Problems Chapter 4.

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Presentation transcript:

Problems Chapter 4

4. Subtract the expression for one temperature from the expression for the other temperature to eliminate the A term, giving, log k283.15 – log k298.15 = (B / R) ( 1/298.15 – 1/283.15) Substituting values for log k298.15, B and R, and taking anti-logs gives, k283.15 = 32 M-1 s-1 From previously, t½ = 1 / ([Al3+]0 kf) = 1 / (10-6M 32 M-1s-1) = 3125 s

7. KH = [A(aq)] / PA = [H2CO3] / PCO2 = 34.06 mol m-3 atm-1 = 0.03405 mol L-1 atm-1 cK1 / (cK2 KH) = {[H2CO3] / ([H+]2 [CO32-])} / {[HCO3-] / ([H+][CO32-])} x PCO2 / [H2CO3] = PCO2 / ([H+][HCO3-]) = 1016.7 / (1010.3 10-1.47) = 107.87 atm M-2 where values have been substituted. Rearrange to give, [H+] = PCO2 / (107.87 [HCO3-]), and substitute for given Conditions to give effect of PCO2 on [H+], [H+] = 10-5.37 @ PCO2 = 10-3.50 and [H+] = 10-3.87 @ PCO2 = 10-2.00

13. [Al3+] = (Al3+) / γ3+ = 10-6.23 / γ3+ I γ3+ 0.0001 0.900564 0.0003 0.835527 0.0010 0.724655 0.0030 0.581919 0.0050 0.504192 0.0100 0.393473 0.0300 0.229707 0.1000 0.107437 0.3000 0.060813 Davies model was used for activity coefficient. [Al3+] increases with increasing I.