Question

If ionisation constant of acetic acid is 1.8 . 10^(-5), at what concentration will it be dissociated to 2%

Answer 1

At a concentration of 0.045[M], acetic acid will be dissociated to 2%

Acetic acid (CH₃COOH) dissociates into two ions- CH₃COO⁻ and H⁺

                         CH₃COOH ⇄ CH₃COO⁻ +  H⁺

Initial conc.:          c                          0            0

At equillibrium:     c(1-α)                  cα           cα

where α is the degree of dissociation = 2% = 2/100 = 0.02

Ionisation constant K = [(cα)(cα)]/[c(1-α)]  =  cα²/(1-α) = cα²      [1-α≈1, as α is very small]

C = K/α²  = (1.8×10⁻⁵)/(0.002)²  = 0.045 [M]

This is the required concentration.