# The 𝐾a of a monoprotic weak acid is 0.00289. What is the percent ionization of a 0.179 M solution of this acid?

### 1 Answer

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- 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

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