Question

If the ph of a 1.00-in. rainfall over 1400 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

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⁺] = $10^{-pH}$
= $10^{-3.7}$
[H⁺] = 2 x 10⁻⁴ M
Since there are 2 moles of H⁺ for every mole of H₂SO₄
[H₂SO₄] = 0.5 * [H⁺]
=0.5 * 2 x 10⁻⁴ M
[H₂SO₄] = 1 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
=1.0 x 10⁻⁴ M * 1.184 x 10⁸ m³
kilo moles H₂SO₄ = 11,839 k mole

Calculate the mass of H₂SO₄

mass H₂SO₄ = kilo moles H₂SO₄ * MW H₂SO₄
= 11839 k mol * 98.07848 kg/kmol
mass H₂SO₄ = 1.16 x 10⁶ kg H₂SO₄