If the pH of a 1.00 in. rainfall over 1700 miles^2 is 3.40, 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. The presence of SO2 in the atmosphere and thesulfuric acid that it produces result in the phenomenonof acid rain. Uncontaminated rainwater is naturallyacidic and generally has a pH value of about 5.6. ThepH of acid rain is about 4. This acidity has affectedmany lakes in northern Europe, the northern UnitedStates, and Canada, reducing fish populations andaffecting other parts of the ecological network withinthe lakes and surrounding forests.The pH of most natural waters containing livingorganisms is between 6.5 and 8.5, but freshwater pHvalues are far below 6.5 in many parts of thecontinental United States. At pH levels below 4.0, allvertebrates, most invertebrates, and manymicroorganisms are destroyed. More than 300 lakes inNew York State contain no fish, and 140 lakes inOntario, Canada, are devoid of life. The acid rain thatappears to have killed the organisms in these lakesoriginates hundreds of kilometers upwind in the OhioValley and Great Lakes regions. Some of theseregions are recovering as sulfur emissions from fossilfuel combustion decrease, in part because of the U.S.Clean Air Act of 1990, which required that power plantsreduce their sulfur emissions by 80%.

Height of rainfall = 1.00 inch = 0.0254 m Area of rainfall = 1700 miles² 1 mile² = 2.59×10⁶ m² Therefore area = 1700×2.59×10⁶ = 4.403×10⁹ m² Volume (V) of rainfall = area (A) * height(h) = 0.0254 × 4.403×10⁹ = 1.118×10⁸ m³ 1 m³ = 1000 L V = 1.118×10¹¹ L It is given that the pH = 3.40 pH = -log[H+] [H+] = 10^-pH = 10⁻3.40 = 3.981×10⁻⁴M Molarity of H2SO4 = moles of H2SO4/volume moles = 3.981*10⁻⁴moles.L/ 1.118*10¹¹ L =2.808×10^7 moles. Molar mass of H2SO4 = 98 g/mol Mass of H2SO4 = 2.808*10⁷ moles * 98 g/mol = 275.184*10⁷ g = 275.184*10⁴ kg