The pH of blood is about 74 whereas that of gastric juices is about 15 Each of

The ph of blood is about 74 whereas that of gastric

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CHEM162-2014 CHAPTER 17 9 PREPARING A BUFFER Mix equal moles of acid and conjugate base (or base and conjugate acid). Neutralize half of a weak acid with a strong base (or half of a weak base with a strong acid).
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CHEM162-2014 CHAPTER 17 10 CHEM 162-2009 FINAL EXAM CHAPTER 15B - APPLIC. OF ACID & BASE EQUILIBRIA 50. Which of the following mixtures willresult in a buffer? W. 0.20 mol CH3COOH + 0.10 mol NaOH in 1.00 L solution X. 0.20 mol NaOH + 0.10 mol CH3COOH in 1.00 L solution Y. 0.20 mol NH3+ 0.10 mol HCl in 1.00 L solution Z. 0.20 mol HCl + 0.10 mol NH3in 1.00 L solution A - 0 A - 0 2 O 0 2 O 0
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CHEM162-2014 CHAPTER 17 11 15.23d Calculate the pH of a mixture containing 0.100 M HC 3 H 5 O 2 and 0.100 M NaC 3 H 5 O 2 . (HC 3 H 5 O 2 , K a = 1.3 x 10 -5 ) *ET: Problem done previously. 0.100 M HP 0.100 M NaP HP + H 2 O H 3 O + + P - HP(aq) + H 2 O( l ) H 3 O + (aq) + P - (aq) Initial 0.100 0 0.100 Change Equilibrium HP(aq) + H 2 O( l ) H 3 O + (aq) + P - (aq) Initial 0.100 0 0.100 Change -X +X +X Equilibrium 0.100 - X +X 0.100+X ([H 3 O + ][P - ])/[HP] = K a ((X) x (0.100 + X))/(0.100 - X) = 1.3 x 10 -5 Simplifying, with small K rule: (X x 0.100)/(0.100) = 1.3 x 10 -5 X = 1.3 x 10 -5 pH = -log 1.3 x 10 -5 = 4.89
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CHEM162-2014 CHAPTER 17 12 27d. Calculate the pH after 0.020 mol HCl is added to 1.00 L of a mixture containing 0.100 M HC3H5O2and 0.100 M NaC3H5O2. Kpropanoic acid= 1.3x10-5Note: In absence of a buffer, pH of 0.020M HCl = 1.70 ET: This slide ties into the previous slide of an HP/P-buffer without a strong acid. ET: Given weak acid, conjugate base, and added strong acid concentrations; find pH. What equilibrium equation should be used for the ICE table? . H HP
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