Lecture5 - Chapter 9: Buffers Consider two different cases...

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Consider two different cases Case 1: H 2 O 1L 1.0 x 10 -4 mol HCl [H + ] = 1.0 x 10 -4 mol/1L = 1.0 x 10 -4 M pH = -log[H + ] = 4.00 pH change = 3.00 Chapter 9: Buffers Neutral water pH = 7
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0.10 M CH 3 COOH 1L 1.0 x 10 -4 mol HCl Case 2: CH 3 COOH ' CH 3 COO - + H + K a = x 2 /(0.10-x) = 1.75 x 10 -5 0.10 -x x x x = [H + ] = 1.31 x 10 -3 M pH = 2.883 After adding 1.0 x 10 -4 mol HCl CH 3 COOH ' CH 3 COO - + H + 0.10 -x x x + 1.0x10 -4 K a = x(x+1.0x10 -4 )/(0.1-x) = 1.75 x 10 -5 x 2 +11.75x10 -5 x -1.75x10 -6 = 0 x =1.27 x 10 -3 M [H + ] = 1.0x10 -4 + 1.27 x10 -3 = 1.37 x10 -3 M pH = 2.863 pH change = 0.02
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Buffer solution : Buffer solution is a mixture of an acid and its conjugate base. it resists changes in pH when acids or bases are added from outside or when it is diluted.
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Le Châtelier’s Principle -- revisited HA ' H + + A - Consider a 0.010M solution of an acid HA whose pK a = 4.00 0.010 -x x x x 2 /(0.010-x) = 1.0 x 10 -4 => x = 9.5x10 -4 M Fraction of dissociation = [A - ]/F= 9.5 x10 -4 /0.010 = 9.5 x 10 -2 Now, we add [A - ] = 1.0 x 10 -4 M to the solution HA ' H + + A - 0.010 -x x x + 1.0 x10 -4 x(x+1.0x10 -4 )/(0.010-x) = 1.0 x 10 -4 => x = 9.0x10 -4 M Fraction of dissociation = x/F= 9.0 x10 -4 /0.010 = 9.0 x 10 -2 Less HA dissociate into H + and A -
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The Henderson-Hasselbalch Equation [HA] ] ][A [H + = = + + [HA]
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Lecture5 - Chapter 9: Buffers Consider two different cases...

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