Question

Dissociation constant of 0.01 M weak acid is $10^{-4}$. What is percent dissociation of acid?

• A. 2%
• B. 6%
• C. 10%
• D. 1.5%

Hint:

Ostwald's shortcut $\alpha = \sqrt{K_a/C}$ is incredibly fast and works perfectly for 99% of competitive exam problems where $K_a \le 10^{-4}$. Always try the shortcut first!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given the initial concentration ($C$) and the acid dissociation constant ($K_a$) of a weak acid. We need to calculate its percent dissociation (degree of dissociation $\alpha$ expressed as a percentage).

Step 2: Detailed Explanation:
According to Ostwald's Dilution Law for weak electrolytes, the relationship between the dissociation constant ($K_a$), the initial concentration ($C$), and the degree of dissociation ($\alpha$) is:
$K_a = \frac{C\alpha^2}{1 - \alpha}$
For weak acids, the degree of dissociation is typically very small ($\alpha \ll 1$), so we can safely approximate $(1 - \alpha) \approx 1$.
The simplified formula becomes:
$K_a = C\alpha^2$
Rearrange to solve for $\alpha$:
$\alpha = \sqrt{\frac{K_a}{C}}$
Substitute the provided values:
$K_a = 10^{-4}$
$C = 0.01 \text{ M} = 10^{-2} \text{ M}$
$\alpha = \sqrt{\frac{10^{-4}}{10^{-2}}}$
$\alpha = \sqrt{10^{-2}}$
$\alpha = 10^{-1} = 0.1$
The degree of dissociation is 0.1. (Note: Since $\alpha = 0.1$ is 10%, which is exactly at the borderline of where the approximation $(1-\alpha) \approx 1$ starts to induce minor errors, it is standard practice in multiple-choice exams to accept this simplified result unless options demand quadratic precision).
To find the percent dissociation, multiply $\alpha$ by 100:
$% \text{ Dissociation} = \alpha \times 100%$
$% \text{ Dissociation} = 0.1 \times 100%$
$% \text{ Dissociation} = 10%$

Step 3: Final Answer:
The percent dissociation is 10%, matching option (c).