Question #a35c1
1 Answer
See explanation.
Explanation:
A common logarithm is a logarithm that has a base of
log_10 = loglog10=log
What a logarithm that has a base of
log_b color(red)(n) = color(darkgreen)(x)logbn=x
if an only if
b^color(darkgreen)(x) = color(red)(n)bx=n
For a common log,
log_(10)color(red)(n) = color(darkgreen)(x) <=> log color(red)(n) = color(darkgreen)(x)log10n=x⇔logn=x
will get you
10^color(darkgreen)(x) = color(red)(n)10x=n
Now, here is how this relates to a solution's pH. As you know, the pH is determined by the concentration of hydronium cations,
The thing to keep in mind here is that the concentration of hydronium cations is usually a very small number, much smaller than
In order to make working with small numbers easier, we tend to express them in scientific notation, which as you know is also based on powers of
So, for example, let's say that you are given a solution that has a concentration of hydronium cations equal to
["H"_3"O"^(+)] = "0.00001 mol L"^(-1)[H3O+]=0.00001 mol L−1
Expressed in scientific notation, this is equal to
["H"_3"O"^(+)] = 1 * 10^(-5)"mol L"^(-1)[H3O+]=1⋅10−5mol L−1
Notice what happens when we take the common log of
log(["H"_ 3"O"^(+)]) = log(1 * 10^(-5)) = log(1) + log(10^(-5))log([H3O+])=log(1⋅10−5)=log(1)+log(10−5)
SIDE NOTE You should also check out
http://www.purplemath.com/modules/logrules.htm
Now,
10^color(darkgreen)(0) = color(red)(1)100=1
For
10^color(darkgreen)(x) = color(red)(10^(-5)) implies color(darkgreen)(x) = -510x=10−5⇒x=−5
Therefore, you can say that
log(1 * 10^(-5)) = 0 + (- 5) = -5log(1⋅10−5)=0+(−5)=−5
Now, we have an easier time working with positive numbers, so look what happens when instead of taking the positive common log of
-log(["H"_3"O"^(+)]) = - [log(1 * 10^(-5))]−log([H3O+])=−[log(1⋅10−5)]
= - [log(1) + log(10^(-5))]=−[log(1)+log(10−5)]
= - [0 + (-5)]=−[0+(−5)]
=-(-5)=−(−5)
=5=5
And there you have, you just found the pH of a solution that has a concentration of hydronium cations equal to
The equation to always keep in mind is
color(blue)(|bar(ul(color(white)(a/a)"pH" = - log(["H"_3"O"^(+)])color(white)(a/a)|)))
We take the negative common log of the concentration of hydronium cations because this concentration is a very small number, usually smaller than
The common log makes it easier for us to think about the acidity of a solution.
