Chem 2 Help -Reworking problems to get ready for test

Question

A 40.9 mL sample of a 0.17 M solution of a weak acid, HA, is placed into a flask. The pH of the solution is 4.38. What is Kb for the conjugate base, A-, of this acid at the temperature of the experiment?

Answer 1

First calculate Ka of the acid

Ka = [A⁻] ∙ [H⁺] / [HA]

Ignoring the rather small effect of self ionization of water, you can assume that all the protons are formed due to ionization of HA. So

[A⁻] = [H⁺] = 10^(-pH)

and

[HA] = [HA]₀ - [H⁺]  

=>

Ka = (10^-4.38 )² / (0.17 - 10^-4.38 ) = 1.0×10^-8

Ionization constant of any acid and its conjugate base are related as:

Ka∙Kb = 10^-14

hence:

Kb = 10^-14 / 1.0×10^-8 = 1.0×10^-6