Logarithmic Properties: Apply It

  • Expand logarithmic expressions.
  • Condense logarithmic expressions.
  • Use the change-of-base formula for logarithms.
Testing of the pH of hydrochloric acid.
The pH of hydrochloric acid is tested with litmus paper. (credit: David Berardan)

In chemistry, pH is used as a measure of the acidity or alkalinity of a substance. The pH scale runs from 0 to 14. Substances with a pH less than 7 are considered acidic, and substances with a pH greater than 7 are said to be alkaline. Our bodies, for instance, must maintain a pH close to 7.35 in order for enzymes to work properly. To get a feel for what is acidic and what is alkaline, consider the following pH levels of some common substances:

  • Battery acid: 0.8
  • Stomach acid: 2.7
  • Orange juice: 3.3
  • Pure water: 7 (at 25° C)
  • Human blood: 7.35
  • Fresh coconut: 7.8
  • Sodium hydroxide (lye): 14

To determine whether a solution is acidic or alkaline, we find its pH, which is a measure of the number of active positive hydrogen ions in the solution. The pH is defined by the following formula, where a is the concentration of hydrogen ion in the solution

[latex]\begin{align}\text{pH}&=-\mathrm{log}\left(\left[{H}^{+}\right]\right) \\ &=\mathrm{log}\left(\frac{1}{\left[{H}^{+}\right]}\right) \end{align}[/latex]

The equivalence of [latex]-\mathrm{log}\left(\left[{H}^{+}\right]\right)[/latex] and [latex]\mathrm{log}\left(\frac{1}{\left[{H}^{+}\right]}\right)[/latex] is one of the logarithm properties we will examine in this section.

We know that
[latex]\text{pH} = -\log\left([H^+]\right)[/latex].

Let’s explore how the logarithm properties help us interpret and compare acidity.

Suppose a solution has [latex][H^+] = 3.2 \times 10^{-4}[/latex]. Let’s use properties of logarithms to simplify its pH.

[latex]\begin{align} \text{pH} &= -\log\big(3.2\times 10^{-4}\big) && \text{substitute into pH formula} \\ &= -\Big(\log(3.2) + \log(10^{-4})\Big) && \text{product rule: }\log(ab)=\log a + \log b \\ &= -\Big(\log(3.2) - 4,\log(10)\Big) && \text{power rule: }\log(10^{-4})=-4\log(10) \\ &= -\log(3.2) + 4\log(10) && \text{distribute the negative sign} \\ &= -\log(3.2) + 4 && \text{since }\log(10)=1  \\ \end{align}[/latex]