1. Problems Chapter 4

2. 4.
Subtract the expression for one temperature from the expression for the other
temperature to eliminate the A term, giving,
log k – log k = (B / R) ( 1/ – 1/283.15)
Substituting values for log k298.15, B and R, and taking anti-logs gives,
k = 32 M-1 s-1
From previously, t½ = 1 / ([Al3+]0 kf) = 1 / (10-6M 32 M-1s-1) = 3125 s

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

4. 13. [Al3+] = (Al3+) / γ3+ = 10-6.23 / γ3+ I γ3+
0.0001
0.0003
0.0010
0.0030
0.0050
0.0100
0.0300
0.1000
0.3000
Davies model was used for activity coefficient.
[Al3+] increases with increasing I.