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Lecture 15 10/05/05 Research Club
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What is the charge and mass balance for 0.l5 M phosphoric acid Charge balance: [H + ] = [H 2 PO 4 - ] + 2[HPO 4 -2 ] + 3[PO 4 -3 ] + [OH - ] Mass balance: 0.15 M = [H 3 PO 4 ] + [H 2 PO 4 - ] + [HPO 4 -2 ] + [PO 4 -3 ]
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Develop a mass balance for chloride and zinc in a solution of zinc chloride complexes: ZnCl 2, Zn +2, ZnCl +1, ZnCl 3 -, ZnCl 4 -2 [Cl-] total = 2 [ZnCl 2 ] + [ZnCl +1 ] + 3 [ZnCl 3 - ] + 4 [ZnCl 4 -2 ] [Zn] total = [ZnCl 2 ] + [Zn +2 ] + [ZnCl +1 ] + [ZnCl 3 - ] + [ZnCl 4 -2 ]
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Reminder: K W = 10 -14 = [OH - ][H + ] pOH + pH = 14 at 25 o C
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pH of 0.1 M NaOH? pH = 13 pH of 0.001 M NaOH? pH = 11 pH of 10 -8 M NaOH? pH = 6 ?????
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Equilibrium equations: H 2 O H + + OH - Charge balance: [Na + ] + [H + ] = [OH - ] Mass Balance: [Na + ] = 1.0 x 10 -8 M Equilibrium constant: K w = 10 -14 = [H + ][OH - ]
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Count equations and unknowns 3 unknowns Na +, OH -, H 3 O + 3 independent equations [Na + ] + [H 3 O + ] = [OH - ] [Na + ] = 1.0 x 10 -8 M K w = [H 3 O + ][OH - ] = 1.0 x 10 -14
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Solve [Na + ] + [H 3 O + ] = [OH - ] 1.0 x 10 -8 + x = [OH - ] K w = [H 3 O + ][OH - ] 1.0 x 10 -14 = (x) (1.0 x 10 -8 + x) 1.0 x 10 -14 = 1.0 x 10 -8 (x) + (x) 2 x 2 + (1.0 x 10 -8 )x – (1.0 x 10 -14 ) = 0
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x = 9.6 x 10 -8 M or -1.1 x 10 -7 M x = [H + ] = 9.6 x 10 -8 M pH = 7.02
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Buffer Mixture of weak acid and conjugate base
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Henderson-Hasselbach equation (only can use with buffers)
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When [A - ] = [HA], pH = pK a
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NH 3 + H 2 O NH 4 + + OH - Still use pK a and still base over acid
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Can find ratio [A - ] / [HA] at any pH [OCl - ] / [HOCl] = ? at pH = 6.20 pK a = 7.53
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Properties of buffers 1.Effect of dilution 2.Effect of added acid/base
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