Question

Part A

If the pH of a 1.00-in . rainfall over 1700 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 = -log10[H+]

[H+] = 10^-pH
= 10^-3.90
[H+] = 1.258 x 10^-4 M

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

[H2SO4] = 0.5 * [H+]
=0.5 * 1.258 x 10^-4 M

[H2SO4] = 6.29 x 10^-5 M

Convert the amount of rain and the area into SI units

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

Area = 1700miles² * 2589988.11m²/miiles² =4.403 x 10⁹ m²

Calculate the volume of water

Volume = Rain * Area
=0.0254m * 4.403 x 10⁹ m²
Volume = 1.118 x 10⁸ m³

Calculate the number of kilomoles of H2SO4

kilomoles H2SO4 = [H2SO4] * Volume
= 6.29 x 10^-5M * 1.118 x 10⁸ m³
kilomoles H2SO4 = 7032.22 kmol

Calculate the mass of H2SO4

mass H2SO4 = kilomoles H2SO4 * MW H2SO4
= 7032.22 kmol * 98.07848kg/kmol
mass H2SO4 = 0.689 x 10⁶ kg H2SO4