Chemistry: Atoms First (2nd Edition)

Published by Cengage Learning
ISBN 10: 1305079248
ISBN 13: 978-1-30507-924-3

Chapter 13 - Exercises - Page 575d: 80

Answer

The concentration of acetic acid in equilibrium in vinegar is = 0.05556M.

Work Step by Step

1. Find the $[H_3O^+]$ value: $[H_3O^+] = 10^{-pH}$ $[H_3O^+] = 10^{- 3}$ $[H_3O^+] = 1 \times 10^{- 3}$ 2. Drawing the equilibrium (ICE) table we get these concentrations at equilibrium:** The image is in the end of this answer. -$[H_3O^+] = [Acetate] = x$ 3. Now, use the Ka and x values and equation to find the initial concentration value. $Ka = \frac{[H_3O^+][Acetate]}{ [Acetic acid]}$ $ 1.8\times 10^{- 5}= \frac{[x^2]}{ [Acetic acid]}$ $ 1.8\times 10^{- 5}= \frac{( 1\times 10^{- 3})^2}{[Acetic acid]}$ $[Acetic acid]= \frac{ 1\times 10^{- 6}}{ 1.8\times 10^{- 5}}$ $[Acetic acid] = 0.05556$
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