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 precipitated as `Mg(OH)_(2)` by adding suitable amount of `NaOH`. the solution, after precipitation of `Mg(OH)_(2)`, is stirred 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 hardness 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)`.

Explanation

Case I (using eriochrome black-T indicator)
`M_1V_1(EDTA)=M_2V_2(Ca^(2+) and Mg^(2+)` in `H_2O)`
`50xx0.005=M_2xx50`
`M_2 of Ca^(2+) and Mg^(2+)=0.005M=5m mol L^(-1)`
Case II (Using calcon indicator)
`M_1V_1(EDTA)=M_2V_2(Ca^(2+)` in `H_2O`)
`10xx0.005=M_2xx50`
`M_2 of Ca^(2+)=10^(-3)M=1.0m mol L^(-1)`
`mmol L^(-1) of Mg^(2+)=5-1=4m mol L^(-1)`
Strength of `Mg^(2+)=4xx24xx10^(-3)=0.096g L^(-1)`
Strength of `Ca^(2+)=1xx40xx10^(-3)=0.04 g L^(-1)`
(b). Hardness due to `Ca^(2+)` ions of the sample in gram of `CaCO_3` in `10^(6)` " mL of " `H_2O`
`=(M(Ca^(2+))xxMw(CaCO_3)xx10^(6))/(10^(3))`
`=(10^(-3)xx100xx10^(6))/(10^(3))=100ppm`
(c). Hardness due to `Mg^(2+)` ions of the sample in g of `CaCO_3`
in `10^(6)` " mL of " `H_2O=(MMg^(2+)xxMw(CaCO_3)xx10^(6))/(10^(3))`
(d). Total hardness `=(M(Ca^(2+) and Mg^(2+))xxMw(CaCO_3)xx10^(6))/(10^(3))`
`=(5xx10^(-3)xx100xx10^(6))/(10^(3))=500ppm`
(Alternatively, Total hardness`=` Hardness due to `Ca^(2+)+` Hardness to `Mg^(2+)=100+400=500ppm)`