Question

In a large sulfide tailing pond, the average pH of overlying water is 2.0 due to the oxidation process of sulfide minerals. According to the environmental regulation, the pH of the mining waste water needs to be neutralized before it can be discharged to the environment. In a mining industrial process, the water is first adjusted to basic solution with saturated lime solution (CaO), then adjusted to a solution to around pH 7.0 with H2SO4. Assuming that the water treatment plant needs to treat an average of 8600 cubic meters of water per hour and the plant works 24 hours per day and 7 days per week, how many tones of CaO is needed per year (at a minimum)? (Hint: in water, CaO +2H2O ó Ca(OH)2, and Ca(OH)2 ó Ca2+ + 2OH-. Ksp of Ca(OH)2 is 5.5 x 10-6.) Atomic weight of Ca = 40.06, of O = 15.9994, of H = 1.0079

Answer 1

In aqueous solution, the addition of CaO involves the following reaction:

But the product formed, Ca(OH)₂ is under the following solubility equilibrium:

Then , we can calculate the solubility of the product formed in a saturated solution of Ca(OH)₂ according to the following table:

Notice that the expression of K_SP in terms of the solubility can be written as:

Solving for solubility:

Substituting known values:

This is the solubility of Ca(OH)₂ in water. Also, is the equilibrium concentration of Ca⁺² ions coming from the addition of CaO to the water. Due to the 1:1 stoichiometry between CaO and Ca⁺² ions, we can establish a value of 0.0111M as the molar concentration of CaO required in water.

From the units of molarity as concentration unit:

In one hour, the water treatment plant needs to treat an average of 8600 cubic meters of water. Converting this discharge into liters units:

If we multiply this discharge by the concentration required of CaO, we will have the molar quantity of CaO required per hour (n_CaO):

To calculate the quantity of CaO require in one day we will apply the following conversion factor:

Notice that we are simplifying units in every conversion factor.

In one year we will have:

The molecular weight of CaO is:

The mass of CaO required per year is given by the product:

Substituting data:

The tons of CaO needed per year are calculated using the following conversion factor:

The answer is