+ + OH^- \]
\[ [OH^-] = C \alpha \]
Step 2: Key Formula or Approach:
1. Degree of ionization (\( \alpha \)) = \( \frac{\text{Percentage Ionization}}{100} \).
2. \( [OH^-] = C \times \alpha \).
3. \( \text{pOH} = -\log[OH^-] \).
4. \( \text{pH} + \text{pOH} = 14 \).
Step 3: Detailed Explanation:
Given:
\( C = 2 \times 10^{-3} \text{ M} \)
Percentage Ionization = 5%, so \( \alpha = \frac{5}{100} = 0.05 \).

First, calculate \( [OH^-] \):
\[ [OH^-] = (2 \times 10^{-3}) \times (0.05) \]
\[ [OH^-] = (2 \times 10^{-3}) \times (5 \times 10^{-2}) \]
\[ [OH^-] = 10 \times 10^{-5} = 10^{-4} \text{ M} \]

Now, calculate pOH:
\[ \text{pOH} = -\log(10^{-4}) = 4 \]

Finally, find pH:
\[ \text{pH} = 14 - \text{pOH} \]
\[ \text{pH} = 14 - 4 = 10 \]
Step 4: Final Answer:
The pH of the weak base solution is 10.