chemistry
posted by Monique on .
calculate the hydronium ion concentration in 0.125M formic acid, HCHO2
Ka= 1.8 x 104
can you show me the steps pls

...........HCOOH ==> H^+ + HCOO^
initial....0.125M.....0......0
change.......x........x......x
equil.....0.125x......x......x
The ICE chart is set up above.
Ka = 1.8E4 = (H^+)(HCOO^)/(HCOOH)
Substitute from the ICE chart and solve for (H^+). 
DrBob is the HCOOH>H^++HCOO^ the HCHO2

Yes. HCOOH is formic acid. The H on the right side comes from the COOH or an organic molecule, called a carboxylic acid, and that can ionize.

I'm stuck
Ka=1.84 x 104= x x/ 0.125
(1.84 x 104)[0.125]
2.3 x 105= x = [H30]
lost which way do I go 
Ka = (x)(x)/(0.125x) = 1.84E4
We make the problem a little easier to solve if we assume x is small and that 0.125x = 0.125. Then
1.84E4 = x^2/0.125
1.84E4*0.125 = x^2
2.3E5 = x^2
x = sqrt(2.3E5) = 0.00480M = (H3O^+).