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Exercises Exponentials and logarithms

Example 12 from Algebra and Trigonometry, 6.5 Logarithmic Properties

Applying of the Laws of Logs

Recall that, in chemistry, $pH = - \log [H^{+}] .$ If the concentration of hydrogen ions in a liquid is doubled, what is the effect on pH?

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Solution
Suppose $C$ is the original concentration of hydrogen ions, and $P$ is the original pH of the liquid. Then $P = - \log (C) .$ If the concentration is doubled, the new concentration is $2 C .$ Then the pH of the new liquid is
$pH = - \log (2 C)$
Using the product rule of logs

$pH = - \log (2 C) = - (\log (2) + \log (C)) = - \log (2) - \log (C)$

Since $P = - \log (C),$ the new pH is

$pH = P - \log (2) \approx P - 0.301$
When the concentration of hydrogen ions is doubled, the pH decreases by about 0.301.

Your turn

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

How does the pH change when the concentration of positive hydrogen ions is decreased by half?

How did it go?

PreviousRewriting as a Single Logarithm Next Changing Logarithmic Expressions to Expressions Involving Only Natural Logs

From Algebra and Trigonometry (1st edition) by OpenStax (Rice University), section 6.5, licensed under CC BY 4.0. Original at openstax.org. We split the solution into steps; the words, numbers and figures are the book's. OpenStax doesn't endorse this site.