The Ka of a weak acid, HA, is 3x10-4.?

Calculate:
a) the OH- concentration in the solution
b) the degree of dissociation of theacid in a 0.15M solution

Answer:
a) the pH+pOH=14
the pH of this solution is equal to -log Ka
thus the pH is 3.52
resulting pOH is 10.48

OH concentration is 10^- pOH
thus the [OH] is 3.333x 10^-11

b) Ka is [H=][A-]/[HA]
thus
3x 10^-4=[x][x]/.15
0.000045=x^2
square root both sides
0.006708=x

.006708/.15
4.47% dissociated
Kaylie,

a) pH does not equal -log Ka, it equals -log [H+]. You cannot simply plug Ka in for the hydrogen ion concentration, you need to know the concentration of the acid and then solve for the equilibrium value of [H+] using Ka = [H+][A-] / [HA].

This part of the problem is not solveable unless you know the starting concentration of acid.

b) Kris's answer is correct.

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