# chemistry

posted by 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.

• chemistry - ,

HCOOH ==> H^+ + HCOO^-

pH = 3.26, solve pH = -log(H^+) for(H^+).
Then [(H^+)(HCOO^-)]/[(HCOOH)-(H^+)]=Ka

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.

• chemistry - ,

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