# Question #a35c1

##### 1 Answer

#### Answer:

See explanation.

#### Explanation:

A **common logarithm** is a logarithm that has a base of *without* an added base, so

#log_10 = log#

What a logarithm that has a base of **powers of**

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

if an only if

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

For a common log,

#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,

The thing to keep in mind here is that the concentration of hydronium cations is usually a **very small number**, much smaller than *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**

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

#log(["H"_ 3"O"^(+)]) = 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)#

For

#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 *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

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.