Question

The pH of a 0.1 M monoacidic base is 11.11. What is the per cent dissociation of base?

Answer 1

Given : 

pH = 11.11; 

percent Dissociation of base = ? 

c = 0.1 M

pH + pOH = 14 

∴ pOH = 14 – pH = 14 – 11.11 = 2.89 

POH = -log₁₀[OH^-]

∴ [OH^-] = Antilog – pOH

= Antilog – 2.89

= Antilog \(\bar 3.11\)

= 1.29 x 10^-3 M

∵ [OH^-] = cα

∴ α = \(\frac{[OH^-]}{c}\)

= \(\frac{1.29 \times 10^{-3}}{0.1}\)

= 1.29 x 10^-2

∴ Percent dissociation = α × 100 

= 1.29 × 10^-2 × 100 

= 1.29

∴ Percent dissociation = 1.29