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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))

Answer» <html><body><p></p>Solution :Case I (Using eriochrome black-T indicator) <br/> (a). M of EDTA solution`=(0.093xx1000)/(372xx250)=0.001M` <br/> Volume of EDTA used `=<a href="https://interviewquestions.tuteehub.com/tag/10ml-267166" style="font-weight:bold;" target="_blank" title="Click to know more about 10ML">10ML</a>` <br/> Volume of water sample `=40mL` <br/> `M_1V_1(EDTA)=M_2V_2(Ca^(2+) and Mg^(2+)` in hard water) <br/> `0.001xx10=M_2xx10` <br/> `M_2=0.001` <br/> Molarities of `(Ca^(2+)+Mg^(2+))` ions `=0.001M` <br/> `=<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>.0 m` moles/L <br/> Case II: (Using murexide indicator) <br/> `M_1V_1(EDTA)=M_2V_2` (Hard water) <br/> `0.001xx10=M_2xx40` <br/> `M_2=0.25xx10^(-3)=0.25mmolL^(-1)` <br/> Total m mol `L^(-1) of Ca^(2+) and Mg^(2+)=1.0` <br/> <a href="https://interviewquestions.tuteehub.com/tag/mmol-2837464" style="font-weight:bold;" target="_blank" title="Click to know more about MMOL">MMOL</a> `L^(-1) of Mg^(2+)=1.0-0.25=0.75mmolL^(-1)` <br/> mmoles `L^(-1) of Ca^(2+)=0.25 mmol L^(-1)` <br/> Amount of `Ca^(2+)L^(-1)implies0.25xx40xx10^(-3)=0.01gL^(-1)` <br/> Amount of `Mg^(2+)L^(-1)implies0.75xx24xx10^(-3)=0.018gL^(-1)` <br/> (<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a>). `[<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)=100gmol^(-1))` <br/> Total m" mol of "`Ca^(2+)` and `Mg^(2+)` ions `L^(-1)` <br/> `=1.0=0.001 (mol)/(L)` <br/> `=0.001M` <br/> 0.1 " mol of "`Ca^(2+) and Mg^(2+)-=0.1mol CaCO_3L^(-1)` <br/> `-=(0.001xx100xx10^(6))/(10^(3))` <br/> `-=100ppm` <br/> `{:[("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))]:}` <br/> Hardness due to `Ca^(2+)` ions of the sample in gram of `CaCO_3` <br/> in `10^(6)` " mL of " `H_2O=(0.25xx10^(-3)xx100xx10^(5))/(10^(3))=25ppm` <br/> Hardness due to `Mg^(2+)` ions of the sample in gram of `CaCO_3` in `10^(6)` " mL of " `H_2O=100-25=75ppm`</body></html>


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