Question

A 0.15 M solution of a weak acid is found to be 1.3% ionized. What is its K_a?

Answer 1

Solution:

We write general formula for weak acid and show the dissociation in water. Set up ICE chart and then find equilibrium unknown concentration. By using initial concentration and percent ionization we get equilibrium concentration of each species. We use ka expression for acid then to calculate ka from equilibrium.

      HA (aq) + H2O (l) -------> A- (aq) + H3O+ (aq)

I    0.15                                    0               0

C -x                                        +x             +x

E (0.15-x)                                  x                   x

Lets write ka expression for this reaction.

Ka = x²/ (0.15-x)

% ionization = x / Initial molarity

1.3 = (x / 0.15 ) * 100

x = 0.00195

lets plug this value in ka expression

ka = (0.00195)²/ (0.15-0.00195)

=2.56 X10^-5

Ka = 2.56 X 10^-5