1.

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))

Answer» <html><body><p></p>Solution :Case I: Using erichrome black-`T` indicator <br/> `M` of `EDTA` solution `=(0.093xx1000)/(372xx250)=0.001 M` <br/> Volume of `EDTA` used `=<a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a> mL` <br/> Volume of water sample=`40 mL` <br/> `M_(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)V_(1)(EDTA)=M_(2)V_(2)(Ca^(2+)` and `Mg^(2+)` in hard water) <br/> `0.001xx10=M_(2)xx10` <br/> `M_(2)=0.001` <br/> `:.` Molarities of `(Ca^(2+)+Mg^(2+))`ions `=0.001 M` <br/> `=0.1` mmoles `L^(-1)` <br/> Case II: Using murexide indicator <br/> `M_(1)V_(1)(EDTA)=M_(2)V_(2)` (Hard water) <br/> `0.001xx10=M_(2)xx40` <br/> `M_(2)=0.25xx10^(-3)=0.<a href="https://interviewquestions.tuteehub.com/tag/25-296426" style="font-weight:bold;" target="_blank" title="Click to know more about 25">25</a> mmol L^(-1)` <br/> `:.` Total `mmol L^(-1)`of `Ca^(2+)` and `Mg^(2+)=1.0` <br/> `mmolL^(-1)` of `Ca^(2+)` and `Mg^(2+)=1.0` <br/> `mmol L^(-1)` of `Mg^(2+)=1.0-0.75 mmol L^(-1)` <br/> mmoles `L^(-1)` of `Ca^(2+)=0.25 mmol L^(-1)` <br/> `:.` Amount of `Ca^(2+) L^(-1)implies0.25xx40xx10^(-3)` <br/> `=0.01gL^(-1)` <br/> Amount of `Mg^(2+) L^(-1)implies0.75xx24xx10^(-3)=0.018 gL^(-1)` <br/> (b) `[<a href="https://interviewquestions.tuteehub.com/tag/mw-550075" style="font-weight:bold;" target="_blank" title="Click to know more about MW">MW</a> (CaCO_(3))=100 g mol^(-1)]` <br/> Total `mmol of Ca^(2+)` and `Mg^(2+)` ions `L^(-1)` <br/> `=1.0=0.001 molL^(-1)` <br/> `=0.001 M` <br/> `0.1 mol of Ca^(2+)` and `Mg^(2+)-=0.1 mol CaCO_(3) L^(-1)` <br/> `-=(0.001xx100xx10^(6))/(10^(3))` <br/> `-=100 p p m` <br/> `:.` total hardness due to `Ca^(2+)` and `Mg^(2+)` ions <br/> of the sample in grams of `CaCO_(3)` in `10^(6) mL of H_(2)O` <br/> `=("Total MxMW"+N19(CaCO_(3))xx10^(6))/(10^(3))` <br/> 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` <br/> 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`</body></html>


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