Question

0.5g of CaCO3 was dissolved in HCl and the solution made up to 500 mL with distilled water. 50 mL of the solution required 48 mL of EDTA solution for the titration. 50 mL of hard
water sample required 15 mL of EDTA and after boiling and filtering required 10 ml of EDTA solution. Calculate the total hardness in ppm.
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Answer 1

Explanation:

Answer:

50 ml SHW required = 48 ml of EDTA solution

∴ 1 ml EDTA solution = 50/48 mgs CaCO3CaCO3 equivalent hardness.

50 ml water sample = 15 ml EDTA solution.

∴ hardness of sample = (15 * 50/48) mgs CaCO3 equivalents for 50 ml sample.

Hardness per litre of sample = (15 * 50/48)* 1000/50 mgs/lit.

Total hardness = 312.50 ppm

50 ml water sample after boiling = 10 ml EDTA solution

Permanent hardness of sample = 10 * 50/48 mgs CaCO3CaCO3 equivalent for 50 ml.

Permanent hardness of one litre sample = (10 * 50/48)*1000/50 mgs/lit.

Answer 2

Given:

• The concentration of solution of HCl in water  = $0.5g CaCO_3/500ml Distilled water$  = 500mgs in 500 ml of water = 1mg/ml
• 50 ml of solution requires 48 ml of EDTA, 1 ml of EDTA = 50/48mgs equivalent $CaCo_3$ hardness.
• 50 ml of hard water sample requires 15 ml of EDTA.
• 50 ml of the sample after boiling requires 10 ml EDTA.

To Find:

• Total Hardness of the solution.

Solution:

• Hardness of sample is given by,
• Hardness of sample = $(15*\frac{50}{48} )mgs CaCO_3$ equivalent of 50ml sample.
• Hardness/litre of the sample = $(15*\frac{50}{48})*1000/50 mgs/litre$  = 312.5 ppm.
• After boiling the sample,
• Permanent hardness of sample = $10*50/48 mgs CaCO_3$ equivalent for 50 ml.
• Permanent hardness/litre of sample = $(10*50/48)*(1000/50)mgs/litre$ = 208.3 ppm.
• Temporary Hardness = Total Hardness - Permanent hardness
• Temporary Hardness = 312.5-208.3 = 104.2 ppm.

The total hardness of the sample is 312.5 ppm.