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A 50 mL sample of hard water containing ...

A `50 mL` sample of hard water containing `Ca^(2+)` and `Mg^(2+)` ions is titrated with `50 mL. 0.005 M EDTA` solution at `pH=10`, using eriochrome balck`-T` indicator to reach equivalence point.
In a equal amount of hard water sample, `Mg^(2+)` ions are precipated as `Mg(OH)_(2)` by adding suitable amount of `NaOH`. the solution, after precipation of `Mg(OH)_(2)`, is strired and then titrated with `EDTA` solution using calcon as indicator, and it requires `10 mL` of above `EDTA` solution to reach equivalence point.
a. Calculate the strength of `Ca^(2+)` and `Mg^(2+)` ions present in hard water.
b. Calculate the hardness due to `Ca^(2+)` ions in p p m of `CaCO_(3)`.
c. Calculate the hardnesss due to `Mg^(2+)` ions in p p m of `CaCO_(3)`.
d. Calculate the total hardness of water in p p m of `CaCO_(3)`.

Text Solution

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(a) Case I: Using eriochrome black-`T` indicator
`M_(1)V_(1)(EDTA)=M_(2)V_(2)(Ca^(2+)` and `Mg^(2+)` in hard water)
`50xx0.005=M_(2)xx50`
`M_(2)` of `(Ca^(2+)` and `Mg^(2+))=0.005 M=5m` moles `L^(-1)`
Case II: Using calcon indicator
`M_(1)V_(1)(EDTA)=M_(2)V_(2)H_(2)O`
`Ca^(2+)=10^(-3) M=1.0 m mol L^(-1)`
`M mol L^(-1)` of `Mg^(2+)=5-1=4 m mol L^(-1)` ltbRgt Strength of `Mg^(2+)=4xx24xx10^(-3)=.096 g L^(-1)`
Strength of `Ca^(2+)=1xx40xx10^(-3)=0.04 gL^(-1)`
(b) Hardness of `Ca^(2+)`ions of the sample in gram of `CaCO_(3)` in `10^(6) mL of H_(2)O`
`=(M(Ca^(2+)xxMW(CaCO_(3))xx10^(6)))/(10^(3))=100 p p m`
Hardness due to `Mg^(2+)` ions of the sample in grams of `CaCO_(3)` in `10^(6)mL of H_(2)O=(M Mg^(2+)xxMW(CaCO_(3))xx10^(6))/(10^(3))`
`=(4xx10^(-3)xx100xx10^(6))/(10^(3))=400 p p m`
(d) Total hardness
`=(M(Ca^(2+)and Mg^(2+))xxMW(CaCO_(3))xx10^(6))/(10^(3))`
`=(5xx10^(-3)xx100xx10^(6))/(10^(3))=500 p p m`
(Alternatively, total hardness =Harndess due to `Ca^(2+)+` Hardness due to `Mg^(2+)=100+400=500 p p m`)
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