MCQ
A $0.1\, M$ solution of $HF$ is $1\%$ ionized. What is the $K_a$
- ✓${10^{ - 5}}$
- B${10^{ - 4}}$
- C$3 \times {10^{ - 5}}$
- D$3 \times {10^{ - 4}}$
✓
Answer
Correct option: A.
${10^{ - 5}}$
a
$\mathrm{K}_{\mathrm{a}}=\frac{\mathrm{C} \alpha^{2}}{1-\alpha}=\frac{0.1\left(\frac{1}{100}\right)^{2}}{1-0.01}=10^{-5}$
$\mathrm{K}_{\mathrm{a}}=\frac{\mathrm{C} \alpha^{2}}{1-\alpha}=\frac{0.1\left(\frac{1}{100}\right)^{2}}{1-0.01}=10^{-5}$
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