ENCE 433 Dr. Alba Torrents

ENVIRONMENTAL ENGINEERING ANALYSIS Fall 1996

Homework Set # 2 SOLUTION

Concentration of Sulfate :

• The concentration of all forms of sulfate in the rain and snow is the number of moles of sulfate divided by the volume of the solution.
• 1 mol SO₄^-2 = 96 g
• Moles of sulfate = 0.25x10¹³ g / 96 = 2.6x10¹⁰ moles
• V=A x h = 1x10¹²m² x 1.5 m = 1.5 x 10¹² m³ = 1.5x10¹⁵ L
  
  and the concentration is: 2.6x10¹⁰/1.5x10¹⁵ = 1.7x10^-5 mol/L
  
  
• More Data :
  The general chemistry way to write the K_a1 for sulfuric acid and water is
  

H₂SO₄(l) + H₂O(l) --> H₃O+(aq) + HSO₄^- (aq)

Here, we save space by omitting H₂O and using H⁺ instead of the hydronium ion.

The equations needed are:

Equilibrium equations:
1) An expression for K_a1
2) An expression for K_a2

Water ionization
3) [H⁺][OH^-] = 10^-14

Electroneutrality equation
4) [H⁺] = [HSO₄^-] + 2[SO₄^-2]+[OH^-]

Mass balance
5) [H₂SO₄] + [HSO₄^-] + [SO₄^-2] = 1.7x10^-5 M

Status : Five unknowns, and five knowns. There seem to be a variety of strategies; let's focus on the charge balance equation. Replace all non-hydrogen terms. Since the sulfate is the "farthest" from H⁺, let's start with it:

Rearrange eq 1:

Substitute eq 1 into 2:

Solve for sulfate concentration:

(6)
Now, substitute eqs 1, 3, and 6 into the charge balance equation, eq 4.

(7)

We need a relationship between H⁺ and H₂SO₄. It's time to turn to eq 5.
Substitute eq 1 and eq 6 into eq 5:

and solve for H₂SO₄:

An Easier way is to assume that as the pH will be acid (<<4) then:

[H₂SO₄] - 0

[OH^-] <<<< [H⁺]

MB : 1.5x10^-5 M = [HSO₄^-] + [SO₄^-2] -- [SO₄^-2] = 1.5x10^-5 M - [HSO₄^-]

PBE : [H⁺] = [HSO₄^-] + 2[SO₄^-2] - [HSO₄^-] = [H⁺] - 2(1.5x10^-5 M - [HSO₄^-])

[HSO₄^-] = 2(1.5x10^-5 M)- [H⁺]

and then

quadratic equation, [H⁺] = 3.4x10^-5 - pH = 4.4