Question

Part A

If the pH of a 1.00-in. rainfall over 1800 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.

Answer 1

The dissociation reaction is

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

Since pH = -log10[H+]

[H+] = 10^(-pH)
= 10^{-3.3)
[H+] = 5.01 * 10^-4 M

As we can see from the reaction there are 2 moles of H+ for every mole of H2SO4

[H2SO4] = 0.5 * [H+]
=0.5 * 5.01 10^-4 M
= 2.505 10^-4 M

Convert the amount of rain and the area into metre and metre^2

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

Area = 1800miles^2 * 2589988.11m^2/miiles^2

= 4.66 * 10^9 m^2

volume of water

Volume = Rain * Area
Volume =0.0254m * (4.66 * 10^9 m^2)
Volume = 1.18 * 10^8 m^3

Calculate the number of kilomoles of H2SO4

kilomoles H2SO4 = [H2SO4] * Volume
= 2.505 10^-4 M * (1.18 * 10^8 m^3)
kilomoles H2SO4 = 29559 kmol

Now,

mass H2SO4 = kilomoles H2SO4 * MW H2SO4
= 29559 kmol * 98.07848kg/kmol
= 2.8 *10^6 kg H2SO4