Question

Ka for a weak acid HA is 2x10^-5. What is the value of Kb for A^-

Answer 1

Ka × Kb = Kw=10^-14
I hope u get it

Answer 2

Answer: $K_b=5\times 10^4$

Explanation:

Equilibrium constant is the ratio of product of concentration of products to the product of concentration of reactants each term raised to their stochiometric ratios.

$HA\rightleftharpoons H^++A^-$

$K_a=\frac{[H^+][A^-]}{[HA]}$

$2\times 10^{-5}=\frac{[H^+][A^-]}{[HA]}$   (1)

$H^++A^-\rightleftharpoons HA$

$K_b=\frac{[HA]}{[H^+][A^-}$   (2)

Dividing 2 by 1

$K_b=\frac{1}{K_a}$

$K_b=\frac{1}{2\times 10^{-5}}$

$K_b=5\times 10^4$