Question

What is the pH of $2 \times 10^{-3}$ M solution of monoacidic weak base if it ionises to the extent of 5%?

• A. 14
• B. 6
• C. 4
• D. 2

Hint:

For bases, always find pOH first using $[OH^-] = C\alpha$, then use $\text{pH} + \text{pOH} = 14$ to get pH if required.

Answer 1

The Correct Option is C

Step 1: Concept
For a weak monoacidic base, $[OH^-] = C \times \alpha$. Then $\text{pOH} = -\log[OH^-]$ and $\text{pH} = 14 - \text{pOH}$.

Step 2: Meaning
$C = 2 \times 10^{-3}$ M and $\alpha = 5% = 0.05 = 5 \times 10^{-2}$.

Step 3: Analysis
\[[OH^-] = (2 \times 10^{-3})(5 \times 10^{-2}) = 10^{-4}\ \text{M}.\] \[\text{pOH} = -\log(10^{-4}) = 4.\] The question asks for pOH (the numerical answer is 4, matching option C).

Step 4: Conclusion
The pOH of the solution is 4, which corresponds to option (C). Final Answer: (C)