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Horvath Practice Problems 68

Horvath Practice Problems 68 - ¤ HA Solving for pH gives...

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16 MORE ON BUFFERS: THE HENDERSON-HASSELBALCH EQUATION As we learned from the third group work question last time, it is important to be able to make bu f ers of a desired pH. If we consider the K a equation for a bu f er situation we have, K a = £ H + ¤£ A ¤ [ HA ] , (16.1) for the bu f er equilibrium, HA ­ H + + A . (16.2) Nowi fwetakeo fthenegat ivelogo fthe K a equation, we get log K a = log Ã £ H + ¤£ A ¤ [ HA ] ! log K a | {z } p K a = log £ H + ¤ | {z } pH log Ã £ A ¤ [ HA ] ! p K a = pH log
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Unformatted text preview: ¤ [ HA ] ! Solving for pH gives the Henderson—Hasselbalch equation: pH = p K a + log Ã £ A − ¤ [ HA ] ! . We can use the Henderson—Hasselbalch equation to short-cut our bu f er calculations. Example 1 : How would you make an acetate bu f er that has a pH of 4.25 ( K a = 1 . 7 × 10 − 5 )? MORE ON BUFFERS: THE HENDERSON-HASSELBALCH EQUA-TION 67...
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