...........HCOOH ==> H^+ + HCOO^-
initial....0.125M.....0......0
change.......-x........x......x
equil.....0.125-x......x......x
The ICE chart is set up above.
Ka = 1.8E-4 = (H^+)(HCOO^-)/(HCOOH)
Substitute from the ICE chart and solve for (H^+).
calculate the hydronium ion concentration in 0.125M formic acid, HCHO2
Ka= 1.8 x 10-4
can you show me the steps pls
5 answers
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 10-4= x x/ 0.125
(1.84 x 10-4)[0.125]
2.3 x 10-5= x = [H30]
lost which way do I go
Ka=1.84 x 10-4= x x/ 0.125
(1.84 x 10-4)[0.125]
2.3 x 10-5= x = [H30]
lost which way do I go
Ka = (x)(x)/(0.125-x) = 1.84E-4
We make the problem a little easier to solve if we assume x is small and that 0.125-x = 0.125. Then
1.84E-4 = x^2/0.125
1.84E-4*0.125 = x^2
2.3E-5 = x^2
x = sqrt(2.3E-5) = 0.00480M = (H3O^+).
We make the problem a little easier to solve if we assume x is small and that 0.125-x = 0.125. Then
1.84E-4 = x^2/0.125
1.84E-4*0.125 = x^2
2.3E-5 = x^2
x = sqrt(2.3E-5) = 0.00480M = (H3O^+).
4 answers