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### Writing Molarity of IONS as Logarithms: Solving Logarithms

Posted: Thu Nov 19, 2015 11:13 pm
In the Chemistry 14a Course Reader on page 139, Lavelle has written:

" -log(1.7 * 10^(-12)) = -(-11.8) = 12 "

I am confused how he got to the value -11.8 from the equation -log(1.7 * 10^(-12)).
Could someone please explain to me how to solve this logarithm (along with logarithms in general) step by step? Thank you.

### Re: Writing Molarity of IONS as Logarithms: Solving Logarith

Posted: Fri Nov 20, 2015 10:36 am
The -11.8 was found by solving the logarithm " -log(1.7 * 10^(-12))". Plugging this into the calculator will give you -11.8. In general, solving logarithms in this form means finding the exponent of 10 that will give you the number after the log. So, this solved for the exponent of ten that will give you 1.7 * 10^(-12). If it was log(1.0 * 10^(-12)), the answer would be negative twelve. But because it is 1.7 is is slightly less.

### Re: Writing Molarity of IONS as Logarithms: Solving Logarith

Posted: Sun Nov 22, 2015 4:50 am
A logarithm question basically asks if $X^{n}=Y$ what is n if you know what X and Y are. The log function puts in your X and Y values to calculate n and is written as $log_{X}(Y)=n$. In standard cases, we say that X=10 and just write it as $log(Y)=n$, omitting X as it is in the course reader.

Now for the example in the course reader, we calculate pH from the concentration of hydrogenium ions, and pH is the "n" or exponent value of 10. So we want to know to what power do we raise 10 in order to get the hydrogenium concentration. In some cases, you don't need a calculator e.g. if [H3O+]=10-10, then we know that the pH is 10 because of the exponent. However, in most cases, you cannot calculate this by hand and will need a calculator.