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Horvath Practice Problems 69

# Horvath Practice Problems 69 - 5.00 K a = 1 7 × 10 − 5 2...

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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 Ã £ C 2 H 3 O 2 ¤ [ HC 2 H 3 O 2 ] ! log Ã £ C 2 H 3 O 2 ¤ [ HC 2 H 3 O 2 ] ! = 4 . 25 4 . 77 = 0 . 52 Ã £ C 2 H 3 O 2 ¤ [ HC 2 H 3 O 2 ] ! = 10 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 ff ers the equilibrium concentrations of the acid and the conjugate base do not di ff er signi fi cantly from the start- ing concentrations. Thus one way to make this bu ff er would be to mix, 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 3 O 2 . Group Work 1. How would you make an acetate bu ff er that has a pH of
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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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