Lecture 13 - C apte 0 o yp ot c ac d base qu b a Chapter 10...

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hapter 10 Polyprotic cid- ase Equilibria Cap t e 0 oypotc ac d base qu b a Diprotic acid and bases Diprotic buffers Polyprotic acid and bases Principle Species, isoelectric point and isoionic pH
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Diprotic acid and bases Calculate the pH of 0.0500 M H2L + .
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Diprotic acid and bases The acid form, H 2 L + Calculate the pH of 0.0500 M H 2 L + .
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he base form L Calculate the pH of 0.0500 M L - . The base form, L -
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The intermediate form, HL? Calculate the pH of 0.0500 M HL.
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Diprotic buffers Henderson-Hasselbalch equitions Example: How many mililiters of 0.800 M KOH should be added to 3.38 g of oxalic acid to give pH of 4.40 when diluted to 500 mL. Oxalic acid (FM 90.035), pKa1 = 1.27; pKa2 = 4.266.
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3. From diprotic to polyprotic acids and bases H 2 A H + + HA - H + + A 2- K a1 K a2 H 3 A H + + H 2 A - H + + HA 2- K a1 K a2 K a3 H + + A 3-
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Example Find pH of 0.10 M H His 2+ , 3 0.10 M H 2 His + , 0.10 M HHis , 0.10 M His -
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Principle Species and Approximate Fraction Buffer Henderson-Hasselbalch pH 7, and pH 11.28
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Isoelectric point and isoionic pH Isoionic pH is the pH obtained when the HA (the neutral zwitterion) is dissolved in water. IsoelectricpH is the pH at which the
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This note was uploaded on 09/10/2009 for the course CHM 3120 taught by Professor Harrison during the Spring '08 term at University of Florida.

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Lecture 13 - C apte 0 o yp ot c ac d base qu b a Chapter 10...

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