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Acid-base equlibrium

  1. Mar 11, 2008 #1
    1. The problem statement, all variables and given/known data
    1) A 0.682-gram sample of an unknown weak monoprotic organic acid, HA, was dissolved in sufficient water to make 50 milliliters of solution and was titrated with a 0.135-molar NaOH solution. After the addition of 10.6 milliliters of base, a pH of 5.65 was recorded. The equivalence point (end point) was reached after the addition of 27.4 milliliters of the 0.135-molar NaOH.

    (a) Calculate the number of moles of acid in the original sample.
    Done. 3.70 * 10^-3 mol HA
    (b) Calculate the molecular weight of the acid HA.
    Done. 184.37 g/mol
    (c) Calc the number of moles of unreacted HA remaining in solution when the pH was 5.65.
    done. 0.0027 mol
    (d) Calculate the [H3O+] at pH = 5.65.
    done. 2.2 * 10^-6
    (e) Calculate the value of the ionization constant, Ka, of the acid HA.
    I am using an ICE chart, with the equation HA <-> H+ + A-
    The initial concentration for HA I used was 0.0027 mol/ 0.050 L
    the initial concentration for H+ was 2.2 * 10 ^-6, which is also equal to the concentration of A-.
    However, when I plug in these values into the equilibrium expression, Ka = ([H+][A-])/[HA], I get 1.1 * 10^ -10, and the answer key says that the correct answer is 1.4 * 10 ^-6.

    Am I doing something wrong here, or is the answer key wrong?
     
  2. jcsd
  3. Mar 11, 2008 #2

    chemisttree

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    Show your work on e). What values of [H+], [A-] and [HA] did you use?
     
  4. Mar 11, 2008 #3
    Ka = [tex]\frac{[H+][A-]}{[HA]}[/tex]

    [H+] = 2.2 * 10^-6 (got this from pH 5.65)
    [A-] = 2.2 * 10^-6
    [HA] = 0.0027 mol/ 0.050 L
     
  5. Mar 11, 2008 #4
    If you write out all of the equations you have and plug them into each other, you'll see that [H+] is not equal to [A-]. Your assumption that they are equal is throwing you off.

    Also don't forget that the volume of the entire solution changed because you added the NaOH solution.
     
  6. Mar 11, 2008 #5
    thanks.
    I forgot about all the [A-] ions that came into the system as a result of the neutralization of the weak acid with NaOH.
     
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