0.093 g of `Na_2H_2` `EDTA.2H_2O` is dissolved in in 250 " mL of " aqueous solution. A sample of hard water containing `Ca^(2+)` and `Mg^(2+)` ions is titrated with the above EDTA solution using a buffer of `NH_4OH+NH_4Cl` using erochrome black-T as indicator. 10 " mL of " hard water at equivalence point. Another sample of hard water is titrated with 10 " mL of " above EDTA solution using KOH solution `(pH=12)`. Using murexide indicator, it requires 40 " mL of " hard water at equivalence point.
(a). Calculate the amount of `Ca^(2+)` and `Mg^(2+)` present in 1 L of hard water.
(b). Calculate the hardness due to `Ca^(2+),Mg^(2+)` ions and the total hardness of water in ppm of `CaCO_3`
(Given: `Mw(EDTA sal t)=372gmol^(-1)`
`Mw(CaCO_3)=100gmol^(-1))`

Explanation

Case I (Using eriochrome black-T indicator)
(a). M of EDTA solution`=(0.093xx1000)/(372xx250)=0.001M`
Volume of EDTA used `=10mL`
Volume of water sample `=40mL`
`M_1V_1(EDTA)=M_2V_2(Ca^(2+) and Mg^(2+)` in hard water)
`0.001xx10=M_2xx10`
`M_2=0.001`
Molarities of `(Ca^(2+)+Mg^(2+))` ions `=0.001M`
`=1.0 m` moles/L
Case II: (Using murexide indicator)
`M_1V_1(EDTA)=M_2V_2` (Hard water)
`0.001xx10=M_2xx40`
`M_2=0.25xx10^(-3)=0.25mmolL^(-1)`
Total m mol `L^(-1) of Ca^(2+) and Mg^(2+)=1.0`
mmol `L^(-1) of Mg^(2+)=1.0-0.25=0.75mmolL^(-1)`
mmoles `L^(-1) of Ca^(2+)=0.25 mmol L^(-1)`
Amount of `Ca^(2+)L^(-1)implies0.25xx40xx10^(-3)=0.01gL^(-1)`
Amount of `Mg^(2+)L^(-1)implies0.75xx24xx10^(-3)=0.018gL^(-1)`
(b). `[Mw(CaCO_3)=100gmol^(-1))`
Total m" mol of "`Ca^(2+)` and `Mg^(2+)` ions `L^(-1)`
`=1.0=0.001 (mol)/(L)`
`=0.001M`
0.1 " mol of "`Ca^(2+) and Mg^(2+)-=0.1mol CaCO_3L^(-1)`
`-=(0.001xx100xx10^(6))/(10^(3))`
`-=100ppm`
`{:[("Total hardness due to Ca^(2+) and Mg^(2+) ions"),("of the sample in gram of CaCO_3in 10^(6) " mL of " H_2O"),(=("Total MxxMw(CaCO_3)xx10^(6))/(10^(3))]:}`
Hardness due to `Ca^(2+)` ions of the sample in gram of `CaCO_3`
in `10^(6)` " mL of " `H_2O=(0.25xx10^(-3)xx100xx10^(5))/(10^(3))=25ppm`
Hardness due to `Mg^(2+)` ions of the sample in gram of `CaCO_3` in `10^(6)` " mL of " `H_2O=100-25=75ppm`