The pH of 0.5 M aqueous solution of HF (Ka = 2 × 10–4) is | Ionic Equilibrium — Chem-Mantra | Chem-Mantra

⚗️ Chem-Mantra NEET Chemistry →

Ionic EquilibriummediumMCQ SINGLE

Question

The pH of 0.5 M aqueous solution of HF (Ka = 2 × 10–4) is

Answer: A

💡 Solution & Explanation

From the given conditions, we know that- 2 2 2 2 2 a 2 10 [H ] C 0.5 2 10 10 pH log[ ] pH log10

💬

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 NEET Chemistry content — all in one place.

Explore Courses →Free on YouTube

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