827486 00006279602 380 0000158489 019892

827486 00006279602 380 0000158489 019892

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5.01187E-11 0.000827486 -0.0006279602 3.80 0.000158489 6.30957E-11 0.001019892 -0.0008614025 3.90 0.000125893 7.94328E-11 0.001250933 -0.0011250410 4.00 0.0001 1E-10 0.001525424 -0.0014254238
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3.381 0.000415911 2.40436E-11 0.000414832 0.0000010786 3.3811 0.000415815 2.40492E-11 0.000414924 0.0000008913 3.3812 0.000415719 2.40547E-11 0.000415015 0.0000007040 3.3813 0.000415623 2.40602E-11 0.000415107 0.0000005166 3.3814 0.000415528 2.40658E-11 0.000415198 0.0000003293 3.3815 0.000415432 2.40713E-11 0.00041529 0.0000001420 3.38 3.3816 0.000415336 2.40769E-11 0.000415382 -0.0000000453 3.3817 0.000415241 2.40824E-11 0.000415473 -0.0000002326
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Non-ideal solutions 1.0 Example Practice Problem-2 : Compare the pH and the concentrations of HF and F- in a solution prepared by adding 10 -3 M HF to pure water, with that of a solution prepared the same way, except using water containing 10 - 1 M NaCl. Assume that the ionic strength is low enough in the NaCl-free solution that the solution behaves ideally. To estimate the activity coefficients in the solution containing NaCl, use Davies’ equation. ) 3 . 0 1 ( log 2 I I I Az i i
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HF in Pure Water - Assume ideal solution ( =1 ) SPECIES WITH UNKNOWN ACTIVITIES AT EQUILIBRIUM Type-a: H, OH- Type-b: HF, F- (HF is a weak acid) CHEMICAL REACTIONS AND EQUILIBRIUM RELATIONSHIPS H 2 O ↔ H + + OH - HF ↔ H + + F - (Eq.1) (Eq.2) MASS BALANCE (MB) TOTF = 10 -3 M = [HF] + [F - ] (Eq.3) CHARGE BALANCE (CB) [H + ] = [OH - ] + [F - ] (Eq.4)
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HF in Pure Water - Assume ideal solution ( =1 ) SOLVING BY USE OF EXCEL SPREADSHEET Based on the convention that at standard state [c i ] =1.0 mol/L , we can rewrite the K eq equations in terms of concentrations and activity coefficients as follows: and MB : TOTF = 10 -3 M = [HF] + [F - ] CB : [H + ] = [OH - ] + [F - ] Since Therefore, CB equation becomes:
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pH [H+]=10 -pH [OH - ] = K w /[H + ] [F - ] = (TOTHF)(K a /(K a +[H + ])) Sum(+)-Sum(-) ([H+]) - ([OH-]-[F-]) 0 1 1E-14 6.75626E-07 0.9999993244 1 0.1 1E-13 6.71543E-06 0.0999932846 2 0.01 1E-12 6.33269E-05 0.0099366731 3 0.001 1E-11 0.000403371 0.0005966292 4 0.0001 1E-10 0.000871148 -0.0007711479 5 0.00001 0.000000001 0.000985425 -0.0009754255 6 0.000001 0.00000001 0.000998523 -0.0009975331 7 0.0000001 0.0000001 0.000999852 -0.0009998521 8 0.00000001 0.000001 0.000999985 -0.0010009752 9 0.000000001 0.00001 0.000999999 -0.0010099975 10 1E-10 0.0001 0.001000000 -0.0010999998 11 1E-11 0.001 0.001000000 -0.0020000000 12 1E-12 0.01 0.001000000 -0.0110000000 13 1E-13 0.1 0.001000000 -0.1010000000 14 1E-14 1 0.001000000 -1.0010000000
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Pure Water : ideal solution =1 (Cont’d) pH [H+]=10 -pH [OH - ] = K w /[H + ] [F - ] = (TOTHF)(K a /(K a +[H + ])) Sum(+)-Sum(-) 3.10 0.000794328 1.25893E-11 0.000459792 0.0003345365 3.20 0.000630957 1.58489E-11 0.000517263 0.0001136948 3.21 0.000616595 1.62181E-11 0.000523010 0.0000935854 3.22 0.00060256 1.65959E-11 0.000528751 0.0000738090 3.23 0.000588844 1.69824E-11 0.000534484 0.0000543597 3.24 0.00057544
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  • Fall '08
  • CHADIK
  • pH, HAc

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