Formic acid (hcooh) is secreted by ants. calculate [h3o+] for a 1.17e-2 m aqueous solution of formic acid (ka = 1.80e-4).

HCOOH + H2O <---> HCOO- + H3O+ 

Ka = 1.8 X 10^-4 = [HCOO-][H3O+] / [HCOOH] 

In the solution, [HCOO-] = [H3O+] = x 
Quite properly, [HCOOH] = (2.30 X 10^-3 - x). 

Since the formic acid solution is pretty dilute, and since Ka is sort of small, we can initially assume that x will be small compared to 2.3 X 10^-3, and so we can ignore it. If we do that, then, 

Ka = x^2 / 2.30X10^-3 = 1.80 X 10^-4 
x = 6.4 X 10^-4 = [H3O+] 
pH = 3.19 

Now, quite properly, our assumption that x would be small compared to 2.3 X 10^-3 is incorrect, and we really cannot ignore x in that expression. So, we should go back to the original expression for Ka: 

Ka = x^2 / (2.30 X 10^-3 - x) = 1.80 X 10^-4 

Quite properly, you should rearrange this into a quadratic form and use the quadratic equation to solve for x. Once you've done that, x = [H3O+], and pH = - log (x).

You need to divided twenty four by a dozen them by the temperature it is getting cook by

Particle exchange energy though elastic collision , hope this .