Question #a35c1

1 Answer
Aug 24, 2016

Answer:

See explanation.

Explanation:

A common logarithm is a logarithm that has a base of #10#. Common logs are usually written without an added base, so

#log_10 = log#

What a logarithm that has a base of #10# basically means is that you're working on a scale based on powers of #10#. As you know, the logarithm base #b# of a number #color(red)(n)# is equal to a number #color(darkgreen)(x)#

#log_b color(red)(n) = color(darkgreen)(x)#

if an only if

# b^color(darkgreen)(x) = color(red)(n)#

For a common log, #b = 10#, so

#log_(10)color(red)(n) = color(darkgreen)(x) <=> log color(red)(n) = color(darkgreen)(x)#

will get you

#10^color(darkgreen)(x) = color(red)(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, #"H"_3"O"^(+)#.

The thing to keep in mind here is that the concentration of hydronium cations is usually a very small number, much smaller than #1#, but always positive.

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 #10#.

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)#

Expressed in scientific notation, this is equal to

#["H"_3"O"^(+)] = 1 * 10^(-5)"mol L"^(-1)#

Notice what happens when we take the common log of #["H"_3"O"^(+)]#

#log(["H"_ 3"O"^(+)]) = log(1 * 10^(-5)) = log(1) + log(10^(-5))#

#color(white)(a)#

SIDE NOTE You should also check out

http://www.purplemath.com/modules/logrules.htm

#color(white)(a)#

Now, #log(color(red)1) = color(darkgreen)(0)#, since

#10^color(darkgreen)(0) = color(red)(1)#

For #log(10^(-5))#, you have

#10^color(darkgreen)(x) = color(red)(10^(-5)) implies color(darkgreen)(x) = -5#

Therefore, you can say that

#log(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 #["H"_3"O"^(+)]#, we take the negative one

#-log(["H"_3"O"^(+)]) = - [log(1 * 10^(-5))]#

#= - [log(1) + log(10^(-5))]#

#= - [0 + (-5)]#

#=-(-5)#

# =5#

And there you have, you just found the pH of a solution that has a concentration of hydronium cations equal to #1 * 10^(-5)"mol L"^(-1)#.

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 #1#, that we can express in scientific notation.

The common log makes it easier for us to think about the acidity of a solution.