Question

The rainfall in a given area totaled 1m/yr and had a pH of 4.27. The lake had an alkalinity of 175 mg CaCO3/L and an average depth of 12m. Assuming that the acidity of the rain was only due to strong acids, what percent of the alkalinity in the lake was neutralized by the rain? Neglect water inflow and outflow.

Answer 1

Answer:

P^H of the acid rain = 4.27

H⁺ concentration = 10^-PH

H⁺ concentration = 10^-4.27

  = 10^-5 x 10^0.73

= 10^-5 x 5.37

The Acid concentration = 5.37 x 10^-5 M

Alkanity (CaCO3) concentration = 175 mg / L

Caluculation of the Alkanity in Molarity:

Weight of CaCO₃ = 175 mg = 0.175 g

Molar mass of CaCO₃ = 100

Molarity (M) = (weight / molar mass) x (1 / V in Liter)

M = (0.175 / 100) x (1 /1)

M = 0.175 x 10^-2

M = 175 x 10^-5

The Alkanity concentration = 175 x 10^-5

The Acid concentration (From rain water) = 5.37 x 10^-5 M

Caluculation of the % of the Neutralization:

From the above concentrations the alkanity concentration is more. We have to consider it as 100%

175 x 10^-5 ..................... 100

5.37 x 10^-5 M................ ?

= (5.37 x 10^-5) x 100 / 175 x 10^-5

= 5.37 x 100 / 175

=537 / 175

= 3.068

Hence the Acid rain can neutralize 3.068% of the alkanity of the lake.