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: Find pOH
$pH + pOH = 14 \implies pOH = 14 - 10 = 4$.
Step 2: Find $[OH^-]$
$[OH^-] = 10^{-pOH} = 10^{-4}$ M.
Step 3: Calculate degree of dissociation ($\alpha$)
For a monoacidic base, $[OH^-] = c \alpha$. $10^{-4} = 0.01 \times \alpha \implies \alpha = \frac{10^{-4}}{10^{-2}} = 10^{-2} = 0.01$.
Step 4: Percentage dissociation
$% \text{ dissociation} = \alpha \times 100 = 0.01 \times 100 = 1%$.
Final Answer:(D)