Introductory Chemistry (5th Edition)

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

Chapter 13 - Solutions - Exercises - Problems - Page 482: 94


0.183 M

Work Step by Step

$HClO_{4}$ + NaOH → $H_{2}O$ + $NaClO_{4}$ From the above equation, it is clear that 1 mole of $HClO_{4}$ completely reacts with 1 mole of NaOH. So, the concentration of $HClO_{4}$ used for neutralization, = (45.3 X 0.101)/25 = 0.183 M
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