Answer:
Here is the full question:
200ml hard water was boiled with 100ml N/50 Na2CO3 reagent. After boilingg the volume was again made to 200mand the solution filtered. 25ml of this filtrate required 8.2ml of N/50 HCl for neutralisation.
Calculate the permanent hardness in ppm.
Explanation:
Remember that the hard water was boiled with 100ml N/50 Na2CO3 reagent. Therefore, the milliequivalent of carbonate ion in 100 ml of N/50 Na2CO3 solution is given as 100 × [tex]\frac{1}{50}[/tex] = 2
The solution was made of 200 ml and 25 ml of the filtrate was neutralised with 8.2 ml N/50 HCl.
The milliequivalent of carbonate ions remaining in the filtrate is [tex]\frac{200}{25}[/tex] × 8.2 × [tex]\frac{1}{50}[/tex] = 1.3 milliequivalent
To get the milliequivalent of carbonate ion precipitated, we need to subtract the milliequivalent of the two carbonate ions. This will give us
2 − 1.3 = 0.7g
The amount of CaCO3 present in 200 ml water = 0.7 × 50 = 35g
Permanent hardness = [tex]\frac{35}{200}[/tex] × 1000 = 175 ppm.
Permanent hardness of water after boiling is 175ppm.
Hardness of water is of two types, one those will cam be removed by boiling is known as temporary hardness and which can not be removed by boiling is known as permanent hardness.
Given that hard water is heated with = 100mL N/50 Na₂CO₃.
Milliequivalent of carbonate ion in 100ml of N/50 Na₂CO₃ = 100×1/50=2
The milliequivalent of carbonate ions remaining in the filtrate is 200/25× 8.2 × 1/50 = 1.3 milliequivalent
Milliequivalent of carbonate ion precipitated = 2 - 1.3 = 0.7g
The amount of CaCO₃ present in 200 ml water = 0.7 × 50 = 35g
Permanent hardness will be calculated as:
P.H. = 35/200 × 1000 = 175ppm.
Hence, permanent hardness of water is 175ppm.
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