JEE Main Chemistry question
The dissociation constant for acetic acid and HCN at 25°C are 1.5×10⁻⁵ and 4.5×10⁻¹⁰ respectively. The equilibrium constant for the equilibrium, CN⁻+CH₃COOH⇌HCN+CH₃COO⁻ would be:
Answer: (A)
Solution
K = Ka(CH₃COOH)/Ka(HCN) = 1.5×10⁻⁵/4.5×10⁻¹⁰ ≈ 3.3×10⁴.