50 " mL of " water required 4 " mL of " (N)/(50)HCl for complete neutralisation. 200 " mL of " this water was then boiled with 10 " mL of " (N)/(10) soda reagent. After filtration, the filtrate and the washing were made up to 200 mL with distilled water 50 " mL of " this solution required 8.0 " mL of " (N)/(50) HCl for complete neutralisation. CAlculate the temporary and permanent hardness in ppm.

50 " mL of " water required 4 " mL of " (N)/(50)HCl for complete neutr

1.	50 " mL of " water required 4 " mL of " (N)/(50)HCl for complete neutralisation. 200 " mL of " this water was then boiled with 10 " mL of " (N)/(10) soda reagent. After filtration, the filtrate and the washing were made up to 200 mL with distilled water 50 " mL of " this solution required 8.0 " mL of " (N)/(50) HCl for complete neutralisation. CAlculate the temporary and permanent hardness in ppm.
Answer» Solution :` 4 " mL of " (N)/(50) Hcl=(4xx0.02xx50)/(1000)g CaCO_(3)` So, the temporary HARDNESS`=(4xx0.02xx50)/(1000xx50)g CaCO_(3)` `=80ppm` After boiling with `Na_(2)CO_(3)` and filtration, 8 " mL of " `(N)/(50)` HCl `-=(8)/(50)=0.16m" Eq of "(HCl)/(50mL) Na_(2)CO_(3)` FILTRATE `-=(0.16xx200)/(50)m" Eq of "(HCl)/(200mL) of Na_(2)CO_(3)` filtrate `-=0.64 mEq` Initially, total m" Eq of "`Na_(2)CO_(3)-=10xx(1)/(10)=1mEq` Total mEw of `Na_(2)CO_(3)` reacted `=1-0.64=0.36 mEq` of weight of `Na_(2)CO_(3)=0.36xx10^(-3)xx50g CaCO_(3)` So PERMANENT hardness`=(0.36xx10^(-3)xx50xx10^(6))/(200)=90ppm` Hence, temporary hardness`=80ppm`, permanent hardness `=90ppm`