The presence of SO_2 in the atmosphere and the sulfuric acid that it produces re

Question

The presence of SO_2 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 1700 miles^2 is 3.70, how many kilograms of sulfuric acid, H_2 SO_4, are present, assuming that it is the only acid contributing to the pH? For sulfuric acid, K_a1 is very large and K_a2 is 0.012.

Explanation / Answer

At ph 3.70, H2SO4 can be regarded as being fully dissociated. The dissociation reaction is

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

Since pH = -log10[H+]

[H+] = 10^(-pH)
= 10^{-3.70)
[H+] = 1.995e-4M

Since there are 2 moles of H+ for every mole of H2SO4

[H2SO4] = 0.5 * [H+]
=0.5 * 1.995e-4M
[H2SO4] = 9.98e-5M

Convert the amount of rain and the area into SI units

Rain = 1.00in * 0.0254m/in
Rain = 0.0254m

Area = 1600miles^2 * 2589988.11m^2/miiles^24.144e9m^2

Calculate the volume of water

Volume = Rain * Area
=0.0254m * 4.414e9m^2
Volume = 1.053e8m^3

Calculate the number of kilomoles of H2SO4

kilomoles H2SO4 = [H2SO4] * Volume
= 9.976311575e-5M * 1.053e8m^3
kilomoles H2SO4 = 10,500kmol

Calculate the mass of H2SO4

mass H2SO4 = kilomoles H2SO4 * MW H2SO4
= 10,500kmol * 98.07848kg/kmol
mass H2SO4 = 1.03e6kg H2SO4

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