Answer
$29.10\ mL$
Work Step by Step
Number of moles of $OH^-$:
$200.\ mg\div78.00\ mg/mmol\times3+200.\ mg\div58.32\ mg/mmol\times2=14.55\ mmol$
Same number of moles of acid to complete neutralization:
$14.55\ mmol\div0.500\ M=29.10\ mL$
Chemistry and Chemical Reactivity (9th Edition)
by Kotz, John C.; Treichel, Paul M.; Townsend, John R.; Treichel, David A.
Published by
Cengage Learning
ISBN 10:
1133949649
ISBN 13:
978-1-13394-964-0
Chapter 4 Stoichiometry: Quantitative Information about Chemical Reactions - 4-8 Spectrophotometry - Applying Chemical Principles - Questions - Page 178: 4
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
