Question

If the pH of a 1.00-in. rainfall over 1800 miles2 is 3.70, 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: Mass of H2SO4 present = 231.5*10⁴ kg

Height of rainfall = 1.00 inch = 0.0254 m

Area of rainfall = 1800 miles²

1 mile² = 2.59*10⁶ m²

Therefore area = 1800*2.59*10⁶ = 4.662*10⁹ m²

Volume (V) of rainfall = area (A) * height(h) = 0.0254 * 4.662*10⁹ = 1.184*10⁸ m³

1 m³ = 1000 L

V = 1.184*10¹¹ L

It is given that the pH = 3.70

pH = -log[H+]

[H+] = 10^-pH = 10⁻³.⁷⁰ = 1.995*10⁻⁴M

Now:

Molarity of H2SO4 = moles of H2SO4/volume

moles = 1.995*10⁻⁴moles.L-1 * 1.184*10¹¹ L = 2.362*10⁷ moles

Molar mass of H2SO4 = 98 g/mol

Mass of H2SO4 = 2.362*10⁷ moles * 98 g/mol = 231.48*10⁷ g = 231.5 *10⁴ kg

Answer 2

There are 2.32 x 10^6 kg sulfuric acid in the rainfall.

Solution:
We can find the volume of the solution by the product of 1.00 in and 1800 miles2:
1800 miles2 * 2.59e+6 sq m / 1 sq mi = 4.662 x 10^9 sq m
1.00 in * 1 m / 39.3701 in = 0.0254 m
Volume = 4.662 x 10^9 m^2 * 0.0254 m
= 1.184 x 10^8 m^3 * 1000 L / 1 m3
= 1.184 x 10^11 Liters

We get the molarity of H2SO4 from the concentration of [H+] given by pH = 3.70:
[H+] = 10^-pH = 10^-3.7 = 0.000200 M
[H2SO4] = 0.000100 M

By multiplying the molarity of sulfuric acid by the volume of the solution, we can get the number of moles of sulfuric acid:
1.184 x 10^11 L * 0.000100 mol/L H2SO4 = 2.36 x 10^7 moles H2SO4

We can now calculate for the mass of sulfuric acid in the rainfall:
mass of H2SO4 = 2.36 x 10^7 moles * 98.079 g/mol
= 2.32 x 10^9 g * 1 kg / 1000 g
= 2.32 x 10^6 kg H2SO4