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Question

Chemistry Question on Law Of Chemical Equilibrium And Equilibrium Constant

The ionization constant of HF, HCOOH and HCN at 298K are 6.8×10–4,1.8×10–4 and 4.8×10–9 respectively. Calculate the ionization constants of the corresponding conjugate base.

Answer

It is known that Kb =KwKa\frac{K_w}{K_a}
Given, Ka of HF = 6.8 × 10-4 Hence, Kb of its conjugate base F = KwKa\frac{K_w}{K_a} = 10−146.8×10−4\frac{10^{-14}}{6.8\times10^{-4}} = 1.5×10-11
Given, Ka of HCOOH = 1.8 × 10-4 Hence, Kb of its conjugate base HCOO = KwKa\frac{K_w}{K_a}= 10−141.8×10−4\frac{10^{-14}}{1.8\times10^{-4}} = 5.6 × 10-11
Given, Ka of HCN = 4.8×10-9 Hence, Kb of its conjugate base CN = KwKa\frac{K_w}{K_a} = 10−144.8×10−9\frac{10^{-14}}{4.8\times10^{-9}} = 2.08 × 10-6