Horvath Practice Problems 69

Horvath Practice Problems 69 - 5.00 ( K a = 1 . 7 10 5 )?...

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The p K a = log 1 . 7 × 10 5 =4 . 77 Using the Henderson—Hasselbalch equation we have 4 . 25 = 4 . 77 + log à £ C 2 H 3 O 2 ¤ [ HC 2 H 3 O 2 ] ! log à £ C 2 H 3 O 2 ¤ [ HC 2 H 3 O 2 ] ! =4 . 25 4 . 77 = 0 . 52 à £ C 2 H 3 O 2 ¤ [ HC 2 H 3 O 2 ] ! =1 0 0 . 52 =0 . 30 This tells use that at equilibrium the ratio must be 0.30. Technically we should back calculate from the equilibrium con- centrations to the initial concentrations. However, for bu f ers the equilibrium concentrations of the acid and the conjugate base do not di f er signi f cantly from the start- ing concentrations. Thus one way to make this bu f e rw ou ldb etom ix ,say ,50 mL of a 0.2 M solution of HC 2 H 3 O 2 with 50 mL of a 0.06 M solution of NaC 2 H
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Unformatted text preview: 5.00 ( K a = 1 . 7 10 5 )? 2. What would the pH of an acetate bu f er made by com-bining 50 mL of a 1.00 M HC 2 H 3 O 2 solution with 50 mL of a 1.25 M NaC 2 H 3 O 2 solution? 3. It is a good idea to choose an acid base conjugate pair that has a p K a that is near to the pH that one is trying to make a bu f er for. Why? 4. Let say you wanted to hold your pH to within 0.1 pH unit around the p K a of acetic acid. What is the acceptable range of the ratio [ C 2 H 3 O 2 ] [ HC 2 H 3 O 2 ] ? 68 MORE ON BUFFERS: THE HENDERSON-HASSELBALCH EQUATION...
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This document was uploaded on 11/01/2011 for the course CHM 2046 at University of Florida.

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