Chemistry 10th Edition

Published by Brooks/Cole Publishing Co.
ISBN 10: 1133610668
ISBN 13: 978-1-13361-066-3

Chapter 18 - Ionic Equilibria I: Acids and Bases - Exercises - The Autoionization of Water - Page 743: 15

Answer

(a) $[OH^-] = 1.587 \times 10^{- 11}M$ (b) $[OH^-] = 1.282 \times 10^{- 6}M$

Work Step by Step

(a) $[H_3O^+] * [OH^-] = Kw = 10^{-14}$ $ 6.3 \times 10^{- 4} * [OH^-] = 10^{-14}$ $[OH^-] = \frac{10^{-14}}{ 6.3 \times 10^{- 4}}$ $[OH^-] = 1.587 \times 10^{- 11}$ (b) $[H_3O^+] * [OH^-] = Kw = 10^{-14}$ $ 7.8 \times 10^{- 9} * [OH^-] = 10^{-14}$ $[OH^-] = \frac{10^{-14}}{ 7.8 \times 10^{- 9}}$ $[OH^-] = 1.282 \times 10^{- 6}$
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