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A 200 mL sample of hard water requires 3...

A 200 mL sample of hard water requires 33.0 " mL of " 0.01 M `H_2SO_4` for complete neutralisation.
200 " mL of " the same sample was boiled with 15.0 " mL of " 0.1 M NaOH solution, filtered and made up to 200 mL again. This sample now requires 53.6 " mL of " 0.01 M `H_2SO_4`. Calculate Mg hardness.

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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