Question

#1 [5 pts) 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.

Answer 1

8. Given - Content of water sample Calcium = 60 mg/Lt. Magnesium : 60 mg/lb. Sodium - 25 mg I lt - Alkalinity of the water = 394 mg/L Caco. eq. wt of :- Calcium = 20.04, Mg - 12.13, Na= 23. >> Calcium, magnesium and sodium in terms of Caco: - calcium hardness - Calcium (mg/Lt) i eq. wt. of CaCo₃ eq. wt. of calcium . e) calcium hardness = 60mg 14 x 50.045 20.04. 2 calcium hardness - 149.835. mg 14 of Calon Similarity toy magnesium and sodius:- Magnesium baudness. 60mg14 Xs0.045 12.15 =) Magnesium hardness - 247.1358 mglut of Cacoa, sodium haudness 2. 25 mg ut X50.045 2 s - 54.397 mg Lt of cacos We know - total hardness hardness due to all component in water sample.

: total hardness - Calcium hardness + Magnesium houdness + Sodium hardness & total hardness = 149.835 mg/4- + 247 1358 mglit - +54.397 mg lu) of cacoa total hardness = 451.3678 mg 1lt of Caco, thus, total hardness is greater than alkalinity of water sample - carbonate hardness - alkalinity of sample. > carbonate handuess - 394 mgle of lacet Now, we know- Total hardness = carbonate hardness & Non- e carbonate hardness D = 451.3678 mg 144 of CaCO₃ = 394 mg 1it of Cacoa + Non-Carbonate hardness - Non-carbonate hardness = (451.3678 - 394) mg 1 of cacoa → Nona Coybonate hauduess : 57.3678 mglu of caco,