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Determining pH problem, help please

  1. Mar 18, 2006 #1
    K(a) for HX is 7.5 x 10^-12. What is the pH of a 0.15-M aqueous solution of NaX?

    a. 7.87
    b. 1.85
    c. 5.97
    d. 8.03
    e. 12.15


    HX --> H+ + X-

    I found H+ concentration, which is 1.06 X 10-6, then found pH of this, and it is 5.97. I wasn't sure where to go from here so I guessed 5.97 and it is wrong (I get another guess).

    How do I go from finding the pH of the HX to finding the pH of the NaX? 14 - 5.97 - 8.03 which is an answer, but I'm not sure why I'd do this? isnt the conjugate base just the X-? Thanks for any help
     
    Last edited: Mar 18, 2006
  2. jcsd
  3. Mar 19, 2006 #2

    siddharth

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    First of all, since NaX is the salt of a strong base and weak acid, the salt will be basic on hydrolysis. So you can eliminate all answers below 7.

    That's wrong, because that's not the hydrolysis equilibrium for the salt. What happens is that the X- will react with water to form HX and OH-.
    You will first need to find the Hydrolysis constant Kh in terms of Ka and Kw.
    Then write down the relation between the concentration of the species at equilibrium. Since you know Kh (becaues you know Ka and Kw), and the initial concentration, you will be able to find the concentration of OH- and hence the pH.
     
    Last edited: Mar 19, 2006
  4. Mar 19, 2006 #3

    Borek

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    Check out this pH cheat sheet. But you can't use it before you will understand what you are doing :smile:
     
  5. Mar 20, 2006 #4

    siddharth

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    Borek, that cheat sheet isn't very useful if the OP doesn't actually understand what's happening. In fact, it would encourage people to blindly apply a formula without actually learning the concepts.

    The sheet will be useful and will save time only for experts like you, who have already derived it by hand and know what's happening.
     
  6. Mar 20, 2006 #5

    Borek

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    Well, it seems to me you know enough to use it :smile:
     
  7. Mar 20, 2006 #6

    siddharth

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    Yeah, but does the original poster know?

    I wanted confusedbyphysics to actually derive that equation by himself/herself, instead of looking up the final formula, because then confusedbyphysics would have understood some of the concepts involved.
     
  8. Mar 20, 2006 #7

    Borek

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    Well, I stated "you can't use it before you will understand what you are doing". So if he tries - he does it on his own risk :wink:

    Sorry for interfering with your pedagogical plan :smile:
     
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