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50 " mL of " water required 4 " mL of " ...

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.

Text Solution

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` 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`
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