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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

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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.

This problem has been solved

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