The following analysis has been reported for a sample of contaminated groundwate

Question

The following analysis has been reported for a sample of contaminated groundwater obtained near an aluminum production facility:

pH = 11.2

Cl- = 1,100 mg/L

Na+ = 6,500 mg/L

F- = 4,350 mg/L

K+ = 2,230 mg/L

DIC = 1,740 mg/L (as C)

Ca2+ = 3,400 mg/L

SO42- = 3;940 mg/L

Mg2+ = 1,350 mg/L

OH- = 10-2.8 mol/L

(a) Check the charge balance for this analysis. Does the total cationic charge (in equiv/L) equal the total anionic charge to within ±5% (the typical target for an acceptable charge balance)? (Note: at pH 11.2, the inorganic carbon is predominantly in the form CO32-.)

(b) Calculate the TDS of the solution in mg/L. Because most of the DIC is present as CO32-, a negligible portion is released as CO2(g) during the TDS analysis.

(c) Calculate the ionic strength of the solution.

(d) Calculate the hardness of the solution in mg/L as CaCO3.

pH = 11.2

Cl- = 1,100 mg/L

Na+ = 6,500 mg/L

F- = 4,350 mg/L

K+ = 2,230 mg/L

DIC = 1,740 mg/L (as C)

Ca2+ = 3,400 mg/L

SO42- = 3;940 mg/L

Mg2+ = 1,350 mg/L

OH- = 10-2.8 mol/L

Explanation / Answer

Total cationic charge= concentration X valency

=[Na+]1+[K+]1+[Ca+2]2+[Mg+2]2

=(6500X1)+(2230X1)+(3400X2)+(1350X2)

=6500+2230+6800+2700=18230

Total anionic charge= concentration X valency

=[Cl-]1+[F-]1+[SO42-]2+[OH-]1+[DIC]2

=1100+4350+(3940X2)+10-2.8+(1740X2)

=1100+4350+7880+0.00158+3480

=16810.00158

Charge balcance= Cationic charge - anionic charge

=18230-16810.00158= 1419.998

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