General Chemistry 10th Edition

Published by Cengage Learning
ISBN 10: 1-28505-137-8
ISBN 13: 978-1-28505-137-6

Chapter 15 - Acids and Bases - Questions and Problems - Page 662: 15.129

Answer

Please see the work below.

Work Step by Step

We know that $1.00g PA\times \frac{1 mol \space PA}{81.994\space g\space PA}\times \frac{2\space mol \space NaOH}{1 mol\space PA}\times \frac{40.00g NaOH}{1 \space mol \space NaOH}=0.976g\space NaOH$
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