Chemistry 12th Edition

Published by McGraw-Hill Education
ISBN 10: 0078021510
ISBN 13: 978-0-07802-151-0

Chapter 16 - Acid-Base Equilibria and Solubility Equilibria - Questions & Problems - Page 765: 16.15

Answer

$\frac{[CH_3COO^-]}{[CH_3COOH]} = 0.56$

Work Step by Step

1. Calculate $[H_3O^+]$: $[H_3O^+] = 10^{-pH}$ $[H_3O^+] = 10^{- 4.5}$ $[H_3O^+] = 3.2 \times 10^{- 5}$ 2. Write the $K_a$ equation, and find the ratio: $K_a = \frac{[H_3O^+][CH_3COO^-]}{[CH_3COOH]}$ $1.8 \times 10^{-5} = \frac{3.2 \times 10^{-5}*[CH_3COO^-]}{[CH_3COOH]}$ $\frac{1.8 \times 10^{-5}}{3.2 \times 10^{-5}} = \frac{[CH_3COO^-]}{[CH_3COOH]}$ $0.56 = \frac{[CH_3COO^-]}{[CH_3COOH]}$
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.