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: 19.22

Answer

pH of buffer as calculated above = 10.55

Work Step by Step

1 The dissociation of Carbonic Acid (H2CO3) occurs in the following two steps : H3PO4(aq) + H2O(l) → H2PO4-(aq) + H3O+(aq) H2PO4-(aq) + H2O(l) → (CO3)2-(aq) + H3O+(aq) (HPO4)2-(aq) + H2O(l) → (PO4)3-(aq) + H3O+(aq) As the buffer contains NaH2PO4 and Na2HPO4, i.e. H2PO4- and (PO4)2-. And, because both of these are present in the second reaction, the Ka of 2nd step, that is Ka2 is more significant with respect to this reaction. 2 (A) Given : 1. [base] = [(PO4)2-] = 0.40 M 2. [acid] = [KHCO3] = 0.50 M 3. Ka2 = 6.3 x 10^-8 3 (B) Formulae : 1. Henderson - Hasselbalch equation of a buffer : pH = pKa + log [conjugate base] / [acid] 2. pKa = -log Ka 4 (C) Solution : 1. pKa = -log Ka = - log (6.3 x 10^-8) = 7.20 .... (1) 2. pH = pKa + log [conjugate base] / [acid] = 7.20 + log (0.40 / 0.50) = 7.20 + (-0.10) = 7.10 Hence, pH of the buffer is 7.10 5 (D) Answer : pH of buffer as calculated above = 10.55
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