MCQ
0.10M CH3COOH is 1.34% ionised, calculate its Ka:
- A1.8 × 10-5
- B1.8 × 10-4
- C5 × 10-4
- D4 × 10-5
✓
Answer
- 1.8 × 10-5
Explanation:
$\text{K}_{\text{a}}=\text{C}\alpha^2=0.1\times\frac{1.34}{100}\times\frac{1.34}{100}=1.795\times10^{-5}$
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