Question

#1 A water sample contains 60 mg/L of calcium, 60 mg/L of magnesium, and 25 mg/L of sodium. Assuming the alkalinity of the water is 394 mg/L CaCO₃. Estimate Total Hardness, Carbonate Hardness, and Noncarbonate Hardness using all polyvalent cations.

#2 What is the % error using only the predominant cations?

#3 Construct a bar chart to determine the speciation of the hardness

Answer 1

orv I ca²+ Mg 2+ | Nat t cation Concentration I ng Id) T60 25 60 25 ATLotuіч - 39ч та ! @ CaCo3 TH (Total hardness) = carbonate hardness + Non-Carbonate hardness LTH = CH + NCH . 1. all CH - minimum {Alkalinita, To} calculating Toteel hardness as CaCo3 using Polyvalent cations, Equivalent wt of cation Ca 2+ = 20 mgle mg2+ = 12 mg/L Nat = 23 mg/l Cacoz = so mgle TH = *50 + 6 x 50 + 25 x 50 TH= 454.35 mg/l CH = minimum { 394, 454:35} CH = 394 ongle NCH = TH CH = 454.35-394 - 60.35 mgle

☺ using au cations TH = 454.35 mglu CH = 394 mgIL NCH = 60.35 mg IL using predorinaut caetion only. predominant cation - cat mg2+ T.H = Ca ²+ + Mg 2t 1 - 60x50 at 60 x 50 = 400 mgle CH = min { 394, 400) CH= 394 ng IL NCH = 400 - 394 = 6 mgle * error TH = 454,35- 400S 4425 11.96 CH 3 394-394 394 07. NCH = 60.35-6 2 60.35 = 90.06% Bar chart - 60 or Mat , Mg2+ Alkalinity as cacos 394 ngh