Question

If the pH of a 1.00-in. rainfall over 1400 miles2 is 3.90, 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 3.90, H2SO4 can be regarded as being fully dissociated.

The dissociation reaction is

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

Since pH = -log[H+]

[H+] = 10^(-pH)
= 10^(-3.90)
[H+] = 1.26e-4 M

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

[H2SO4] = 0.5 * [H+]
= 0.5 * 1.26 e-4 M
[H2SO4] = 6.3e-5 M

Convert the amount of rain and the area into SI units

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

Area = 1400 miles^2 * 2589988.11 m^2/miiles^2 = 3.625e9m^2

Calculate the volume of water

Volume = Rain * Area
=0.0254m * 3.625 e9m^2
Volume = 0.92e8m^3

Calculate the number of kilomoles of H2SO4

kilomoles H2SO4 = [H2SO4] * Volume
                          = 6.3e-5 M * 0.92e8 m^3
= 5796 kmol

Calculate the mass of H2SO4

mass H2SO4 = kilomoles H2SO4 * MW H2SO4
                    = 5796kmol * 98.07848kg/kmol
                   = 568471 kg H2SO4

Therefore,

mass of H2SO4 present = 568471 kilograms