09.04 - Horowitz - Buffers and pH

09.04 - Horowitz - Buffers and pH - 14 Derivation for pH of...

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14 Derivation for pH of weak acid; pH = ½ pKa + ½ pC HA H + + A K a = [H + ][A ] [HA] If add C moles of acid to 1L H 2 O, • material balance= [C] = [HA] + [A ] • electroneutrality = [H+] = [A ]+[OH ] Since [OH ] is very small in solution of common weak acid, you can simplify by neglecting [OH ], [H + ] = [A ] • ex. 0.1M HA in pH 2.8 where [H + ] is 1.6x10 –3 pOH is 11.2 or [OH ] = 6.3x10 –12 Also, [C] = [HA]+[H + ] or [HA] = [C] – [H + ] and [H + ] is small, so [HA] = [C] K a = [H + ][A ] [[C]–[H + ]] ––– [H+] = K a C pH = ½ (pKa + pC) ½ (pKa – log C)
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15 Derivation of the Henderson-Hasselbalch Equation Consider a weak acid [HA], where HA = Brønsted acid (proton donor, conjugate acid) A = Brønsted base (proton acceptor, conjugate base) HA H + + A K a = [H + ][A ] (for buffers, [H + ] [A ]) [HA] Take the logarithm and rearrange: log K a = log [H+] + log [A-] [HA] –log [H+] = –log K a + log [A-] [HA] pH = pK a + log [A-] [HA] 1. Express concentrations as molar, millimolar, etc. 2. Buffering range is pK
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This note was uploaded on 04/18/2008 for the course BIOC 1010 taught by Professor Zhang during the Fall '07 term at New York Medical College.

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09.04 - Horowitz - Buffers and pH - 14 Derivation for pH of...

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