General Chemistry: Principles and Modern Applications (10th Edition)

Published by Pearson Prentice Hal
ISBN 10: 0132064529
ISBN 13: 978-0-13206-452-1

Chapter 16 - Acids and Bases - Exercises - Strong Acids, Strong Bases, and pH - Page 739: 17

Answer

It is necessary to have 0.0539L of HCl to neutralize this quantity of $NH_3$

Work Step by Step

1. Find the volume necessary for the neutralization: $1 * {C_1 * V_1} = 1 * {C_2 * V_2}$ $ 0.265* 1.25= 6.15 * V_2$ $ 0.33125 = 6.15 * V_2$ $V_2 = 0.05386L$ of HCl.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After 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.

GradeSaver will pay $15 for your literature essays
GradeSaver will pay $25 for your college application essays
GradeSaver will pay $50 for your graduate school essays – Law, Business, or Medical
GradeSaver will pay $10 for your Community Note contributions
GradeSaver will pay $500 for your Textbook Answer contributions