Calculate the temporary hardness and permanent hardness of a sample of water containing: Mg(HCO3)2 =7.3 mg/L; Ca(HCO3)2 = 16.2 mg/L; MgCl2 = 9.5 mg/L; CaSO4 = 13.6 mg/LMolecular weights: Ca(HCO3)2 = 162; Mg(HCO3)2=146; CaSO4= 136; MgSO4 = 120; MgCl2 = 95; Al2(SO4)3 = 114;Ca(NO3)2=164
Question
Calculate the temporary hardness and permanent hardness of a sample of water containing: Mg(HCO3)2 =7.3 mg/L; Ca(HCO3)2 = 16.2 mg/L; MgCl2 = 9.5 mg/L; CaSO4 = 13.6 mg/LMolecular weights: Ca(HCO3)2 = 162; Mg(HCO3)2=146; CaSO4= 136; MgSO4 = 120; MgCl2 = 95; Al2(SO4)3 = 114;Ca(NO3)2=164
Solution
To calculate the temporary and permanent hardness of water, we first need to understand what these terms mean. Temporary hardness is due to the presence of bicarbonate minerals (Ca(HCO3)2 and Mg(HCO3)2) which can be removed by boiling the water. Permanent hardness is due to sulfate and chloride minerals (MgCl2, CaSO4) which cannot be removed by boiling.
Step 1: Calculate the hardness due to each compound. Hardness is usually expressed in terms of equivalent of CaCO3. The formula to calculate hardness is:
Hardness (as CaCO3) = (Concentration of the ion (mg/L) / Equivalent weight of the ion) x Equivalent weight of CaCO3
The equivalent weight of CaCO3 is 50.
Step 2: Calculate the temporary hardness. This is due to the bicarbonates of calcium and magnesium.
Temporary hardness due to Ca(HCO3)2 = (16.2 mg/L / 162) x 50 = 5 mg/L as CaCO3 Temporary hardness due to Mg(HCO3)2 = (7.3 mg/L / 146) x 50 = 2.5 mg/L as CaCO3
Total temporary hardness = 5 + 2.5 = 7.5 mg/L as CaCO3
Step 3: Calculate the permanent hardness. This is due to the sulfates and chlorides of calcium and magnesium.
Permanent hardness due to MgCl2 = (9.5 mg/L / 95) x 50 = 5 mg/L as CaCO3 Permanent hardness due to CaSO4 = (13.6 mg/L / 136) x 50 = 5 mg/L as CaCO3
Total permanent hardness = 5 + 5 = 10 mg/L as CaCO3
So, the temporary hardness of the water sample is 7.5 mg/L as CaCO3 and the permanent hardness is 10 mg/L as CaCO3.
Similar Questions
Calculate the hardness in terms of CaCO3 equivalent if 32.1 mg/l CaSO4 is present in water sample.a.24 mg/lb.13 mg/lc.32.10 mg/ld.23.60 mg/l
Calculate the total hardness of water in °Fr and °Cl from the following analysis of water sample. Mg(HCO3)2=20.8 mg/L, Mg(NO3)2=28.8 mg/L, MgCl2=21.0 mg/L, MgSO4=20 mg/L, CaCO3=21.2 mg/L, KCl=73.2 mg/L
Example 1: Calculate the hardness of water sample having following analysis : Mg(HCO3)2 = 7.3ppm, Ca(HCO3)2 = 16.2 ppm, MgCl2 = 9.5 ppm, CaSO4 = 13.6 ppm, Fe2O3 = 8.5 ppm, Mg(NO3)2=14.8 ppm. Also calculate carbonate and non carbonate hardnes
Temporary hardness in water is mainly due to the presence of:a.Calcium sulfateb.Calcium chloridec.Calcium bicarbonated.Magnesium bicarbonate
A 3L water sample has a hardness of 300ppm by mass CaCO3. The grams of Calcium Carbonate CaCO3 is:Question 34Select one:a.1.0gb.0.20gc.0.30gd.1.20g
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.