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Chrome File View History Bookmarks People Window Help 11% Mon Mar 5 7:17 PM , a MasteringChemistry: Assignment 07- Chapter 17 Secure https://session.masteringchemistry.com/myct/itemView?assignmentProblemID-95939378 Assignment 07-Chapter 17 Item 6 2017 ) 6of22 ))) Part A The Henderson-Hasselbalch equation relates the pH of a buffer solution to the pKa of its conjugate acid and the ratio of the concentrations of the conjugate base and acid. The equation is important in laboratory work that makes use of buffered solutions, in industrial processes where pH needs to be controlled, and in medicine, where understanding the Henderson-Hasselbalch equation is critical for the control of blood pH. Sc 2017- As a technician in a large pharmaceutical research firm, you need to produce 100. mL of a potassium dihydrogen phosphate buffer solution of pH-6.91. The pK, of H2POs is 7.21. You have the following supplies: 2.00 L of 1.00 M KH2 P04 stock solution, 1.50 L of 1.00 M K2HPO4 stock solution, and a carbay of pure distiled H20 How much 1.00 M KH POs wil you need to make this solution? (Assume acditive volumes.) Express your answer to three significant digits with the appropriate units View Available Hint(s) Volume of KH POs needed ValueUnits Submit Previous Answers Request Answer incorrect; Try Again The Henderson-Hasselbalch equation in medicine Carbon dioxide (CO2) and bicarbonate (HCOs concentrations in the bloodstream are physiolopicaly controlled to keep blood pll constant at a normal value of 7.40 5

Explanation / Answer

pH = pKa + log [HPO4^2-]/ [H2PO4-]

6.91 = 7.21 + log [HPO4^2-]/ [H2PO4-]

log [HPO4^2-]/ [H2PO4-] = -0.30

[HPO4^2-]/ [H2PO4-] = 0.501 ; 0.501 also represents a mole ratio

100 mL x 1 liter/1000 mL= 0.10 liters

Let x = volume of H2PO4-

Then (0.10 –x) = volume of HPO4-2

(0.10-x) 1.00 moles HPO4-2/ liter / x(1.00 mole H2PO4-/ liter = 0.501

0.10 – x / x = 0.501

0.10 – x = 0.501 x

x = 0.066 liters = 6.6mL

Volume of KH2PO4 = 6.6 mL

Volume of K2HPO4 = 100 mL – 6.6 mL = 93.4 mL

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