Question

Show full calculation of HCO3- calculation by balance of anion charges and cation charges.

The anion concnetration (mg/L) that were given was:

Fluoride= 1.286

Chloride=43.014

Nitrate=12.016

Phosphate=0.513

Sulfate=113.326

And the cation concentration (mg/L) that were given was:

Sodium=47.788

Potassium=5.741

Calcium=56.796

Magnesium=17.019

Answer 1

Aqueous solution of electrolytic solution is always elctrically neutral i.e. the sum of milliequivalent/litre of anion and cation should always balance

therefore

meq of cation = meq of anion

so, we are to calculate the meq of anion

meq of Na⁺ = (wt (in mg)* valence) / mol.wt = (47.788*1)/23 = 2.08

meq of K⁺ = (wt (in mg)* valence) / mol.wt = (5.741*1)/39.1 = 0.147

meq of Ca⁺⁺ = (wt (in mg)* valence) / mol.wt = (563.796*2)/40.1 = 2.83

meq of Mg⁺⁺ = (wt (in mg)* valence) / mol.wt = (17.019*2)/24.3 = 1.43

therefore total meq of cation = 2.08+.0147+2.83+1.40 = 6.457

now, we are to calcullate total no. of anion

meq of F^- = (wt (in mg)* valence) / mol.wt = (1.286*1)/19 = 0.067

meq of Cl^- = (wt (in mg)* valence) / mol.wt = (43.014*1)/35.5 = 1.21

meq of NO₃^- = (wt (in mg)* valence) / mol.wt = (12.016*1)/62 = 0.194

meq of PO₄^-- = (wt (in mg)* valence) / mol.wt = (0.513*2)/94.97 = 0.010

meq of SO₄^-- = (wt (in mg)* valence) / mol.wt = (113.326*2)/96.06 = 2.36

total meq of anion = 0.067+1.21+0.194+0.010+2.36 = 3.841 + HCO₃^-

so,

6.457 = 3.841 + HCO₃^-

HCO₃- = 6.457-3.841 = 2.616

meq ofHCO₃- = (wt (in mg)* valence) / mol.wt = (wt*1)/61.0168 = 2.616

wt of HCO₃- = 159.62