Question

The presence of SO2 in the atmosphere and the sulfuric acid that it produces result in the phenomenon of acid rain. Uncontaminated rainwater is naturally acidic and generally has a pH value of about 5.6. The pH of acid rain is about 4. This acidity has affected many lakes in northern Europe, the northern United States, and Canada, reducing fish populations and affecting other parts of the ecological network within the lakes and surrounding forests.

The pH of most natural waters containing living organisms is between 6.5 and 8.5, but freshwater pH values are far below 6.5 in many parts of the continental United States. At pH levels below 4.0, all vertebrates, most invertebrates, and many microorganisms are destroyed. More than 300 lakes in New York State contain no fish, and 140 lakes in Ontario, Canada, are devoid of life. The acid rain that appears to have killed the organisms in these lakes originates hundreds of kilometers upwind in the Ohio Valley and Great Lakes regions. Some of these regions are recovering as sulfur emissions from fossil fuel combustion decrease, in part because of the U.S. Clean Air Act of 1990, which required that power plants reduce their sulfur emissions by 80%.

Part A

If the pH of a 1.00-in. rainfall over 1800 miles2 is 3.70, how many kilograms of sulfuric acid, H2SO4, are present, assuming that it is the only acid contributing to the pH? For sulfuric acid, Ka1 is very large and Ka2 is 0.012.

Express your answer to two significant figures and include the appropriate units.

Answer 1

the dissociation reaction is

H2SO4(l) = 2H^+(aq) + SO4^2-(aq)

pH = -log(H+)

= 10^(-3.70)

[H+] = 1.995*10^-4 M

since there are 2 moles of H^+ for every mole of H2SO4

[ H2SO4] = 0.5*[H+]

= 0.5*(1.995*10^-4)

[ H2SO4 ] = 9.98*10^-5 M

convert the amount of rain and area into SI units

rain = 1.0 in*0.0254 m /in

rain = 0.0254 m

area = 1800 miles^2 = 4.662*10^9 m^2

calculate the total volume of water

volume = rain*area

= 0.0254 m * 4.662*10^9 m^2 = 1.184*10^8 m^3

calculate the number of kilomoles of H2SO4

kilomoles  H2SO4 = [ H2SO4] * volume

= 9.98*10^-5 M * 1.184*10^8 m^3

kilomoles H2SO4 = 11,816 Kmol

calculate the mass of H2SO4

mass H2SO4 = Kilomoes H2SO4 * MW H2SO4

= 11,816 Kmol * 98 kg/kmol

mass H2SO4 = 1.16*10^6 Kg H2SO4