ComplicatedAcidBaseExampleSoln

- 7.66 61927.97 3.74E-19 7.68 58443.71 5.47E-19 7.7 55177.6 7.32E-19 ANSWER pH is 7.61[PO 4 3 = f[H[PO 4 3 EN EN check 3.05E-07 2.92E-13 8.83E-08

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Lake Mendota ferric phosphate Ka1 7.59E-03 Ka2 6.17E-08 Ka3 4.79E-13 Ksp 1.26E-18 Kw 0 should be zero at answer pH [H+] alpha -2 100 ### 0 -3.78E-18 -1 10 ### 0 -3.78E-18 0 1 ### 0 -3.78E-18 1 0.1 ### 0 -3.74E-18 2 0.01 ### 0 -3.36E-18 3 0 ### 0 -2.85E-18 4 0 ### 0 -2.72E-18 5 0 ### 0 -2.70E-18 6 0 ### 0 -2.63E-18 7 0 547779.24 0 -2.04E-18 8 0 24282.41 0 5.69E-18 9 0 2124.18 0 2.42E-16 10 0 210.27 0 7.74E-15 11 0 21.9 0 2.40E-13 12 0 3.09 0 6.38E-12 13 0 1.21 0 1.02E-10 14 0 1.02 0 1.11E-09 7 0 547779.24 0 -2.04E-18 7.2 0 266724.08 0 -1.69E-18 7.4 0 136880.84 0 -1.11E-18 7.6 0 73861.44 0 -8.34E-20 7.8 0 41625.51 0 1.87E-18 8 0 24282.41 0 5.69E-18 7.6 0 73861.44 0 -8.34E-20 7.62 0 69618.23 0 5.92E-20 7.64 0 65646.86 0 2.11E-19
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Unformatted text preview: 7.66 61927.97 3.74E-19 7.68 58443.71 5.47E-19 7.7 55177.6 7.32E-19 ANSWER: pH is 7.61 [PO 4 3-] = f([H+], [PO 4 3-]) EN EN check 3.05E-07 2.92E-13 8.83E-08 2.17E-07 2.02E-08 2.96E-07 2.62E-13 8.29E-08 2.13E-07-1.39E-08 [Fe 3+ ] [H 3 PO 4 ] [H 2 PO 4-] [HPO 4 2-] [PO 4 3-] [Fe 3+ ] [H 3 PO 4 ] [H 2 PO 4-] [HPO 4 2-] [PO 4 3-]...
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This note was uploaded on 05/04/2010 for the course CEE 320 taught by Professor Mcmahon during the Spring '10 term at Wisconsin.

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- 7.66 61927.97 3.74E-19 7.68 58443.71 5.47E-19 7.7 55177.6 7.32E-19 ANSWER pH is 7.61[PO 4 3 = f[H[PO 4 3 EN EN check 3.05E-07 2.92E-13 8.83E-08

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