If the pHpH of a 1.00-inin. rainfall over 1400 miles2miles2 is 3.60, how many kilograms of sulfuric acid, H2SO4H2SO4, are present, assuming that it is the only acid contributing to the pHpH? For sulfuric acid, Ka1Ka1K_a1 is very large and Ka2Ka2K_a2 is 0.012.

mass H₂SO₄ = 1.45 x 10⁶ kg H₂SO₄

Explanation:

At pH 3.70, H2SO4 can be regarded as being fully dissociated. The dissociation reaction is

H₂SO₄(l) ⇄ 2H⁺(aq) + SO₄²⁻(aq)

Since pH = -log₁₀[H⁺]

[H⁺] = 2.5 x 10⁻⁴ M

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

[H₂SO₄] = 0.5 * [H⁺]

=0.5 * 2.5 x 10⁻⁴ M

[H₂SO₄] = 1.25 x 10⁻⁴ M

Convert the amount of rain and the area into SI units

Rain = 1.00 in * 0.0254 m/in

Rain = 0.0254 m

Area = 1600 miles² * 2589988.11 m²/miles²    = 4.661 x 10⁹  m²

Calculate the volume of water

Volume = Rain * Area

=0.0254 m * 4.661 x 10⁹

Volume = 1.184 x 10⁸ m³

Calculate the number of kilo moles of H₂SO₄

kilo moles H₂SO₄ = [H₂SO₄] * Volume

=1.25 x 10⁻⁴ M * 1.184 x 10⁸ m³

kilo moles H₂SO₄ = 14,800 k mole

Calculate the mass of H₂SO₄

mass H₂SO₄ = kilo moles H₂SO₄ * MW H₂SO₄

= 14800 k mol * 98.07848 kg/kmol

mass H₂SO₄ = 1.45 x 10⁶ kg H₂SO₄