15. An isolated artificial containment lake from an industrial plant has the fol
ID: 583351 • Letter: 1
Question
15. An isolated artificial containment lake from an industrial plant has the following characteristics: pH = 10.4, [HCO]-5 x 10" mol/L, depth (average) = 15 m, area-8 ha, and has no other source of alkalinity than the ions present in solution. It receives 2 m of rain per year. The average pH of the rain is 4.25, and an industrial effluent containing 10 ppm of HCl at a rate Of 80,000 m per year. (a) (6p) Calculate the alkalinity of this lake (b) (4p) Calculate how many years will it take for this lake to become neutralExplanation / Answer
a
1 hectare= 10000 m2
area of river = 80000 m2
volume of river = 80000 * 15 = 1.2*10^6 m3
initial pH= 10.4
amount of rain = 2m
So total water added by rain = 2 * 80000 = 1.6*10^5 m3
pH of rain = 4.25
conc of OH- ions in rain water = 10^-9.75
= 1.778*10^-10 mol/l
= 1.778*10^-7 mol/m3
total OH- ions in rain = 1.778*10^-7 mol/m3 * 1.6*10^5 m3
= 0.028 moles
initial conc of OH- ions= 10^-3.6
= 0.00025 mol/l
= 0.25 mol/m3
Total no of OH- ions present = 0.25 mol/m3 * 1.2*10^6 m3
= 300000 moles
Total OH- ions = 0.028 moles+ 300000 moles
= 300000.028
Total volume of water= 1.2*10^6 +1.6*10^5
=1.36 *10^6 m3
conc of OH-ions = 300000.028 /1.36 *10^6 mol/m3
=0.2205 mol/m3
=0.00022 mol/l
pH= 14- (-log(0.00022) )
= 14 + log(0.00022)
= 10.34 pH
b.
1ppm = 1 mg/l
10 ppm = 10 mg/l
= .01 g/l
= 2.7 * 10^-4 mol/l is the conc of HCl
=.27 mol/m3 is the conc of HCl
amount of H+ entering per year
=.27 mol/m3 * 80000 m3
=21600 moles per year
Currently present moles of OH- = 300000.028
No of years for nuetralization =300000.028 /21600 moles per year
= 13.89 year
So roughly14 years
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.