A standard water sample of volume 500 ml is prepared by dissolving 0.5 g of CaCO3 in HCl. 20 ml of the solution required 16.6 ml of EDTA solution. 20 ml of a given water sample consumed 17.2 ml, EDTA solution and 2.2. ml after ion exchange. Determine the normality of EDTA solution, total hardness and residual hardness.
Answers
Answer:
Given:
The concentration of solution of HCl in water = 0.5g CaCO_3/500ml Distilled water0.5gCaCO
3
/500mlDistilledwater = 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_3CaCo
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(15∗
48
50
)mgsCaCO
3
equivalent of 50ml sample.
Hardness/litre of the sample = (15*\frac{50}{48})*1000/50 mgs/litre(15∗
48
50
)∗1000/50mgs/litre = 312.5 ppm.
After boiling the sample,
Permanent hardness of sample = 10*50/48 mgs CaCO_310∗50/48mgsCaCO
3
equivalent for 50 ml.
Permanent hardness/litre of sample = (10*50/48)*(1000/50)mgs/litre(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.