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Dissociation constant from absorbance

  1. Oct 21, 2008 #1
    1. The problem statement, all variables and given/known data

    A 0.120 M solution of sodium picrate ina 1 M sodium hydroxide solution was observed to have an absorbance of 0.335, due only to the absorption by the picrate anion. In the same spectrophotometer cell and at the same wavelength as in the previous measurement, a 0.300 M solution of picric acid was found to have an absorbance of 0.581. Calculate the dissociation constant for picric acid

    2. Relevant equations

    A=abc

    Ka = [A]/[AB]

    3. The attempt at a solution

    I am not sure where to begin but this is what I thought so far:

    0.335 = 0.120 a a=2.792
    0.581 = .300 a a = 1.937

    I am not sure how to use the absorption values in the dissociation constant equation.
     
  2. jcsd
  3. Oct 21, 2008 #2

    Borek

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    Staff: Mentor

    Convert them to concentrations.

    Hint: how much undissociated weak acid can be present in the 1M NaOH solution?
     
  4. Oct 21, 2008 #3
    I thought I was given the concentrations in the question?

    will the [OH]- not matter because [H]+ >> [OH]-?
     
  5. Oct 21, 2008 #4

    Borek

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    Staff: Mentor

    You need equilibrium concentrations, you are given total concentrations.

    You have 1M NaOH and 0.12M weak acid and you think [H+] >> [OH-]? In terms of simple stoichiometry you have around 8 times more NaOH that is needed to completely neutralize the acid.
     
  6. Oct 21, 2008 #5
    So to find the equilibrium concentrations I did:

    NaA + NaOH ---> A + OH + 2Na
    0.012 M 1.0 M 0 0 -
    -x -x +x +x -
    0.12 - x 1.0 -x x x -


    Kc = [A][OH]/[NaH][NaOH] = (0.12 -x)(1.0 -x)/(x)(x)

    =(0.12 -0.88x + x2)/x2

    Where do I go from here? how do I incorporate the absorbances?
     
  7. Oct 21, 2008 #6

    Borek

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    Staff: Mentor

    Looks like you have no idea what you are doing :frown:

    First experiment is to find the absorbance coefficient of A-. You use very high pH solution to be sure that all acid is in the neutralized form.
     
    Last edited: Oct 21, 2008
  8. Oct 21, 2008 #7
    the molar absorptivity of the picric acid would be

    0.581 = a(0.300 M) a= (0.581/0.300) = 1.937 M

    and for the mixture of sodium picrate and NaOH it would be

    0.335 = a1(0.120 M) + a2(1 M)

    This gives me two unknowns for one equation. can I use the molar absorptivity of the picric acid in the equation for the mixture?
     
  9. Oct 21, 2008 #8

    Borek

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    Staff: Mentor

    It is not absorptivity of picric acid, but of the picrate anion. NaOH doesn't matter - you may assume its absorptivity coefficient is 0.
     
  10. Oct 21, 2008 #9
    So if the absorbance of 0.335 is only due to the picric anion the concentration of the picric anion of the solution is as follows:

    0.335 = (1.973)CA CA= 0.173 M

    Ka = [A+]/[NaA][NaOH] = [0.173]/[.12][1] = 1.44 M
     
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