Question

2. If an acid HA has a Ka value of 1.5 x 10-5, what will be the value of equilibrium constant for the following reaction? Hint: You need to use Rxns 1 and 3 and Eqs 1 and 3 from the Background material. HA(aq) + OH-(aq) ↔ A-(aq) + H2O(l) Will this reaction be reactant favored or product favored? What does this tell you about the “completeness” of the titration reactions used in this laboratory?

Answer 1

Given,

HA + H2O ------> A- + H3O+

Ka for the above reaction = 1.5 x 10^-5

We know that,

Ka * Kb = 10^-14

=> Kb for A- = 10^-14 / Ka = 10^-14 / 1.5 x 10^-5 = 6.667 x 10^-10

The reaction of A- as base is

A- + H2O -----> HA + OH- ------------- Kb = 6.667 x 10^-10 ------------ (1)

We need to find the Equilibrium COnstant value for the reaction,

HA(aq) + OH-(aq) ↔ A-(aq) + H2O(l) which is inverse of the reaction (1)

=> Keq = 1 / Kb = 1.5 x 10^9,

Since Keq >> 1, Products are favoured