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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)) |
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Answer» Solution :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` |
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