Determine PH of this solution

[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.

Thanks for your help.

 

[TD]

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

 

[dav2008]

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

Think of the following equation:
10^y=40

Now you can see how to write y in terms of logarithms. Then when you find what y is you will have 10^y=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.

 

[petuniac]

Thanks! Not sure what I was thinking.