Question

What is molar concentration of weak monobasic acid if dissociation constant is $5 \times 10^{-8}$ and undergoes $0.5\%$ dissociation?

• A. $0.03\ \text{M}$
• B. $0.002\ \text{M}$
• C. $0.001\ \text{M}$
• D. $0.005\ \text{M}$

Hint:

When dealing with Ostwald's approximation equations, isolate the arithmetic steps: dividing $5$ by $25$ leaves a factor of $\frac{1}{5}$ or $0.2$. This step lets you confidently pick option (B) without over-analyzing the scientific powers of ten.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The problem asks for the total molar concentration ($c$) of a weak monobasic acid given its acid dissociation constant ($K_a$) and its corresponding degree of dissociation ($\alpha$) expressed as a percentage value.

Step 2: Key Formula or Approach: According to Ostwald's Dilution Law for a weak electrolyte, the equilibrium dissociation constant is linked to its concentration and degree of dissociation by the expression:
$$ K_a = \frac{\alpha^2 c}{1 - \alpha} $$ Since the acid undergoes a very low extent of dissociation ($0.5\%$), $\alpha$ is much smaller than 1. Thus, the denominator $(1 - \alpha)$ safely simplifies to approximately 1, yielding:
$$ K_a = \alpha^2 c \implies c = \frac{K_a}{\alpha^2} $$

Step 3: Detailed Explanation:
Let's first convert the percentage degree of dissociation into its absolute decimal fraction form:
$$ \alpha = \frac{0.5}{100} = 5 \times 10^{-3} $$ Now, substitute the given values of $K_a = 5 \times 10^{-8}$ and $\alpha$ into the rearranged relationship:
$$ c = \frac{5 \times 10^{-8}}{(5 \times 10^{-3})^2} $$ $$ c = \frac{5 \times 10^{-8}}{25 \times 10^{-6}} $$ $$ c = \frac{1}{5} \times 10^{-2} $$ $$ c = 0.2 \times 10^{-2} = 0.002\ \text{M} $$

Step 4: Final Answer:
The molar concentration of the weak acid solution is $0.002\ \text{M}$, which matches option (B).