Question

f the pH of a 1.00-in. rainfall over 1600 miles2 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, Ka1 is very large and Ka2 is 0.012.

Answer 1

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+] = 2.551e-4M

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

[H2SO4] = 0.5 * [H+]
=0.5 * 2.551 e-4M
[H2SO4] = 1.2559e-4M

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
= 1.2559e-4M * 1.053e8m^3
kilomoles H2SO4 = 13,224 kmol

Calculate the mass of H2SO4

mass H2SO4 = kilomoles H2SO4 * MW H2SO4
= 13,224 kmol * 98.07848kg/kmol
mass H2SO4 = 1.297051 e6 kg H2SO4