Determine PH of this solution

1. Jun 8, 2006

petuniac

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.

2. Jun 8, 2006

TD

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

3. Jun 8, 2006

dav2008

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

Think of the following equation:
10y=40

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
4. Jun 8, 2006

petuniac

Thanks! Not sure what I was thinking.