Chemistry: The Molecular Nature of Matter and Change 7th Edition

Published by McGraw-Hill Education
ISBN 10: 007351117X
ISBN 13: 978-0-07351-117-7

Chapter 19 - Problems: 19.13

Answer

pH of buffer = 3.28 Hydronium ion concentration = 5.2 x 10^-4 M

Work Step by Step

1 (A) Given : 1. [conjugate base] = [NO2-] = 0.75 M 2. [acid] = [HNO2] = 0.55 M 3. Ka of HNO2 = 7.1 X 10^-4 2 (B) Formulae : 1. Henderson - Hasselbalch equation of a buffer : pH = pKa + log [conjugate base] / [acid] 2. pKa = -log Ka 3. pH = -log [H+] 3 (C) Solution : 1. pKa = -log Ka = - log (7.1 X 10^-4) = 3.15 .... (1) 2. pH = pKa + log [conjugate base] / [acid] = 3.15 + log (0.75 / 0.55) = 3.15 + 0.13 = 3.28 Hence, pH of the buffer is 3.28. 3. pH of the solution = -log [H+] So, [H+] = 10^(-pH) = 10^(-3.28) = 5.2 x 10^-4 M 4 (D) Answer : pH of buffer = 3.28 Hydronium ion concentration = 5.2 x 10^-4 M
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