Question

If the pH of a 1.00-in. rainfall over 1300 miles2 is 3.30, 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. Express your answer to two significant figures and include the appropriate units.

Answer 1

At pH = 3.30

H2SO4 is fully dissociated as follows:

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

Since pH = -log10[H+]

[H+] = 10^(-pH)
        = 10^(-3.30)
[H+] = 0.0005 M

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

[H2SO4] = ½ * [H+]
                =1/2 * 0.0005 M
[H2SO4] = 0.00025 M

Now convert the amount of rain in meter as follows:

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

Now calculate the area as follows:

Area = 1300miles^2 = 3.37*10^9 m^2

Calculate the volume of water

Volume = Rain * Area
              =0.0254m * 3.37 e9 m^2
Volume = 8.55*10^7m^3

Calculate the number of kilomoles of H2SO4

kilomoles H2SO4 = [H2SO4] * Volume
                              = 0.00025 M * 8.55*10^7m^3

kilomoles H2SO4 = 21380.25 kmol

Calculate the mass of H2SO4

mass H2SO4 = kilomoles H2SO4 * MW H2SO4
                    = 21380.25 kmol * 98.07848 kg/kmol
mass H2SO4 = 2.10 *10^6 kg H2SO4