Neutralization Question!

Discussion in 'MCAT Study Question Q&A' started by RHC1987, May 12, 2008.

  1. RHC1987

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    What is the Keq of the neutralization reaction between oxalic acid (H2C2O4; KA1 = 5.9 × 10^-2) and sodium hydroxide?

    ANSWER: 5.9 x 10^12

    Can't seem to get this one...
     
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  3. TypeSH07

    TypeSH07 Junior Member

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    no other info is given?
     
  4. scottyT

    scottyT Real Member

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    I'm just taking a guess here but I think this is going on:

    H2C2O4 + NaOH --> H2O + HC2O4
    Ka=5.9*10^-2 Kw=1*10^-14

    We know Ka*Kb=Kw. If you have achieved neutralization then you can reword the question as follows: What is the Kb of the base dissociation reaction?

    HC2O4 + H2O --> H2C2O4 + NaOH

    Please correct me if I'm wrong but I think this should help.
     
  5. Zerconia2921

    Zerconia2921 Bring your A-game!

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    Neutralization reaction requires a strong base and strong acid resulting product water plus salt.

    Ka = 5.9x10^-2
    Kb= x
    Kw= 1.0x10-^14

    Solve for Kb
    1.0x10-^14/5.9x10^-2= 5.9x10-^12
     
  6. RHC1987

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    the answer to this question is 5.9x10^+12 not 5.9 x10^-12... which is what threw me off.. i originally did it your way too...
     
  7. Zerconia2921

    Zerconia2921 Bring your A-game!

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    Raised to the power of positive 12? I am thinking about it and if the Kb was positive power than means its a VERY VERY VERY strong base so it will completely dissocate so I am not sure if that is an typo error or what. Sorry i wasnt helpful.
     
  8. Scarletblack

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    Uh well here's my attempt


    H2C2O4 + NaOH --> H2O + HC2O4

    so

    [HC2O4-][H20] ................. (Call this eq 1)
    ----------------- = Keq
    [-OH][H2C2O4]

    ok but

    [HC2O4-][H+]
    ---------------- = Ka = 5.9*10^-2
    [H2C2O4]

    You can do algebra to get:

    5.9*10^-2.....[HC2O4-] (2)
    ------------ = -----------
    [H+]............ [H2C2O4]

    Substitute (2) into (1), you get:

    5.9*10^-2 [H2O]
    ------------------- = Keq
    [-OH] [H+]

    But at equilibrium [H+][OH-]/[H2O] = 10^-14 so

    5.9*10^-2
    ------------ = 5.9 * 10^12
    10^-14
     
    #7 Scarletblack, Jun 8, 2008
    Last edited: Jun 8, 2008
  9. minhaj

    minhaj Awesomeness Incarnate

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    I am not sure if that is true. Neutralization reaction occurs regardless of the strength of the acid or the base.
     
  10. ahs4n

    ahs4n MS0

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    bringing back an old thread... does this look right.. I'm trying to figure this problem out from the Kaplan FL 9 on the PS problem 12
     
  11. Fort

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    Here's another way. It's a little less mathematical, but you still arrive at the final answer.

    We know the dissociation constant for oxalic acid to be 5.9e-12.

    So we can write

    H2C2O4 --> HC2O4- + H+

    We want to get the final equation to look like

    H2C2O4 + OH- --> HC2O4- + H2O

    Well, we can actually add another equation to the first one:

    H2C2O4 --> HC2O4- + H+ K1 = 5.9e-12
    OH- + H+ --> H2O K2 = 1.0e14 (it's e14 because we've reversed the water dissociation reaction)

    If we add these two equations, we get the desired reaction. To find Keq, we simply multiply K1 and K2.

    K1*K2 = (5.9e-12)(1.0e14) = 5.9e12

    Hope this helps.
     
  12. ahs4n

    ahs4n MS0

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    That helps tremendously! Thank you for doing that.
     
  13. gunj122

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    i'm sorry if i'm missing something here but how is 10^-12 * 10^14 = 10^12?
    shouldn't it equal 10^2?

     

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