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0.0093 g of Na(2)H(2)EDTA.2H(2)O is diss...

`0.0093 g of Na_(2)H_(2)EDTA.2H_(2)O` is dissolved 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_(4)OH+NH_(4)Cl` using eriochrome balck-`T` as indicator. `10 mL` of the above `EDTA` solution requires `10 mL` of hard water at equivalent 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 ammount of `Ca^(2+)` and `Mg^(2+)` present in `1L` of hard water.
b. Calculate the hardness due to `Ca^(2+)`, `mG^(2+)` ions and the total hardness of water in p p m of `CaCO_(3)`. (Given MW(EDTA sal t)=372 g `mol^(-1),MW(CaCO_(3))=100 gmol^(-1)`)

Text Solution

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Case I: Using erichrome black-`T` indicator
`M` of `EDTA` solution `=(0.093xx1000)/(372xx250)=0.001 M`
Volume of `EDTA` used `=10 mL`
Volume of water sample=`40 mL`
`M_(1)V_(1)(EDTA)=M_(2)V_(2)(Ca^(2+)` and `Mg^(2+)` in hard water)
`0.001xx10=M_(2)xx10`
`M_(2)=0.001`
`:.` Molarities of `(Ca^(2+)+Mg^(2+))`ions `=0.001 M`
`=0.1` mmoles `L^(-1)`
Case II: Using murexide indicator
`M_(1)V_(1)(EDTA)=M_(2)V_(2)` (Hard water)
`0.001xx10=M_(2)xx40`
`M_(2)=0.25xx10^(-3)=0.25 mmol L^(-1)`
`:.` Total `mmol L^(-1)`of `Ca^(2+)` and `Mg^(2+)=1.0`
`mmolL^(-1)` of `Ca^(2+)` and `Mg^(2+)=1.0`
`mmol L^(-1)` of `Mg^(2+)=1.0-0.75 mmol L^(-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)=0.018 gL^(-1)`
(b) `[MW (CaCO_(3))=100 g mol^(-1)]`
Total `mmol of Ca^(2+)` and `Mg^(2+)` ions `L^(-1)`
`=1.0=0.001 molL^(-1)`
`=0.001 M`
`0.1 mol of Ca^(2+)` and `Mg^(2+)-=0.1 mol CaCO_(3) L^(-1)`
`-=(0.001xx100xx10^(6))/(10^(3))`
`-=100 p p m`
`:.` total hardness due to `Ca^(2+)` and `Mg^(2+)` ions
of the sample in grams of `CaCO_(3)` in `10^(6) mL of H_(2)O`
`=("Total MxMW"+N19(CaCO_(3))xx10^(6))/(10^(3))`
Hardness due to `Ca^(2+)` ions of the sample in gram of `CaCO_(3)` in `10^(6) mL H_(2)O=(0.25xx10^(-3)xx100xx10^(6))/(10^(3))=25 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=100-25 =75 p p m`
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