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A solution is prepared by dissolving NH3(g) in water according to the following equation. NH3(g) + H2O(l)=NH4^+(aq)+ OH^-(aq)

If the pH= 9.60 what is the [OH-] in the solution.

  • Chemistry -

    pH + pOH = pKw = 14. You know pH, solve for pOH and convert to OH^-.
    pOH = -log(OH^-).

  • Chemistry -

    is this right
    pH+pOH =14
    [OH-] =10^-pOH
    [OH-] = 10^-4.4 NOW IM STUCK?

  • Chemistry -

    I'm stuck there too Ryan

  • Chemistry -

    pH + pOH = 14
    9.60 + pOH = 14
    pOH = 14 - 9.60 = 4.40
    pOH = 4.4 = -log(OH^-)
    -4.40 = log(OH^-)
    Key in -4.4 on your calculator and hit the 10x button. It should return 3.981E-5 which I would round to 3.98E-5.

  • Chemistry -

    I get error when I type in -4.4 log and when I put in 4.4 I get 0.6434

  • Chemistry -

    I got error also

  • Chemistry -

    You aren't following instructions. I did NOT say take the log of -4.4. Doing that gives you an error message. You want the NUMBER whose LOG is -4.4. To get that you DON'T punch in the log of -4.4. You key in -4.4 and hit the 10x button (that's 10^x button). That give you the number whose log is -4.4. Your display should read 3.98E-5. You may have seen this notation, which is the same thing, as (OH^-) = 10-4.40.
    (Note that if you punch in 3.98E-5 and hit the log key, you will get -4.4).

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