If the pH of a 1.00-in. rainfall over 1300 miles2 is 3.30, 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.

Question

contestada

Respuesta :

ilzinhootigee ilzinhootigee · Answer 1 · 2018-11-11T20:07:51+01:00

Hey there!:

Volume of rainfall = Area X height

Height = 1 in = 0.254 dm

1mile2 = 2.59 X 10⁸ dm²

so 1300 miles² = 3367 *10⁸ dm²

Volume = 0.254 * 3367 * 10⁸ = 855.218 * 10⁸ dm³

Also,

H2SO4 ionizes as :

H2SO4 --> H⁺ + HSO4⁻

Ka1 is very high so it completely dissociates , due to first dissociation [H⁺] = Concentration of H2SO4

HSO4⁻ --> H⁺ + SO₄⁻²

We also consider it as high dissociation :

so each mole of H2SO4 will give 2 moles of H⁺ (proton)

pH = 3.30 = -log[H⁺]

So [H+] = 0.000501 M = 0.000501 moles / L or moles / dm³

Volume = 855.218 * 10⁸dm³

so moles of H+ present in this volume:

= 0.000501 * 855.218 *10⁸ = 4.28* 10⁷moles

So moles of H2SO4 = 4.28 * 10⁷ / 2 = 2.14 * 10⁷ moles

Molecular weight of H2SO4 = 98 g / mol

Mass of H2SO4:

Moles X molecular weight = 98 * 2.14 X 10⁷ grams = 209.72 * 10⁷ grams in kg:

= 2.09 X 10⁶Kg

Hope that helps!