The 𝐾a of a monoprotic weak acid is 0.00289. What is the percent ionization of a 0.179 M solution of this acid?
- hcbiochemLv 74 weeks ago
HA <--> H+ + A-
Ka = 2.89X10^-3 = [H+][A-]/[HA]
Let [H+] = [A-] = x, and [HA] = 0.179 - x. As a first approximation, assume that x<<0.179 and so can be ignored in the denominator. Then,
2.89X10^-3 = x^2 / 0.179
x = 0.0227
Ignoring x compared to 0.179 is not appropriate in this situation. So, you will have to go back to:
2.89X10^-3 = x^2 / (0.179-x)
Rearrange this into the form ax^2 + bx + c = 0 and use the quadratic formula to solve for x.
Once you find x, the % ionization is calculated as:
% ionization = (x / 0.179) X 100