Question

If the pH of a 1.00-in. rainfall over 1800miles2 is 3.40, 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

ans=2.31E6 kg of H2SO4

solution:

             ph= 3.40, 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^{-3.40)
[H+] = 3.981E-4 M

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

[H2SO4] = 0.5 * [H+]
=0.5 * 3.981E-4M
[H2SO4] = 1.991E-4M

Convert the amount of rain and the area into SI units
Rain = 1.00in * 0.0254m/in
Rain = 0.0254m

Area = 1800miles^2 * 2589988.11m^2/miiles^2   = 4.662e9m^2

Calculate the volume of water
Volume = Rain * Area
=0.0254m * 4.662e9m^2
Volume = 1.184E8m^3

Calculate the number of kilomoles of H2SO4

kilomoles H2SO4 = [H2SO4] * Volume
= 1.99E-4M * 1.184 E8m^3
kilomoles H2SO4 = 23564kmol

Calculate the mass of H2SO4

mass H2SO4 = kilomoles H2SO4 * MW H2SO4
= 23564kmol * 98.07848kg/kmol
mass H2SO4 = 2.31 E6   kg of H2SO4