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Environmental Chemistry of Aqueous Systems

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Introduction

Environmental Chemistry of Aqueous Systems 1. Introduction Water is an incredibly important part of our environment. * Biological systems cannot survive without it. The recently funded ESA programme, Darwin, which is to look for signs of life on planets from other Solar Systems targets water along with CO2 and O2 (detected as O3) as key indicators for life. * Evaporation and condensation of water allows the transport of heat within our atmosphere, ultimately driving the wind system, while currents in the oceans provide another means of heat transfer. * The oceans can hold gases in solution, supplying the atmosphere with gases and acting as a buffer against atmospheric change. * It is involved in the formation of OH radicals in the atmosphere, which clean up our atmosphere. Distribution of Water 97 % is in the oceans 2.4 % is in snow and the polar ice sheets 0.6 % is in rivers and freshwater lakes 0.001 % is present in the atmosphere Need to understand Clearly, there are aspects of physics beyond the scope of these lectures that need to be addressed; e.g., latent heat of evaporation, condensation and freezing; atmospheric circulation; ocean currents, etc. For this course, we need to understand the chemistry of species dissolved in water. Two aspects are important: * Natural water systems. ...read more.

Middle

CaCO3 ( Ca2+ + CO32- Ksp = 5 ? 10-9 M2 S = Solubility = [Ca2+] = [CO32-] = S = 7 ? 10-5 M Effect of reaction of carbonate CO32- + H2O ( HCO3- + OH-; K3 = Kw/K2 = 2 ? 10-4 M. Combine to get CaCO3 + H2O == Ca2+ + HCO3- + OH- K = KspK3 = 1 ? 10-12 M3 = [Ca2+][HCO3-][OH-] Assuming this is the only important process, and it proceeds to completion, S = [Ca2+] = [HCO3-] = [OH-] = 1 ? 10-4M ie about 50 % higher than when the reaction of carbonate is ignored. In fact, this calculation underestimates the solubility, because not all carbonate is converted to bicarbonate (we will cover this in a problem). For [OH-] = 1 ? 10-4 M, pOH = 4, and hence, pH = 10. CaCO3 is thus a moderate base. 4. CaCO3/H2O/CO2(g) CaCO3 ( Ca2+ + CO32- Ksp = 5 ? 10-9 M2 HCO3- ( H+ + CO32- K2 = 5 ? 10-11 M? CO2(aq) + H2O ( H+ HCO3- K1 = 5 ? 10-7 M H2O ( H+ + OH- Kw = 10-14 M2 Maintaining Charge Neutrality 2[Ca2+] + [H+] = 2[CO32-] + [HCO3-] + [OH-] (I) Hence: Substitute in (I) which gives Solution can be obtained using a spreadsheet, as shown below: From the [H+] concentration, the concentrations of the other species can be obtained. ...read more.

Conclusion

The pH of non-calcareous waters can drop well below 5. Aluminium. One of the side effects of increased acidity is to bring more Aluminium into solution, through the dissolution of Al(OH)3. Al(OH)3 ? Al3+ + 3OH-; Ksp = 10-33 M4 In neutral water, the solubility = [Al3+] = Ksp/(10-7)3 = 10-12 M. However, its solubility increases dramatically as the pH reduces. [Al3+] = Ksp([H+]/Kw)3 The solubility is proportional to [H+]3! One effect of Al3+ is that when it reaches the more alkaline gills of the fish, it precipitates as a gel, thus preventing the fish from breathing properly. In addition, Al3+ has also been implicated in senile dementia. 7. Summary The aim of this short course has been to provide an introduction to the environmental chemistry of aqueous systems. This has been achieved by examining the equilibrium behaviour of natural waters under the influence of CO2, CaCO3 and anthropogenic acid rain. The principles introduced are more generally applicable to other aqueous systems; e.g., examining the effect of other ions on natural waters. 1 [Note that: CO2 + H2O ( HCO3- + H+ should really be written as CO2 + H2O ( H2CO3 H2CO3 ( H+ + HCO3-] ? This is equivalent to CO32- + H2O ( HCO3- + OH-; K3 = Kw/K2 = 2 ? 10-4 M. ?? ?? ?? ?? ...read more.

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