posted by jin on .
how do you solve this problem??
What is the original molarity of a solution of formic acid (HCOOH) whose pH is 3.26 at equilibrium?
the answer is 2.3 x 10^-3M.
HCOOH ==> H^+ + HCOO^-
pH = 3.26, solve pH = -log(H^+) for(H^+).
I don't know what you are using for Ka. Using 1.77 x 10^-4 for Ka, which I obtained from an old book of mine, I arrived at 0.00226 M.
First, the idea of pH = -log [H+] where [H+] is the molarity of H+
This can be rearranged to:
[H+] = 10^-pH
so from the pH you know the molarity of [H+] = 10^-3.26, which equals 5.4954 * 10^-4 M
Next, the idea of a weak acid and Ka. A weak acid has the form:
HA --> H+ + A-
and Ka = [H+][A-]/[HA]
The Ka for formic acid is 1.7 * 10^-4
Set-up the table
HA H+ A-
Initial x 0 0
Final x - 0.0005954 0.0005954 0.0005954
The x is the original molarity of the acid. The 0.0005954 is the molarity of the H+ as told by the pH. Whatever amount of H+ is made is the same as the amount of A-, and the amount of HA gone. So:
1.7 * 10^-4 = [0.0005954] * [0.0005954]/ [x -0.0005954]
1.7*10^-4 * x - 9.89*10^-8 = 3.545 * 10^-7
1.8*10^-4 * x = 4.534 * 10^-7
x = 0.0023 M
Pretty close to the answer they got. I may have messed up using this computer's calculator, but that's the process:
1. Use pH to get molarity of H+
2. Set-up a table to get the values for HA, H+ and A-
3. Plug values into Ka equation