1.

50 " mL of " water on titration with standard soap solution gave the following results: Lather factor `=0.4` mL, total hardness `(TH)=8.2mL`, permanent hardness (PH)`=2.5mL` and standard hard water (containing `0.2 g CaCO_3L^(-1))=19.9mL`. Calculate each type of hardness in ppm.

Answer» Strength of `CaCO_3` in hard water
`=(0.2xx50)/(1000)g L^(-1)[Ew(CaCO_3)=50]`
`=0.01g=10mg of CaCO_3 Eq`
Deducting lather factor,
`TH=8.2-0.4=7.8mL`
`PH=2.5-0.4=2.1mL`
Volume of soap solution corresponding to 50 " mL of " standard hard water `=19.9-0.4=19.5mL`
`19.5` " mL of " soap solution`-=10CaCO_3Eq`
1 " mL of " soap solution `=(10)/(19.5)`
`=0.5128 mg CaCO_3` Eq
Volume of soap solution for total hardness
`=7.8mL`
`=7.8xx0.5128mgCaCO_3` equivalents
`=(7.8xx0.5128xx10^(3)mL)/(50mL)`
`=80mg L^(-1)=80ppm (80g 10^(6)mL)`
`TH=80ppm`
Volume of soap solution for PH
`=2.1mL`
`=2.1xx0.5128mg`
`=(2.1xx0.5128xx10^(3)mL)/(50mL)`
`=21.53mgL^(-1)=21.53(g)/(10^(6))mL`
`=21.53ppm`
Hence, `PH=21.53ppm`
`TH=80.21.53=58.47ppm`


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