1. Hydrogen sulphide gas at a partial pressure of 10-4.8bar is in contact with a solution at pH8.5.a. Prepare a log C-pH diagram for H2S/HS-/S2-system in equilibrium with 10-4.8bar H2S(g).b. Assuming that the S-containing species are the only contributors to alkalinity, what is the alkalinity of the solution? Assume that the endpoint of the alkalinity titration is pH 4.5 and that no H2S enters or leaves the solution during the titration.You have to use the full alkalinity equation (Hf+-Hi+) –(OH)f-etc. You can then eliminate terms that are trivial to simplify it. Note that I went throughthis for the carbonate systemin class, where the ALK simplified to [HCO3-]i.See slide 31 in pdf #7.c. Assuming that the system in part a. remains in continuous equilibrium with the gas phase, write out the proton condition and find the pH after 5 x 10-4M NaOH has been added to the solution. What is the change in total dissolved sulphide concentration when the NaOH is added?You can use the same log C-pH diagram that you drew for part a, but you have to add a line for TOTHin. Note that we are dealing with an opensystem, so you mustuse the species that goes in to the gas phase (H2S in this case) as your reference species. This is different from a closed system (what we have been dealing with up till now) where the reference species is the dominantspecies at equilibrium).2. The microorganisms in a waste treatment pond are exposed to 35 mg/L of easily degradable organic compounds with a typical composition C5H10O5(completely soluble).The standard molar Gibbs energy of formation of these compounds is -2797 kJ/mol (Gfofor H2CO3= -623.2kJ and for O2= 16.32 kJ). The pond is at pH 7.3, its temperature is 25ºC, and its alkalinity is 2 x 10-3eq/L. The overall reaction for the oxidation of organic matter is:C5H10O5+ 5O2(aq) 5H2CO3(aq)Note that this system is notat equilibrium, so you can’t use Henry’s Law to determine the H2CO3(need Ka1and HCO3-).
Assume that the alkalinity is entirely due to the carbonate system (remember how we can simplify the equation then (done in class)).
a. Due to oxygen consumption by the above reaction, the dissolved oxygen concentration in the water is only 2 mg/L. What is the maximum amount of energy (i.e. G) that an organism could obtain by oxidizing 1 mol of the organic matter?
b. The operator of the treatment facility has suggested that the treatment efficiency could be improved by bubbling air through the water to encourage biological activity. Would the organisms find the environment in the untreated water more favourable, less favourable, or identical (from an energy standpoint) after equilibration with air? Explain briefly –a calculation wouldn’t hurt to make your point(i.e. recalculate G).(Bringing the system into equilibrium with air –can now use Henry’s Law to calculate to determine O2and CO2
c. Use the log C-pH diagram for an open carbonate system (you have the spreadsheet already) to determine the pH of the system after it has been bubbled with air.
