Question

Calculate the molar concentration of weak monobasic acid if \(K_a = 1.8\times 10^{-5}\) and degree of dissociation is \(0.01\) ?

• A. \(1.8\times 10^{-1}\)
• B. \(1.8\times 10^{-2}\)
• C. \(5.55\times 10^{-3}\)
• D. \(5.55\times 10^{-4}\)

Hint:

Use Ostwald dilution law, Ka = C alpha squared, for a weak acid.

Answer 1

The Correct Option is A

Step 1: Understand the concept
For a weak monobasic acid HA of concentration \(C\), the Ostwald dilution law says \(K_a = \dfrac{C\alpha^2}{1-\alpha}\). When \(\alpha\) is very small, \(1 - \alpha \approx 1\) and the law becomes \(K_a = C\alpha^2\).

Step 2: Rearrange
\[ C = \frac{K_a}{\alpha^2} \]

Step 3: Substitute
Given \(K_a = 1.8 \times 10^{-5}\) and \(\alpha = 0.01\):
\[ C = \frac{1.8 \times 10^{-5}}{(0.01)^2} = \frac{1.8 \times 10^{-5}}{10^{-4}} = 1.8 \times 10^{-1}\ \text{M} \]

Step 4: Check the options
The value 1.8e-2 would need \(\alpha^2\) to be 1e-3, and 5.55e-3 and 5.55e-4 come from inverting the ratio or mishandling the powers of ten. Only 0.18 M matches the correct formula.

Final Answer:
The molar concentration is 0.18 M. This is option (A). \[ \boxed{\text{(A) }1.8 \times 10^{-1}\ \text{M}} \]