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boyakko [2]
3 years ago
12

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 tha

t it is the only acid contributing to the pH? For sulfuric acid, Ka1 is very large and Ka2 is 0.012.
Chemistry
1 answer:
Jobisdone [24]3 years ago
7 0

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!

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