Ionic EquilibriummediumMCQ SINGLE

Ka for HF is 3.5 × 10–4. Calculate Kb for the fluoride ion.Ionic Equilibrium Chemistry Question

Question

Ka for HF is 3.5 × 10–4. Calculate Kb for the fluoride ion.

Answer: C

💡 Solution & Explanation

HF and F are conjugate Acid-Base pair (CABP) for CABP, a b w K K = K x 14 11 w b 4 a K 10 K 2.86 10 K 3.5 10 x x IONIC EQUILIBRIUM 43 Salt Hydrolysis

💬
Still have doubts about this question?
Send it to our AI chemistry tutor on WhatsApp — gets answered in minutes
Ask on WhatsApp →

Practice 22,000+ questions like this

AI-adaptive practice, video lectures, and full JEE Advanced Chemistry content — all in one place.

JEE Advanced · JEE Mains · NEET · IChO · AP Chemistry