Question

The dissociation constant of a weak acid HA and weak base BOH are 2×10^-5 and 5×10^-6 respectively.The equilibrium constant for the neutralization reaction of the two is ( ignore hydrolysis of resulting salt)

Answer 1

Dissociation constant for weak acid (Ka) = 2 x 10^-5
Dissociation constant for weak base (Kb) = 5 x 10^-6
Consider the weak acid can be represented as HA and weak base can be represented as BOH.
Neutralisation reaction: HA + BOH ⇌​ A^- + BH⁺
HA +H₂O ⇌​ A^- + H₃O⁺ K_a
BOH + H₂O ⇌​ BH⁺ + OH^- K_b
H₃O⁺ + OH^- ⇌​ 2H₂O 1/K_w (Kw is tthe dissociation constant of water)
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Net Reaction: HA + BOH ⇌​ A^- + BH⁺
Thus the equilibrium constant of the neutralisation reaction will be the product of the equilibrium constant of the above three reactions and it can be written as
K = $\frac{K_a{K}_b}{K_w}$


Equilibrium constant for the neutralisation reaction (K) = $\frac{K_a{K}_b}{K_w}=\frac{2\times {10}^{-5}\times 5\times {10}^{-6}}{{10}^{-14}}={10}^4$