Chemistry 12th Edition

Published by McGraw-Hill Education
ISBN 10: 0078021510
ISBN 13: 978-0-07802-151-0

Chapter 15 - Acids and Bases - Questions & Problems - Page 710: 15.16


Please see the work below.

Work Step by Step

$[NaOH]=0.62M$ We know that $[H^+]\times [OH^-]=1\times 10^{-14}$ Thus $[H^+]=\frac{1\times 10^{-14}}{[OH]^-}=\frac{1\times 10^{-14}}{ 0.62}=1.61\times 10^{-14}$
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