If the pH of a 1.00-in. rainfall over 1300miles is 3.60, how many kilograms of s
ID: 729961 • Letter: I
Question
If the pH of a 1.00-in. rainfall over 1300miles is 3.60, how many kilograms of sulfuric acid, H2SO4, are present, assuming that it is the only acid contributing to the pH?For sulfuric acid, Ka is very large and Ka2 is 0.012.
Explanation / Answer
At ph 3.60, 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^{-3.60) [H+] = 0.000251189 Since there are 2 moles of H+ for every mole of H2SO4 [H2SO4] = 0.5 * [H+] =0.5 * 2.511e-4M [H2SO4] = 1.2505e-4M Convert the amount of rain and the area into SI units Rain = 1.00in * 0.0254m/in Rain = 0.0254m Area = 1300miles^2 * 2589988.11m^2/miiles^2 =3366984543m^2 Calculate the volume of water Volume = Rain * Area =0.0254m * 3366984543m^2 Volume = 8.55214073922 × 10^7 Calculate the number of kilomoles of H2SO4 kilomoles H2SO4 = [H2SO4] * Volume = 1.2505 x 10 ^-4 M * 8.55214073922 × 10^7 kilomoles H2SO4 = 10694.45 kmol Calculate the mass of H2SO4 mass H2SO4 = kilomoles H2SO4 * MW H2SO4 = 10694.45 kmol * 98.07848kg/kmol mass H2SO4 = 1.0488955960431919×10^6 kg H2SO4