In room temperature, Br2 has been spread through in H2O and in a organic compound called L which is not mixable with H2O. In equilibrium state,10 cm^3 of brown colour aqueous layer has been taken out and put through aqueous SO2 solution and then the solution has been boiled down and the excess amount of SO2 has been taken off from the system and then put in to methyl orange and the burette reading from titration with 0.01 mol dm^-3 NaOH is 16.7 cm^3.
2 cm^3 of orange coloured organic layer has also been gone through the above process and the burette reading from the titration with 0.05 mol dm^-3 NaOH is 30 cm^3.
From the above data find the Distribution constant of Br2 in L and H2O solvents.
Note : Br2 oxidizes SO2 to SO4^(2-) and itself gets reduced to HBr. Also, when taking out the remaining SO2, HBr would not get out. In the testing state HBr does not get oxidized by H2SO4 and the organic compound won't be an obstacle for titration.
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Br2 oxidizes an aqueous solution of SO2 to H2SO4, itself being reduced to HBr. The balanced reaction is:
`SO_2+Br_2+2H_2O rarr 2HBr+H_2SO_4`
After boiling off excess SO2, the remaining solution contains HBr and H2SO4, which is titrated in the mixture, with NaOH using methyl orange as indicator.
`2HBr+H_2SO_4 + 4NaOH rarr Na_2SO_4+2NaBr+4H_2O`
Overall, 4 eqv. NaOH corresponds to 2 eqv., i.e. 1 mole of bromine.
`16.7*0.01*10^(-3)` eqv. NaOH corresponds to `(1*16.7*0.01*10^(-3))/4` , i.e. `4.175*10^(-5)` moles of bromine.
Therefore, 10 cm^3 of aq. layer contain `4.175*10^(-5)` moles of bromine.
Similarly 2 cm^3 of org. layer, L contains `(0.05*30*10^(-3))/4` , i.e. `3.75*10^(-4)` moles of bromine.
So, 10 cm^3 of org. layer, L contains `(3.75*10^(-4)*10)/2` i.e. `1.875*10^(-3)` moles of bromine.
Distribution constant of Br2 in L and H2O solvents=`(Conc. of Br2 _ L)/(Conc. of Br2_ (H2O))`
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