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InterviewSolution
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In a chrome plating plant, `CrO_4^(2-)` (chromate) ions are present in waste water. The chromate ions are reduced to insoluble chromium hydroxide, `Cr(OH)_3`, by dithionate ion, `S_2O_4^(2-)` in basic medium. `CrO_4^(2-)+S_2O_4^(2-)+overset(ɵ)(O)H+H_2OtoCr_2(OH_3)+SO_3^(2-)` 10 L of water requires 522 g of `Na_2S+2O_4`. Calculate the normality and molarity of `CrO_4^(2-)` in waste water. Also express the concentration of `Na_2CrO_4` in ppm. |
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Answer» `{:[undersetunderset(2x=6)(x-8=-2)(S_2O_4^(2-))toundersetunderset(2x=8)(2x-12=-4)(2SO_3^(2-))+2e^(-):}](n=2)` `{:[undersetunderset(x=6)(x-8=-2)(CrO_4^(2-)+3e^(-))toundersetunderset(x=3)(x-3=0)(Cr(OH))]:}(n=3)` `Mw(Na_2S_2O_4)=2xx23+2xx32+16xx4=174g` `(Ew=(174)/(2))` `S_2O_4^(2-)-=CrO_4^(2-)` `Eq-=Eq` `("Weight")/("Equivalent weight")-=Eq` `(522)/((174)/(2))-=Eq` `therefore` 6 " Eq of "`S_2O_4^(2-)-=(6" Eq of "CrO_4^(2-))/(10L of "water")` `-=(6)/(10)=0.6EqL^(-1)` `therefore` Normality of `CrO_4^(2-)=0.6N` Molarity of `CrO_4^(2-)=(N)/("n-factor")=(0.6)/(3)=0.2M` Strength of `Na_2CrO_4=MxxMw` `(Mw Na_2CrO_4=2xx23+52+16xx4=162g)` `0.2xx162` `=32.4gL^(-1)` Concentration in ppm `=("Weight of" Na_2CrO_4" in "1000 mL xx10^(6))/(1000mL)` `=(32.4xx10^(6))/(10^(3))=32400p p m` |
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