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Lecture 10 Acid-Base Equilibria -II
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K a = 10 -4 1. C = 0.01 M [H + ] = (10 -4 10 -2 ) ½ = 10 -3 M pH=3 90% acid, 10% base 2 C = 10 -9 M ??
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K b = 2.6 10 -6 C=10 -3 M [OH - ]=5 10 -5 [H + ]=2 10 -10 pH=9.7
![K b = 2.6 C=10 -3 M [OH - ]=5 [H + ]=2 pH=9.7](http://images.slideplayer.com./16/5110269/slides/slide_4.jpg "K b = 2.6 C=10 -3 M [OH - ]=5 [H + ]=2 pH=9.7")
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Sodium lactate C=0.01 M K a = 1.4 10 -4 = 10 -3.85 K b = 10 -10.15 [OH - ] =(10 -10.15 10 -2 ) ½ = 10 -6.08 [H + ] = 10 -7.92 pH=7.92 8
![Sodium lactate C=0.01 M K a = 1.4 = K b = [OH - ] =( ) ½ = [H + ] = pH=7.92 8](http://images.slideplayer.com./16/5110269/slides/slide_5.jpg "Sodium lactate C=0.01 M K a = 1.4 = K b = [OH - ] =( ) ½ = [H + ] = pH=7.92 8")
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Polyprotic acids: 1.4 6.0 3.4 8.5 Zwitter-ion Aminoacids:
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pK 1 =2.35 pK 2 =9.87 H 2 A HA + H + HA A + H +
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H2AH2AHAA pH of H 2 CO 3 -same as any weak acid; use K 1 pH of Na 2 CO 3 Na 2 CO 3 2 Na + + CO 3 2- CO 3 2- same as any weak base, use K 2
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H2AH2A HAA NaHCO 3 NaHCO 3 Na + + HCO 3 - In HA, [A]=[H 2 A] because 2HA A + H 2 A [H + ] 2 = K 1 K 2
![H2AH2A HAA NaHCO 3 NaHCO 3 Na + + HCO 3 - In HA, [A]=[H 2 A] because 2HA A + H 2 A [H + ] 2 = K 1 K 2](http://images.slideplayer.com./16/5110269/slides/slide_10.jpg "H2AH2A HAA NaHCO 3 NaHCO 3 Na + + HCO 3 - In HA, [A]=[H 2 A] because 2HA A + H 2 A [H + ] 2 = K 1 K 2")
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NaHCO 3 pK 1 =6.35 pK 2 = 10.33
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H 3 PO 4 H 2 PO 4 - HPO 4 2- PO 4 3- 2.157.2012.15
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