Fundamentals of Biochemistry: Life at the Molecular Level 5th Edition

Published by Wiley
ISBN 10: 1118918401
ISBN 13: 978-1-11891-840-1

Chapter 2 - Water - Exercises - Page 40: 15

Answer

PH = 7.15

Work Step by Step

Final volume = 200 mL Hence, Final molarity of K2HPO4 = 50 mL x 2M / 200 mL = 0.50 M Final molarity of KH2PO4 = 25 mL x 2 M/ 200 mL = 0.25 M pH of the solution is calculated by Henderson equation as, PH = pka + log [ K2HPO4] / [KH2PO4] Since, pKa of KHPO4 = 6.85 [K2HPO4] = 0.50 M [KH2PO4] = 0.25 M Hence, PH = 6.85 + log [0.50 / 0.25] PH = 6.85 + log 2 PH = 6.85 +0.3010 = 7.151 PH = 7.15
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