Question

* An analysis of a sample of water with PH-7.5 has produced t following concentrations (mg/l). (Answer the following two questions) Cations Anions 102 Cl HCO3150 SO4 Na 50 4 260 Ca 46 Mg 26 4. The total dissolved solids in mg/l a) 588 b) 540 38 d) 238 e)None 5. The Non-carbonate hardness in mg/l as CaCO is: a) 98.75 c) 122.95 d) Zero e)None

Answer 1

4) Total dissolved solids(TDS) is the sum of mass concentration of each ion

TDS=50+4+46+26+102+150+260 = 638 mg/L

5) Total hardness (TH) = Ca²⁺ x (50/20) + Mg²⁺ x (50/12)

=46 x (50/20) +26 x (50/12) = 223.33

Total alkalinity (TA) = {[HCO₃^-] x Equivalent weight of CaCO₃} / Equivalent weight of HCO₃^-

= (150 x 50) / 61 =122.95

Here, since TH>TA

Carbonate hardness (NCH) = TH-CH =100.38 mg/mL

This is option (e) None