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Determine PH of this solution

  1. Jun 8, 2006 #1
    Question is:

    An unknown solution is 40 times more alkaline than neutral water which has a PH of 7. Determine the PH of the unknown solution.

    Here is what I have:

    40 = (log base 10 x)/(log base 10 7)
    40 = 10^(x-7)
    10^1.4 = 10^(x-7)
    1.4 = x - 7
    x = 8.4

    The answer is supposed to be 8.6??? Not sure where my mistake is.

    Thanks for your help.
  2. jcsd
  3. Jun 8, 2006 #2


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    Homework Helper

    Is the first line correct? Assuming the second is, it's wrong to go from 40 to 10^1.4, recheck that.
  4. Jun 8, 2006 #3


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    Gold Member

    Like td said, your exponent of 1.4 in that line is wrong.

    Think of the following equation:

    Now you can see how to write y in terms of logarithms. Then when you find what y is you will have 10y=10(x-7).

    Note that this is kind of the work-around method. It's easier to look at the line 40=10(x-7) and take the log base 10 of both sides.
    Last edited: Jun 8, 2006
  5. Jun 8, 2006 #4
    Thanks! Not sure what I was thinking.
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