Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 9 - Section 9.5 - Common and Natural Logarithms - 9.5 Exercises - Page 620: 33

Answer

poor fen

Work Step by Step

Using $ pH=-\log[H_3O^+] $ or the formula for the pH of a solution, with $H_3O^+= 3.1\times10^{-5} $, then \begin{align*}\require{cancel} pH&=-\log(3.1\times10^{-5}) \\&= -\left(\log3.1+\log10^{-5}\right) &(\text{use }\log_b (xy)=\log_b x+\log_b y) \\&= -(\log3.1-5\log10) &(\text{use }\log_b x^y=y\log_b x) \\&= -\left(\log3.1-5(1)\right) &(\text{use }\log10=\log_{10}10=1) \\&= -\log3.1+5 .\end{align*} Using a calculator, the value of $ \log3.1 $ is approximately $ 0.4914 $. Hence, the equation above is equivalent to \begin{align*} pH&\approx-0.4914+5 \\&\approx 4.5086 .\end{align*} Note that a pH value of $6.0$ to $7.5$ is a rich fen, a pH value of $3.0$ to $6.0$ is a poor fen, and a pH value of less than $3.0$ is a bog. Since the pH solution of water from the given wetland area is $ 4.5086 $, then the wetland area is considered a poor fen.
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