Introductory Chemistry (5th Edition)

Published by Pearson
ISBN 10: 032191029X
ISBN 13: 978-0-32191-029-5

Chapter 14 - Acids and Bases - Exercises - Problems - Page 523: 63

Answer

a. [OH-] = 6.7 $\times 10^{-6}$ M b. [OH-] = 1.1 $\times 10^{-6}$ M c. [OH-] = 4.5 $\times 10^{-6}$ M d.[OH-] = 1.4 $\times 10^{-11}$ M

Work Step by Step

Kw=[$H_{3}O$+][OH-] [OH-] = $\frac{Kw}{[H_{3}O+]}$ *** Kw= 1 $ \times 10^{-14}$ a. [OH-] = $\frac{Kw}{[H_{3}O+]}$ [OH-] = $\frac{1 \times 10^{-14}}{1 \times 10^{-9}}$ [OH-] = 6.7 $\times 10^{-6}$ M b. [OH-] = $\frac{Kw}{[H_{3}O+]}$ [OH-] = $\frac{1 \times 10^{-14}}{9.3 \times 10^{-6}}$ [OH-] = 1.1 $\times 10^{-6}$ M c. [OH-] = $\frac{Kw}{[H_{3}O+]}$ [OH-] = $\frac{1 \times 10^{-14}}{2.2 \times 10^{-6}}$ [OH-] = 4.5 $\times 10^{-6}$ M d. [OH-] = $\frac{Kw}{[H_{3}O+]}$ [OH-] = $\frac{1 \times 10^{-14}}{7.4 \times 10^{-4}}$ [OH-] = 1.4 $\times 10^{-11}$ M
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