# A solution of 1 mol dm^-3 and 10 cm^3 NH3(aq) has been added to a flask which also has 50cm^3 CHCl3 and 40cm^3 water and then mixed them all together. 0.08 mol dm^-3 and 10 cm^3 H2SO4 has been used...

A solution of 1 mol dm^-3 and 10 cm^3 NH3(aq) has been added to a flask which also has 50cm^3 CHCl3 and 40cm^3 water and then mixed them all together. 0.08 mol dm^-3 and 10 cm^3 H2SO4 has been used to cold down 10 cm^3 of the water layer.

Find the Distribution constant of NH3 for CHCl3/H2O.

Asked on by shihan

llltkl | College Teacher | (Level 3) Valedictorian

Posted on

Number of moles of NH3 present in `10 cm^3` aq. layer (after mixing)=`0.08*2*10*10^(-3)` [one mole H2SO4 consumes 2 moles of NH3]

`=1.6*10^(-3)`

Number of moles of NH3 present in total aq. layer, after thorough mixing (i.e. in whole 50 cm^3)

`=(50*1.6*10^(-3))/10=8.0*10^(-3)`

Total number of moles of NH3 added to the aq.-organic system=`1.0*10*10^(-3)`

`=10*10^(-3)`

Therefore, number of moles of NH3 going to the CHCl3 layer, assuming no loss of NH3 to the atmosphere during mixing `=(10-8)*10^(-3)`

`=2*10^(-3)`

This is present in `50 cm^3` of CHCl3 layer.

Hence, distribution constant of NH3 for CHCl3/H2O, Kd = `C_(CHCl3)/C_(aq.)= (2*10^(-3))/(8.0*10^(-3))=0.25`

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