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    Okay, so I've got a diprotic weak acid question. The question states that both protons have the same tendency to dissociate from the initial molecule.

    Then, for the final part in the question, you get asked this:

    'In reality, the pK values for the two dissociating protons in the molecules are not likely to be identical. How does this fact alter the calculation in this question?'

    So we can use  ka = [H^+]^3 / [H2A] because in the question I'm give the value of ka (4.78x10^-3) and that it's at 1.0M of H2A.

    I'm guessing that the value for Ka will increase because you remove a hydrogen ion concentration value from the equilibrium equation to set up a second equation for the second removal of hydrogen (deprotonation?). Because the H+ ion conc. is less than 1, removing that value increases the value of the numerator and thus Ka.

    This is a blatant, wild guess, but I don't have any threads to hang on. My main flaw with the argument I presented is that how do you factor in this H+ ion conc. which disappears from the 2nd equation? It isn't reacting with HA+ so it can't be included in the equation.
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    What is [H+]3?

    Can't say I understand the question (especially without seeing whole problem), but this part cries out loud "I am wrong!"

    You won't have one Ka value, for diprotic acid we use two separate Ka values (Ka1 and Ka2).

    What you tried to use is so called overall dissociation constant.
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    (Original post by Borek)
    What is [H+]3?

    Can't say I understand the question (especially without seeing whole problem), but this part cries out loud "I am wrong!"

    You won't have one Ka value, for diprotic acid we use two separate Ka values (Ka1 and Ka2).

    What you tried to use is so called overall dissociation constant.
    Here you go. It explains it a lot better than I did! Question 5.Name:  ChemQuestion.jpg
Views: 72
Size:  209.2 KB

    So my answers are...

    a) mol^2 dm^-6
    b) cube root of 4.78x10^-3 = 0.168 = [H+] therefore multiply this value by 2 to get the total conc of H3O+. You get 0.336 for that and so 0.168 for the A-.
    c) The pH (using 0.336) would be 0.47 (which seems unusually low for a weak acid, even if it is diprotic and has a relatively high Ka value)
    d) Pisghaihfihgafg?

    I *think* what I've done is correct, but please let me know if I'm just being stupid. Also, bear in mind that it's not on the spec to use Ka1 and Ka2 and so I haven't been given this information. I don't think it wants a numerical value at all though, just how I'd calculate it.

    Do I just say that you'd need to do 2 separate equations?
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    This is a crazy problem IMHO - poorly defined, difficult to solve and doing more harm than good. Basically it asks "how does the world behave according to the laws of chemistry, if the laws of chemistry don't hold". Sigh.

    Let's assume what they mean is that the dissociation constant is

    K_a = \frac {[H^+]^2[A^{2-}]}{[H_2A]} = 4.78\times 10^{-3}

    (so the answer to question about units is concentration2, you got it right)

    You can't just take a cube root of Ka, as it will mean ignoring some obvious conclusions from the dissociation stoichiometry. If you write the dissociation reaction:

    H2A <=> 2H+ + A2-

    it is obvious that

    [H^+]=2[A^{2-}]

    and (as the initial concentration was 1 M, and each A2- produced means one H2A consumed)

    [H_2A] = 1 - [A^{2-}]

    When we plug these into Ka, we get

    K_a = \frac {(2[A^{2-}])^2[A^{2-}]}{1 - [A^{2-}]} = \frac{4[A^{2-}]^3}{1 - [A^{2-}]}

    which is a cubic equation, and it is not clear how to simplify it. We can assume the dissociation didn't went far (so the denominator is equal almost 1), but - sparing you some math - I have checked that it is not true, and it doesn't yield a reasonable result (more precisely it doesn't follow 5% rule - I can explain it in more details if you are interested).

    So, we are already in the blind alley. But let's try brute force. Let's use Wolfram Alpha to solve the equation: ((2*x)^2*x)/(1-x)=4.78e-3

    It lists the only real answer as 0.102 - that's concentration of A-, so concentration of H+ is twice that, and pH is 0.69.

    And here comes d... The only serious answer one can give is that "pH will be different than the one we calculated". Most likely higher.

    I don't see how to answer this problem without violating the logic.
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    (Original post by Borek)
    This is a crazy problem IMHO - poorly defined, difficult to solve and doing more harm than good. Basically it asks "how does the world behave according to the laws of chemistry, if the laws of chemistry don't hold". Sigh.

    Let's assume what they mean is that the dissociation constant is

    K_a = \frac {[H^+]^2[A^{2-}]}{[H_2A]} = 4.78\times 10^{-3}

    (so the answer to question about units is concentration2, you got it right)

    You can't just take a cube root of Ka, as it will mean ignoring some obvious conclusions from the dissociation stoichiometry. If you write the dissociation reaction:

    H2A <=> 2H+ + A2-

    it is obvious that

    [H^+]=2[A^{2-}]

    and (as the initial concentration was 1 M, and each A2- produced means one H2A consumed)

    [H_2A] = 1 - [A^{2-}]

    When we plug these into Ka, we get

    K_a = \frac {(2[A^{2-}])^2[A^{2-}]}{1 - [A^{2-}]} = \frac{4[A^{2-}]^3}{1 - [A^{2-}]}

    which is a cubic equation, and it is not clear how to simplify it. We can assume the dissociation didn't went far (so the denominator is equal almost 1), but - sparing you some math - I have checked that it is not true, and it doesn't yield a reasonable result (more precisely it doesn't follow 5% rule - I can explain it in more details if you are interested).

    So, we are already in the blind alley. But let's try brute force. Let's use Wolfram Alpha to solve the equation: ((2*x)^2*x)/(1-x)=4.78e-3

    It lists the only real answer as 0.102 - that's concentration of A-, so concentration of H+ is twice that, and pH is 0.69.

    And here comes d... The only serious answer one can give is that "pH will be different than the one we calculated". Most likely higher.

    I don't see how to answer this problem without violating the logic.
    Thanks for your suggestion. I agree, it just doesn't work. I'll have to pester my teacher and see what he has to say about this shambles of a question.

    Thanks for your help and I'll quote you in when he gets back to me.
 
 
 
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