Question

Sewage exists a force main and enters a gravity sewer. At the exit of the force main all sulfur in the sewage is found as sulfide (H₂S or HS^-)*. When the sewage enters the gravity sewer it is exposed to oxygen that may cause the sulfide to transform to sulfate (SO₄^2-). Assume that the pH is 7, the total sulfur concentration of the sewage where the force main meets the gravity sewer is 2.5 x 10^-6 M, and the sewage flow is 1.00 x 10⁷ liter/day. Hydrogen sulfide is a weak acid. The pKa for it’s first dissociation constant (H₂S = HS^- + H⁺) is 7.0.

A. Write a balanced equation for the oxidation of sulfide to sulfate

B. How much oxygen is required to transform all of the sulfide in a liter of sewage to sulfate?

C. How much limestone (CaCO₃) would have to dissolve each day to keep the alkalinity of the sewage constant? i.e., neutralize the H+ generated by the oxidation of sulfide

D. What would the equilibrium [O²]aq be if the [SO4^2-]/[HS-] ratio is 1000?

The S^2- ion may also be present but its concentration will be negligible at all relevant pHs and can be ignored

In a force main the water/sewage is pumped and the fluid fills the whole pipe. Gravity sewers slope downhill so no pumping is needed. The sewers flow partially full; i.e., there is air above the water/sewage

Answer 1

A) Sulfide to Sulfate : H₂S + 2 O₂ H₂SO₄

B) From the above balanced reaction we can see that, 1 mole Sulfide requires 2 moles of Oxygen for the formation of Sulfate as we know that concentration of Sulfur in the sewage is 2.5 x 10^-5 M , 1 liter of this solution implies 2.5 x 10^-5 moles of Sulfur. And the Oxygen requirement is double the Sulfur requirement. Therefore, the amount of Oxygen required for conversion of 1 Liter Sulfide into Sulfate is

= 2 x 2.5 x 10^-5 mole = 5 x 10^-5 moles of Oxygen.

C) For this question we need write the equilibrium dissociation equation for H₂S and then we would decide upon the amount of CaCO₃ .

H₂S H⁺ + HS^- , For 1 mole of H₂S , we have as follows :

1-x x x , pKa = 7, This implies that pH = 3.5 and [H⁺] = 10^-3.5

Now, according to the given Sulfur content and the rate of flow through the sewage, we get total

[H⁺] = 10^-3.5 x 1 x 10⁷ = 10^3.5 , Amount of CaCO₃ required is determined by the below reaction : CaCO₃ + H₂SO₄   CaSO₄ + H₂O + CO₂ . We can see that, 1 mole of H₂SO₄ ( or 2 moles of H⁺). So this implies that, amount of  CaCO₃ required = 2 x 10^3.5 moles = 2 x 10^3.5 x 100 grams = 632.45 Kg

D) HS^- + 2 O₂ SO₄^2- + H⁺ , K = [SO₄^2- ] [H⁺ ] / [ HS^- ] [O₂ ]² , given that [SO₄^2- ] / [ HS^- ] = 1000

Therefore, [O₂ ] = 5.62 x 10^-3