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Homework Help: Finding pH

  1. Feb 8, 2006 #1
    I'm being asked to find the pH of a 0.050 M solution of NaCN.
    I figured out the expression to be x^2/ 0.050 -x, now all i need is a value for k, however I'm not given that value. I know the pH depends on the molarity of the hydrogen ion at equilibrium, however to find that I need to find x, and in order to find x I need a value for k. I would appreciate any help, thanks.
     
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  3. Feb 8, 2006 #2

    Gokul43201

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    The expression for what ?
    where 'x' is ...?
     
  4. Feb 8, 2006 #3
    He is talking about the equilibrium constant equation.
    Btw, i just finished coding a little simple program on finding the equilibrium constant.
    Just saves a few keystroked on the annoying calculator :) .


    water dissociation const.
    http://en.wikipedia.org/wiki/Dissociation_constant
     
    Last edited: Feb 9, 2006
  5. Feb 9, 2006 #4

    Borek

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    pKa for HCN is 9.31.

    Best,
    Borek
     
    Last edited by a moderator: Aug 13, 2013
  6. Feb 9, 2006 #5

    GCT

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    yeah, you should have been given the K value
     
  7. Feb 9, 2006 #6
    I didn't check my book very well, and so later I found a chart that had the Kb value, so I could just used that to divide 10^-14 to find Ka, since Kb x Ka = Kw (10^-14). Once that's done I could have just used that value, plug it into the equation, solve for x, if it passes the 5% test then take the -log of x and had the pH.
     
  8. Feb 9, 2006 #7
    I have no idea what pKa is... something with pressure? Thanks anyway though Borek. I got some help earlier, and was given the Ka constant, which was 6.2 x 10^-10.
     
  9. Feb 9, 2006 #8
    There is a very good chance that all of this is wrong. I am basically just trying this problem out for myself as a challenge, for I don't know what pKa is either. I'm guessing...

    [tex]pKa = - log Ka[/tex]

    [tex]9.31 = - log Ka[/tex]

    [tex]10^{-9.31} = Ka[/tex]

    [tex]Ka = 4.9*10^{-10}[/tex]

    [tex]4.9*10^{-10} = \frac{x^{2}}{.050 - x}[/tex]

    [tex]x^{2} + 4.9*10^{-10}x - 2.5*10^{-11}[/tex]

    [tex]x = 5.0*10^{-6}[/tex]

    [tex]5.0*10^{-6} = [HCN] = [H^{+}][/tex]

    [tex]pH = - log (5.0*10^{-6})[/tex]

    [tex]pH = 5.30[/tex]

    Once again, I know that I am most likely wrong. I assumed that pKa = - log Ka. I also assumed that [HCN] = [H+]. I haven't learned about that yet; we just started acids and bases earlier this week in my class.

    I'm interested to see how this problem is really meant to be worked out :wink:.
     
    Last edited: Feb 9, 2006
  10. Feb 9, 2006 #9
    Sorry, that was actually the Kb constant...so when you divide it out you get 1.6 x 10^-5 as Ka.
     
  11. Feb 10, 2006 #10
    pKa = -log(Ka)
     
  12. Feb 10, 2006 #11
    So my assumption was correct :cool:. So did I do the problem correctly after all?
     
  13. Feb 10, 2006 #12

    Borek

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    You guessed OK :)

    Write the reaction equation - you don't start from HCN, you start with CN- - CN- will react with water to produce OH-. You are close, but you missed ;)

    Best,
    Borek
     
    Last edited by a moderator: Aug 13, 2013
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