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: Plan
Write the equilibrium concentrations and substitute into $K_a$.

Step 2: Equilibrium
Start with $C$ M of HA. At equilibrium $[\text{H}^+] = [\text{A}^-] = C\alpha = 0.01C$, and $[\text{HA}] = C(1-\alpha) \approx 0.99C$.

Step 3: Solve
$K_a = \dfrac{(0.01C)^2}{0.99C} \approx 10^{-4}C$. So $C = 1.8\times10^{-5}/10^{-4} = 0.18$ M.

Step 4: Result
The answer is $1.8 \times 10^{-1}$ M. A tiny $\alpha$ of 1% justifies dropping $\alpha$ next to 1.

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