MCQ
$0.10M \ \ce{CH_3COOH}$ is $1.34\%$ ionised, calculate its $\ce{K_a}$ :
- ✓$ 1.8 \times 10^{-5} $
- B$ 1.8 \times 10^{-4} $
- C$ 5 \times 10^{-4} $
- D$ 4 \times 10^{-5} $
✓
Answer
Correct option: A.
$ 1.8 \times 10^{-5} $
$\text{K}_{\text{a}}=\text{C}\alpha^2=0.1\times\frac{1.34}{100}\times\frac{1.34}{100}=1.795\times10^{-5}$
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