Q26 / 150 · MHT CET 2026 (19 April Shift 2) · Chemistry
Calculate the molar concentration of weak monobasic acid if $K_a = 1.8 \times 10^{-5}$ and degree of dissociation is $0.01$ ?
- Correct answerA$1.8 \times 10^{-1}$
- B$1.8 \times 10^{-2}$
- C$5.55 \times 10^{-3}$
- D$5.55 \times 10^{-4}$
Solution
For a weak monobasic acid, the dissociation constant is given by:
$K_a = \dfrac{C\alpha^2}{1-\alpha}$
Since $\alpha = 0.01$, which is very small compared to $1$, we can approximate $1-\alpha \approx 1$.
$K_a = C\alpha^2$
Rearranging for $C$:
$C = \dfrac{K_a}{\alpha^2}$
Substituting the given values:
$C = \dfrac{1.8 \times 10^{-5}}{(0.01)^2}$
$C = \dfrac{1.8 \times 10^{-5}}{10^{-4}}$
$C = 1.8 \times 10^{-1}$ M
Answer: $1.8 \times 10^{-1}$
$K_a = \dfrac{C\alpha^2}{1-\alpha}$
Since $\alpha = 0.01$, which is very small compared to $1$, we can approximate $1-\alpha \approx 1$.
$K_a = C\alpha^2$
Rearranging for $C$:
$C = \dfrac{K_a}{\alpha^2}$
Substituting the given values:
$C = \dfrac{1.8 \times 10^{-5}}{(0.01)^2}$
$C = \dfrac{1.8 \times 10^{-5}}{10^{-4}}$
$C = 1.8 \times 10^{-1}$ M
Answer: $1.8 \times 10^{-1}$
Hence, the correct answer is option: 'A'.