Question

1. Calculate the temporary and permanent hardness of a water sample having the following data: Mg(HCO3)2: 36.5 mg/l, Ca(HCO3)2: 81 mg/l, CaSO4: 68 mg/l, MgSO4: 30 mg/l, MgCl2: 47.5 mg/l, CaCl2: 55.5 mg/l.
2. Calculate the carbonate hardness (CH) and non-carbonate hardness (NCH) of a water sample having the following impurities: Mg(HCO3)2: 73 mg/l, Ca(HCO3)2: 40.5 mg/l, CaSO4: 34 mg/l, MgCl2: 23.75 mg/l and NaCl: 90 mg/l.
3. The given water sample contains the following hardness producing salts in terms of CaCO3 equivalent. Mg(HCO3)2: 50 ppm, Ca(HCO3)2: 40 ppm, CaSO4: 60 ppm, MgSO4: 35 ppm, MgCl2: 25 ppm, CaCl2: 30 ppm. Calculate individually the proportionate amount of hardness producing salts present in the water sample.
4. 50 ml of a sample of water required 10 ml of 0.01 M EDTA for the titration with Eriochrome Black-T indicator. 50 ml of the same sample after boiling and filtering required 5 ml of 0.01 M EDTA solution. Calculate the temporary, permanent and total hardness of the water sample.
5. 1 g of CaCO3 was dissolved in dil. HCl and diluted to 500 ml of water. 50 ml of this solution required 27 ml of EDTA solution for the titration. 50 ml of a hard water sample required 20 ml of same EDTA solution for titration. Calculate the total hardness of a water of the given water sample.
6. 0.2 g of CaCO3 was dissolved in dil. HCl and diluted to 200 ml of water. 100 ml of this solution required 25 ml of EDTA solution for titration. 100 ml of a hard water sample required 35 ml of same EDTA solution for titration. 100 ml of the same sample after boiling and filtration required 21 ml of EDTA solution for titration. Calculate the total, permanent and total hardness of the given water sample.
7. 1.5 g of CaCO3 was dissolved in dil. HCl and diluted to 1000 ml of water. 100 ml of this solution required 25 ml of EDTA solution for titration. 50 ml of a hard water sample required 23 ml of same EDTA solution. 50ml of the same sample after boiling and filtration required 12 ml of EDTA solution. Calculate the Total hardness, CH and NCH in ppm.

Answer 1

Answer:

Calculate the temporary and permanent hardness of a water sample having the following data: Mg(HCO3)2: 36.5 mg/l, Ca(HCO3)2: 81 mg/l, CaSO4: 68 mg/l, MgSO4: 30 mg/l, MgCl2: 47.5 mg/l, CaCl2: 55.5 mg/l.

Answer 2

Answer:

Mole of MgCL₂ = 0.001 moles.

Permanent hardness = 200 ppm

Given :

Permanent hardness of water :

When the soluble salts of magnesium and calcium are gift withinside the shape of chlorides and sulphides in water, we name it everlasting hardness due to the fact this hardness can't be eliminated via way of means of boiling. We can take away this hardness via way of means of treating the water with washing soda.

Mg (HCO₃)₂ = 36.5 mg/l

Ca(HCO₃)₂ = 81 mg/l

CaSO₄ = 68 mg/l

MgSO₄ = 30 mg/l

MgCL₂ = 47.5 mg/l

CaCL₂ = 55.5 mg/l

Solution :

Mole of Ca(HCO₃)₂ = $\frac{81 * 10^-^3}{81}$ = 0.01 moles

mole of CaSO₄ = $\frac{68*10^-^3}{68}$ = 0.001 moles

Total mole of ca = 2 x 10 ⁻³

Mass of CaCO₃ = 2 x 10 ⁻³ x 100

                          = 0.2g

Permanent hardness = $\frac{6.2}{1000} * 10^6$ =200 ppm

Mole of MgCL₂ = $\frac{47.5*10^-^3}{47}$ = 0.001 moles

Permanent hardness = 200 ppm

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