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Acid mixture pH problem

  1. Feb 7, 2016 #1
    1. The problem statement, all variables and given/known data

    30 mL CH3COOH 0.2 M is mixed with 60 ml CH3COONa 0.1 M and then added with 10 mL HCl 0.1 M, then what is the pH of the solution ? (Ka = 10-5)

    A. 3
    B. 5 - log 1.4
    C. 5 - log 2
    D. 4 - log 1.4
    E. 5 + log 2

    2. Relevant equations

    [H+] = Ka * [HX] / [X- ] ---- buffer
    [H+] = ([H+]V1 + [H+]V2) / (V1 + V2) ---- mixture

    3. The attempt at a solution

    The hydrogen molarity from the buffer is 10^-5 * 6 / 6 = 10^-5

    Hydrogen concentration of the mixture = (10^-5 * 90 + 10^-1 * 10) / ( 90 + 10 ) = (9 * 10^-4 + 1 ) / 100 = 9 * 10^-6 + 10^-2

    pH = - log ( 9 * 10^-6 + 10^-2 ) which doesn't appear in the options..
    Please help where I got wrong
     
  2. jcsd
  3. Feb 7, 2016 #2
    Here is what I did:

    First off, we can neglect the H+ ions provided by the weak acid to the solution as they are negligible compared to the H+ provided by the HCl.
    Then the concentration can be replaced with the moles, because the entire mixture is in the same solution and the volume terms will get cancelled.
    On doing so, we get 4 + log 6. This is pretty close to the value of option (c).
    Whats the answer?
     
  4. Feb 7, 2016 #3
    The answer is B. 5 - log 1,4
    But, the explanation of the answer is just 10^-5 * 7 / 5 which is very confusing...
    I really don't get it..
     
  5. Feb 8, 2016 #4
    The solution to the answer has assumed that the CH3COONa reacts with the HCl providing you with more CH3COOH. That is not completely correct. The salt will completely dissociate, but not all the CH3COO(minus) ions will interact with HCl to give CH3COOH.
     
  6. Feb 8, 2016 #5
    Why not all the ions will interact? How do you do the problem?
     
  7. Feb 8, 2016 #6
    OK wait. I am sorry. the answer given is correct. It is (b).

    This is because there exists an equilibrium for the dissociation of CH3COOH.

    CH3COOH ↔ CH3COO(-) + H(+)

    Now if there was no HCl or CH3COONa, the dissociation would proceed without any hindrance. But the prescence of these two generates CH3COO(-) ions and H(+) ions. Now these will shift the equlibrium reaction (use Le-Chatlier's principle). Thus CH3COOH will be produced. Obviously in the reaction, HCl is the limiting reagent.
     
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