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 CO₃^2-.)

(b) Calculate the TDS of the solution in mg/L. Because most of the DIC is present as CO₃^2-, a negligible portion is released as CO_2(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.

Answer 1

(a).

Given

Strength of Na+ = 6500 mg/L

No. of gram milli equivalent equivalents = given mass / equivalent mass = 6500/23 = 282.6 meq

Normality of Na+ = 282.6 meq/L

Similarly

Normality of K+ = 2230/38 meq/L = 58.68 meq/L

Normality of Ca2+ = 3400 / 20 meq/L = 170 meq/L

Normality of Mg2+ = 1350/12 meq/L = 112.5 eq/L

Normality of Cl- = 1100/35 meq/L = 31.43 meq/L

Normality of F- = 4350/19 meq/L = 228.94 meq/L

DIC as CO32- = 1740/30 meq/L = 58 meq/L

Normality of SO42- = 3940/48 meq/L = 82 meq/Land

Normality of OH- = 10^-2.8 mol/L = 0.00158 mol/L = 0.0269 g/L = 26.9 / 17 mg/L = 1.58 meq/L

Now

For cations

282.6 + 58.68 + 112.5 + 170 = 623.78 meq/L

For anions

31.43 + 228.94 + 58 + 82 + 1.58 = 401.95 meq/L

The total cationic charge does not comes to ±5% of anionic charge.

(b).

TDS of the solution = 6500 + 2230 + 3400 + 1350 + 1100 + 4350 + 1740 + 3940 + 26.9 = 246.36.9 mg/L

(c).

Ionic strength = 1/2 ∑ C_iZ_i = 1 /2 [ (6.5/23 mol/l x 1²) + (2.23/39 mol/l x 1²) + (4.35/19 x 1²) + (1.74/60 x 2²) + ( 3.94/96 x 2²) + (0.00158 x 1²) ]

Ionic strength = 1/2 [ 0.28 + 0.057 + 0.34 + 0.225 + 0.031 + 0.229 + 0.116 + 0.164 + 0.00158 ] = 0.722

(d).

[Na+] = 282.6 meq/L x 50 mg CaCO3 / meq = 141.30 mg /L of CaCO3

[ K+] = 57.18 meq/L x 50 mg / meq of CaCO3 = 2859 mg/L of CaCO3

[Ca2+] = 170 x 50 = 8500 mg/L of CaCO3

[Mg2+] = 112.5 x 50 = 5625 mg/L of CaCO3

[Cl-] = 31.43 x 50 = 1571.5 mg/L of CaCO3

[F-] = 228.94 x 50 = 11447 mg/L of CaCO3

DIC = 58 x 50 = 2900 mg/L of CaCO3

[SO₄^2-] = 82 x 50 = 4100 mg/L of CaCO3

[OH-] = 1.58 x 50 = 79 mg/L of CaCO3

Total hardness = 14130 + 2859 + 5625 + 8500 + 1571.5 + 11447 + 2900 +4100 + 79 = 51211.5 mg/L