[art1]

Hi, I am having much difficulty with weak acid ionization constants. As far as i knwo it is done the same way the equilibrium constant is done but multiplied by the %of ionization and the concentration of the solution of the acid.

I was given this question and I do not understand why there is an X in the denominator.

calculate the concentration of H3O+ in .2M of methanoic acid (HCOOH) at 25%

HCOOH(aq) + H20(l) ----> H30(+1)(aq) +HCOO-(aq)

Ka= [H30(+)][HCOO]/HCOOH
then my teacher told me I can re write the equation as such

(1.(1^-4)=X^2 /.2M - X

I can see why everything is in that second equation except the -x in the denominator

[Borek]

http://www.chembuddy.com/?left=pH-calculation&right=pH-weak-acid-base

[mrdeadman]

ICE. initial change equilibrium Ka=[H+][HCOO-]/[HCOOH] at equilibrium the [H+] and [HCOO-] increase while [HCOOH] decreases the same amount as the increase therefore you solve for x. does that make sense or no?

[rctrackstar2007]

In actuallity, the x in the denominator, in this case, does not matter because the concentration (.2) and the Ka ((1^-4) which i think you mean 1 x 10^-4, but i'm not sure) are far enough apart that the concentration that the x in the denominator stands for will change an insignificant amount. I hope i didn't sound too confusing with that. If you don't understand reply and i will try to clear up any questions.