Question

If enough of a monoprotic acid is dissolved in water to produce a 0.0172 M solution with a pH of 6.68, what is the equilibrium constant, Ka, for the acid?

Answer 1

Let the monoprotic acid be denoted by HA

HA dissociates in aqueous solution as follows,

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

Given,

[HA] = 0.0172 M

pH = 6.68

We know that,

[H3O+] = 10^(-pH)

=> [H3O+] = 10^(-6.68) = 2.09 x 10^-7 M

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

0.0172-X ...........X......X

X = [H3O+] = 2.09 x 10^-7 M

Ka = [A-] [H3O+] / [HA]

=> Ka = (2.09 x 10^-7) x (2.09 x 10^-7) / (0.0172 - 2.09 x 10^-7)

=> Ka = 2.54 x 10^-12 = Equilibrium Constant for the Acid