The molar ratio of dimer to monomer for 0.1 M acetic acid in water (neglecting the dissociation of a β Ionic Equilibrium Chemistry Question
Question
The molar ratio of dimer to monomer for 0.1 M acetic acid in water (neglecting the dissociation of acetic acid in water) is equal to:
Answer: C
π‘ Solution & Explanation
In water, the equilibrium constant for dimerization is K = 3.6 Γ 10^-2 M^-1. Since K is very small, we can approximate [M] approx C = 0.1 M. Then, dimer concentration [D] approx K Γ [M]^2 = 0.036 Γ (0.1)^2 = 3.6 Γ 10^-4 M. Molar ratio [D]/[M] = 3.6 Γ 10^-4 / 0.1 = 3.6 Γ 10^-3 = 36 / 10000 = 9 / 2500.
π¬Ask on WhatsApp β
Still have doubts about this question?
Send it to our AI chemistry tutor on WhatsApp β gets answered in minutes