Question

An organic monobasic acid has dissociation constant \(1.96\times 10^{-8}\). What is its percentage dissociation in \(0.01\) M solution ?

• A. \(1.40\,\%\)
• B. \(0.14\,\%\)
• C. \(2.4\,\%\)
• D. \(0.19\,\%\)

Hint:

For a weak acid alpha = sqrt(Ka/C). Then multiply by 100.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
For a weak monobasic acid, the degree of dissociation is small, so Ostwald's dilution law can be used in its approximate form.

Step 2: Key Formula or Approach:
\[ K_a = C\alpha^2 \;\Rightarrow\; \alpha = \sqrt{\frac{K_a}{C}} \]

Step 3: Detailed Explanation:
Given \(K_a = 1.96\times10^{-8}\) and \(C = 0.01\) M.
\[ \alpha = \sqrt{\frac{1.96\times10^{-8}}{10^{-2}}} = \sqrt{1.96\times10^{-6}} = 1.4\times10^{-3} \]
Percentage dissociation:
\[ \alpha \times 100 = 1.4\times10^{-3}\times100 = 0.14\,\% \]
The value is small, so the approximation \(1-\alpha \approx 1\) is valid. Option (A) 1.40 % comes from missing a factor of ten. Options (C) 2.4 % and (D) 0.19 % do not match \(\sqrt{K_a/C}\).

Final Answer:
The acid is 0.14 % dissociated, option (B). \[ \boxed{0.14\,\% \text{ (B)}} \]