Question

This seems like a complicated one. can someone help me with this question?

If the pH of a 1.00-in. rainfall over 1400miles2 is 3.60, 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

[H+] [SO4--] / [HSO4-] = Ka2

At pH 3.2,
[SO4--] / [HSO4-] = Ka2 / 10^-3.2
= 19

All of the H2SO4 is negatively charged already, so
the total amount of sulfuric acid is:
[SO4--] + [HSO4-] = 19 [HSO4-] + [HSO4-]
=20 [HSO4-]

Proportion of the pH level to the H2SO4 that is negatively charged is:
19 [SO4--] / 20 [HSO4-]
= 19/20

proportion that is in a single deprotonation is:
[HSO4-] / 20 [HSO4-]
= 1/20

The moles of hydrogen ion per mole of sulfuric acid at this pH is
(19/20)*2 + (1/20)*1
= 39/20

The amount of sulfuric acid is:

[H+] (1 mol sulfuric acid / 39/20 mol H+)
= (20/39) 10^-3.2 mol/L
=0.51 x 10^-3.2 mol/L