Answer
Please see the work below.
Work Step by Step
$[NaOH]=0.62M$
We know that
$[H^+]\times [OH^-]=1\times 10^{-14}$
Thus $[H^+]=\frac{1\times 10^{-14}}{[OH]^-}=\frac{1\times 10^{-14}}{ 0.62}=1.61\times 10^{-14}$
Chemistry 12th Edition
by Chang, Raymond; Goldsby, Kenneth
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
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