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March 27, 2017

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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

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