ACID - RAIN

Acid rain means that rain having a pH lower than the natural rain pH (5.6). The natural rain dissolved in carbon dioxide from the atmosphere and it's  give a mildly acidic solution. This rain acidity is primarily the result of dissolving in rain water of sulfur oxides and nitrogen oxides from human activities. In the especially united states the strong acids components in acid rain are about 62% H₂SO₄ ,32% HNO₃ ,and 6% Hcl.

     The sulfuric acid in acid rain have been traced to the burning of fossil fuels and to burning of sulfide ores in the production of metals, such as copper and zinc. Coal for contains some sulfur mainly as pyrite. Iron (ll) disulfide and FeS₂ when this burns in air, it produces sulfur dioxide.

4FeS₂, (s) + 11O₂(ɡ) → 2Fe₂O₃(s)+ 8SO₂(ɡ)

             

        Presence of dust particles and other substances in pollutants air and Sulfur dioxide oxidize further to give sulfur trioxide which is react with the water molecule to form sulfuric acid.

 

2SO₂(ɡ) + O₂(ɡ) →2 S₃(ɡ)

SO₃(ɡ) +H₂O (l) →H₂SO₄(aq)

       

                        Generally all of the sulfuric acids are dissociate into the ions H₃O+, SO₄²-and HSO₄- .The first ionization of sulfuric acid is complete the second ionization HSO₄-  partial is

       ( Ka₂= 1.1× 10-₂)

The acid rain can be a harmful for fish to plants by changing the PH of lake water and some structural materials and monuments marble, the composed of calcium carbonate (CaCO₃) which dissolves in water of low PH level.

H₃O+( aq) + CaCO₃(s)→ Ca²+( aq)+ HCO₃-( aq) + H₂O (l)

The calculate the PH of natural rain is water saturated with CO₂ from the atmosphere. First, we can need the equilibrium for the solution of carbondioxide in water.

CO₂(ɡ) + H₂O (l) ⇔ H₂CO₃(aq)

K= [H₂CO₃]∕Pco₂ =3.5× 10-²

                   The [H₂CO₃]  represents of the concentration of molecule formed when CO₂ in the atmosphere is average 0.00033 atmosphere. If we substitute this for Pco₂ and solve for H₂CO₃ we obtain 1.2× 10-⁵M

     We need for acid ionization of H₂CO₃, we look only at the first ionization, which produces almost the all of the H₃O±

H₂CO₃(aq) + H₂O(l) ⇔ H₃O+ (aq) + HCO₃-(aq)

Ka1 = [ H₃O+] [ HCO₃-]∕ [ H₂CO₃]= 4.3×10-⁷

Denoted [ H₃O+] = [ HCO₃−] by x, we can write the equilibrium equation

                     x²           = 4.3 ×10-⁷

          1.2 × 10-⁵−x    

             Generally the usual assumption about neglecting x in the nominator, we obtain x= H₃O+ = 2.3× 10-⁶ actually these assumptions and the neglect of the self- ionization of water and affecting the answer in the second significant figure. Retaining of the one significant figure of, pH= - log(2× 10-⁶) = 5.7 which is essentially what we gave earlier for the pH is always natural in rains.