Answer
$pH=2.432$
Work Step by Step
$(H^{+})=3.7\times 10^{-3}$M
Putting in the equation:
$pH=-\log(H^{+}) $
$pH=-\log(3.7\times10^{-3})$
$pH=-(\log3.7+\log10^{-3})$
$pH=-(\log3.7-3\log10)$
$pH=-\log3.7+3$
$pH=2.432$
Intermediate Algebra: Connecting Concepts through Application
by Clark, Mark; Anfinson, Cynthia
Published by
Brooks Cole
ISBN 10:
0-53449-636-9
ISBN 13:
978-0-53449-636-4
Chapter 6 - Logarithmic Functions - Chapter Review Exercises - Page 543: 44
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