Chemistry: The Central Science (13th Edition)

Published by Prentice Hall
ISBN 10: 0321910419
ISBN 13: 978-0-32191-041-7

Chapter 16 - Acid-Base Equilibria - Exercises - Page 718: 16.34


The difference is approximately 2.398.

Work Step by Step

1. As mentioned in the question, the [H+] in A is 250 greater than in B. $[H^+]^A = [H^+]^B \times 250$ 2. Find the difference between the pH of solutions A and B. $pH^B - pH^A$ $-log[H^+]^B - (-log[H^+]^A)$ *** Substitute the concentration of H+ in A, by the concentration of H+ in B multiplied by 250. $-log[H^+]^B - (-(log[H^+]^B \times 250))$ $-log[H^+]^B - (-(log[H^+]^B + log(250)))$ $-log[H^+]^B + log[H^+]^B + log(250)$ $log(250) \approx 2.398$
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