Question

You make a 0.1 M solution with acid HA, a monoprotic acid. The Ka for this acid is 1.0E-8. If HA is mixed with a 50 M NaCl solution (no common ions with HA), determine the percent ionization of HA.

So I calculated that the ionic strength of the solution is equal to 0.2 and found the activity coefficent to be 0.7469.

When I plugged in the values into the Ka=(activity of A- * activity of H3O+)/activity of HA equation, I got x = 4.233E-5

Finally, I calculated the percent ionization to be (4.233E-5/0.1) * 100 = 0.0423 but it’s telling me that that is the wrong answer.

Can someone explain why?

Answer 1

Ka = CX^2

Ka of a monoprotic weak acid = 1.0*10^-8

C = concentration of monoprotic weak acid = 0.1 M

(1*10^-8) = 0.1*x^2

x = degree of dissociation = 0.00032

percent ionization = x*100

                   = 0.00032*100

                   = 0.032%