Question

If the pH of a 1.00-in. rainfall over 1500 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.

I keep following the steps from similar problems with different values and its still wrong.

Can you please solve step by step.

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 = -log10[H+]

[H+] = 10^(-pH)
= 10^{-3.90)
[H+] = 1.2589e-4M

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

[H2SO4] = 0.5 * [H+]
=0.5 * 1.2589e-4M
[H2SO4] = 6.2945e-5M

Convert the amount of rain and the area into SI units

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

Area = 1500miles^2 * 2589988.11m^2/miiles^2= 3.885e9m^2

Calculate the volume of water

Volume = Rain * Area
=0.0254m * 3.885e9m^2
Volume = 0.9868e8m^3

Calculate the number of kilomoles of H2SO4

kilomoles H2SO4 = [H2SO4] * Volume
= 6.2945e-5M * 0.9868e8m^3
kilomoles H2SO4 = 6,211.35kmol

Calculate the mass of H2SO4

mass H2SO4 = kilomoles H2SO4 * MW H2SO4
= 6,211.35kmol * 98.07848kg/kmol
mass H2SO4 = 6.092e5kg H2SO4