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A small high school problem

  1. May 10, 2007 #1
    A 0.0025 mol/L solution of KHCrO4 has a pH of 3.50.
    Calculate the acid dissociation constant (Ka) for the equilibrium between HCrO4(-) and CrO4(2-).

    Thanks!:smile:
     
  2. jcsd
  3. May 10, 2007 #2
    Any attempt at the problem?
     
  4. May 10, 2007 #3

    chemisttree

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    You took the words right off of my keyboard!

    Why don't you try writing down the expression for Ka in general form and see where that leads you.

    Hint: You know the [H+] concentration from the pH.
     
  5. May 10, 2007 #4
    Haha beat you right to it! At school too! Akousmatikos, have you done any problems like this before chemisttree is right just use log rules and you'll be all set.
     
  6. May 10, 2007 #5

    Borek

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    He'll need dissociation reaction and its stoichiometry as well to calculate HCrO4- and CrO4-2 concentrations.
     
  7. May 13, 2007 #6
    Ok. This is what I know.
    1. The concentration of [H+]= 3.16*10^(-4) mol/L
    2. KHCrO4 -> H(+) + KCrO4(-)
    KHCrO4 -> H(+) + K(+) + CrO4(2-)
    3. Add them?
    2KHCrO4 -> 2H(+) + K(+) + Cro4(2-) + KCrO4(-)
    Is this right so far?
    4. Ka
    = [H+]^2 * [K(+)] * [CrO4(2-)] * [KCrO4(-)] / [KHCrO4]^2
    = [3.16*10^(-4)]^2 * [0.5*3.16*10^(-4)]^3 / [0.0025-3.16*10^(-4)]^2
    = 8.257 *10^(-14) mol/L
    But this is wrong. The answer in the back of textbook is 4.1 * 10^...
    What did I do wrong?
     
    Last edited: May 13, 2007
  8. May 14, 2007 #7

    chemisttree

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    For the reaction:

    HCrO4- <---------> H+ + CrO4-2

    You can calculate the Ka with the equation:

    Ka = [H+] [CrO4-2]/[HCrO4-]

    You are given the pH (pH = -log[H+]) and you can calculate the [H+]. This is also the concentration of [CrO4-2]. Can you see why? The original concentration of [HCrO4-] has been decreased by the same amount of [H+] that forms.

    Can you put this information into the expression for Ka and solve?
     
  9. May 15, 2007 #8
    Wait...
    Does [HCrO4-]=[KHCrO4]=0.0025mol/L ?

    Ka = [H+] [CrO4 2-] / [HCrO4-]
    = [3.16*10^(-4)]^2 / [0.0025 - 3.16*10^(-4)]
    = 4.57*10^(-5) mol/L

    Well that's close to the answer in the textbook, but I don't think it's right.
     
  10. May 16, 2007 #9

    chemisttree

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    Last edited: May 16, 2007
  11. May 16, 2007 #10
    Thanks for the sites.
    What about the equilibrium between HCrO4(-) and CrO4(2-)?
    Sorry if these questions are too basic. I'm just having trouble understanding.
    :P
     
  12. May 17, 2007 #11

    chemisttree

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    for Ka = [H+][CrO4-2]/[HCrO4-], rearranging we get:

    [H+] = Ka[HCrO4-]/[CrO4-2]

    applying log to both sides gives us:

    log[H+] = log(Ka) + log[HCrO4-]/[CrO4-2]

    multiplying by -1 gives us

    -log[H+] = -log(Ka) + log[CrO4-2]/[HCrO4-]

    which is pH = pKa + log[CrO4-2]/[HCrO4-]

    pH is 3.5 and [CrO4-2] is equal to [H+] or 3.16X10^-4.

    [HCrO4] is 0.0025 - 3.16 X 10^-4 or 0.00218

    The ratio of [CrO4-2]/[HCrO4-] is 3.16X10^-4/0.00218 = 0.145

    Soooo,

    3.5 = pKa + log(0.145)
    3.5 - log(0.145) = pKa
    4.339 = pKa
    4.58X10-5 = Ka

    Same answer by Henderson-Hasselbalch equation as by the method you described. How far off is the answer from that in your book?
     
  13. May 17, 2007 #12
    The answer in the book is 4.1 * 10^..
    But now I think their answer is wrong.
    Anyway, I'll ask my teacher next week.
    Thank you for all the help. :approve:
     
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