# Question a35c1

Aug 24, 2016

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 $\textcolor{red}{n}$ is equal to a number $\textcolor{\mathrm{da} r k g r e e n}{x}$

${\log}_{b} \textcolor{red}{n} = \textcolor{\mathrm{da} r k g r e e n}{x}$

if an only if

${b}^{\textcolor{\mathrm{da} r k g r e e n}{x}} = \textcolor{red}{n}$

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

${\log}_{10} \textcolor{red}{n} = \textcolor{\mathrm{da} r k g r e e n}{x} \iff \log \textcolor{red}{n} = \textcolor{\mathrm{da} r k g r e e n}{x}$

will get you

${10}^{\textcolor{\mathrm{da} r k g r e e n}{x}} = \textcolor{red}{n}$

Now, here is how this relates to a solution's pH. As you know, the pH is determined by the concentration of hydronium cations, ${\text{H"_3"O}}^{+}$.

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 $\left[{\text{H"_3"O}}^{+}\right]$

$\log \left(\left[{\text{H"_ 3"O}}^{+}\right]\right) = \log \left(1 \cdot {10}^{- 5}\right) = \log \left(1\right) + \log \left({10}^{- 5}\right)$

$\textcolor{w h i t e}{a}$

SIDE NOTE You should also check out

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

$\textcolor{w h i t e}{a}$

Now, $\log \left(\textcolor{red}{1}\right) = \textcolor{\mathrm{da} r k g r e e n}{0}$, since

${10}^{\textcolor{\mathrm{da} r k g r e e n}{0}} = \textcolor{red}{1}$

For $\log \left({10}^{- 5}\right)$, you have

${10}^{\textcolor{\mathrm{da} r k g r e e n}{x}} = \textcolor{red}{{10}^{- 5}} \implies \textcolor{\mathrm{da} r k g r e e n}{x} = - 5$

Therefore, you can say that

$\log \left(1 \cdot {10}^{- 5}\right) = 0 + \left(- 5\right) = - 5$

Now, we have an easier time working with positive numbers, so look what happens when instead of taking the positive common log of $\left[{\text{H"_3"O}}^{+}\right]$, we take the negative one

$- \log \left(\left[{\text{H"_3"O}}^{+}\right]\right) = - \left[\log \left(1 \cdot {10}^{- 5}\right)\right]$

$= - \left[\log \left(1\right) + \log \left({10}^{- 5}\right)\right]$

$= - \left[0 + \left(- 5\right)\right]$

$= - \left(- 5\right)$

$= 5$

And there you have, you just found the pH of a solution that has a concentration of hydronium cations equal to $1 \cdot {10}^{- 5} {\text{mol L}}^{- 1}$.

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.