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Calculating pH

  1. Dec 19, 2011 #1
    1. The problem statement, all variables and given/known data
    x mole of a strong monoacidic base is dissolved in one litre of water. The pH of the solution will be
    (a)-log x
    (b)14-log x
    (c)14+log x
    (d)-log (14-x)


    2. Relevant equations



    3. The attempt at a solution

    For instance i assumed that the base is NaOH but it does not react with water to form any compound. So i am confused how should i start?

    Thanks! :smile:
     
  2. jcsd
  3. Dec 19, 2011 #2

    chemisttree

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    I'm still working on "monoacidic base"...
    What's that?

    OK. I just Googled it. So you have a base that generates 1 mole of OH- per mole of base in water. You also know the following:
    1) pH = -log[H+]
    2) [H+][OH-] = 1.01 X 10-14

    Can you get it from there?
     
    Last edited: Dec 19, 2011
  4. Dec 19, 2011 #3
    First calculate the pOH, which equals to -log[OH-].
    Then, pH=14-pOH. That's it.
     
  5. Dec 19, 2011 #4
    I think it isn't stated that it generates one mole of OH-. :smile:
     
  6. Dec 23, 2011 #5
    Monoacidic base is written in the form of MOH where M is any univalent metal.

    Hints :
    1 mole of MOH in water furnishes one OH- ion.
    Then x moles of MOH will furnish.... ?

    Write an ionic equation in terms of "x" and think about Mole Concept once again. You can then use the formulas given by chemisttree.
     
    Last edited: Dec 23, 2011
  7. Dec 23, 2011 #6

    Borek

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    Staff: Mentor

    Huh? What do the laws concerning gases have to do with the pH of the solution?
     
  8. Dec 23, 2011 #7
    This isn't a hard question as i thought of and thanks to chemistree for the idea of pOH.

    I found out the pOH which is -log x. Then calculated pH, which is completely easy pH=14-pH.

    @sankalp:- I think i am not that bad at chemistry that you need to remind me of Mole concept.

    I just forgot to reply in this thread after solving the problem. :P
     
  9. Dec 23, 2011 #8

    Borek

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    Staff: Mentor

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