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 522: 59

Answer

a. NaOH is a strong base and dissociates completely into Na+ and OH-. So, the concentration of [OH-] is the same as NaOH. Therefore [OH-] = 0.25M. b. $NH_{3}$ is a weak base and dissociates partially. Therefore [OH-] < [$NH_{3}$] c. It is a strong base and dissociates completely. Therefore [OH-]= [$Sr(OH)_{2}$] d. KOH is a strong base and dissociates completely. Therefore [K+] = [OH-]

Work Step by Step

a. NaOH is a strong base and dissociates completely into Na+ and OH-. So, the concentration of [OH-] is the same as NaOH. Therefore [OH-] = 0.25M. b. $NH_{3}$ is a weak base and dissociates partially. Therefore [OH-] < [$NH_{3}$] c. It is a strong base and dissociates completely. Therefore [OH-]= [$Sr(OH)_{2}$] d. KOH is a strong base and dissociates completely. Therefore [K+] = [OH-]
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