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

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 × = 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 × 8.2 × = 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 = × 1000 = 175 ppm.

catya that's not even a b or c