Please help with parts a through e of this problem. Free chlorine disinfection i
ID: 1024385 • Letter: P
Question
Please help with parts a through e of this problem.
Free chlorine disinfection is accomplished by the weak acid hypochlorous acid, which is subject to dissociation to hypochlorite ion according to the following equilibria:
HOCl <-----> H+ + OCl-
Hypochlorous acid is a much stronger oxidant/disinfectant than hypochlorite, so total disinfectant dosages will depend on the degree of HOCl dissociation in solution, which is dependent on solution pH.
(a) Assume that 0.1 mM HOCl is added to solution. Using the thermodynamic information provided in Appendix A.1, estimate the value of Keq for the hypochlorous acid dissociation reaction at T = 25 °C, and P = 1 atm(b)
(b) Estimate the ratio of [OCl-]/[HOCl] at pH = 7 (i.e., {H+} = 10-7) if we assume infinite dilution conditions.
(c) Estimate the ratio of [OCl-]/[HOCl] at pH = 7 if we assume that ionic strength = 0.1 M
(d) Estimate the ratio of [OCl-]/[HOCl] at pH = 7 if we assume infinite dilution conditions and T = 10 °C
(e) Discuss how changes in solution composition (e.g., pH, ionic strength) or temperature of water in a treatment plant would affect design disinfectant dosages if we assume that the water is highly buffered at pH 7.
APPENDIX INFO : free energies of soecies in water
HOCL : -79.9 kj/mol
H+: 0kj/mol
OCL-: -36.8 kj/mol
Explanation / Answer
a) We know that
Delta G0 rxn = -RT ln Keq ....(1)
Also
Delta G0 rxn = sum [Delta G0 products] - sum [Delta G0 reactants]
APPENDIX INFO : free energies of soecies in water [hope the signs are correct as mentioned by you]
HOCL : -79.9 kj/mol
H+: 0kj/mol
OCL-: -36.8 kj/mol
Delta G0 rxn = [-36.8 + 0 ] - [-79.9] = 43.1 KJ / mole
Putting the value in equation (1)
Delta G0 rxn = 43.1 = - 8.314 X 10^-3 X 298 X ln Keq
ln Keq = -17.39 t
taking antilog
Keq = 2.81 X 10^-8
b) Keq = [[H+] [ OCl-] / [HOCl]
[H+] = 10^-7
2.81 X 10^-8 = 10^-7 X [ OCl-] / [HOCl]
[OCl-]/[HOCl] = 0.281
c) ionic strength = 1/ 2 (sum [ci zi2]
ci = concentration of ion
zi = charge on ion
ci for H+ = 10^-7 zi = +1
ci for OCl- = Ci zi =-1
Ionic strength = 1/2 ( 10^-7 (1)^2 + Ci (-1)^2 ) = 0.1
0.2 = 10^-7 + Ci
Ci = 0.2 (approx)
Keq = [[H+] [ OCl-] / [HOCl] = 10^-7 X 0.2 / [HOCl]
[HOCl] = 10^-7 X 0.2 / 2.81 X 10^-8 = 7.12
[OCl-] / [HOCl] = 0.21 / 7.12 = 0.0294
d) The equilibrium constant will change
DeltaG = -RT ln Keq
Delta G0 rxn = 43.1 = - 8.314 X 10^-3 X 283 X ln Keq
Keq = 1.106 X 10^-8
Keq = 1.106 X 10^-8 = [H+] [ OCl-] / [HOCl]
Keq = 1.106 X 10^-8 = 10^-7 [ OCl-] / [HOCl]
[ OCl-] / [HOCl] = 0.1106