Chemistry-(general)
NEET Chemistry (general) MCQ Question
Type: MCQ-numerical-Medium-Class 11
Given that the ionization constant of HF at 298K is 6.8 × 10⁻⁴, calculate the ionization constant of its conjugate base, F⁻.
A
1.47 × 10⁻¹¹
B
1.47 × 10⁻¹⁰
C
1.47 × 10⁻⁷
D
6.8 × 10⁻⁴
Correct Answer
Option A
Detailed Explanation
The ionization constant of a conjugate base is given by Kw divided by the ionization constant of the acid. Kw at 298K is 1.0 × 10⁻¹⁴. Therefore, for F⁻, it is 1.0 × 10⁻¹⁴ / 6.8 × 10⁻⁴ = 1.47 × 10⁻¹¹.
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