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The vapour pressure of a 0.01m solution of a weak base BOH in water at 20 C is 17.536 mm. Calculate Kb for the base. Aq tension at 20 C=17.54 mm. Ans=9.7*10^-4
As you know ,
relative lowering of vapour pressure is proportional to mole fraction of the solute.
Since it is a weak base, in aquoeus solution at equilibrium
BOH --> B+ + OH-
initial: 1 0 0
FInal: 1-x x x
Therefore the vanthoff factor i = 1-x+2x = 1+x.
Now, substitute in formula
(P-p)/P = i * molefraction
(17.54 - 17.536)/17.4 = (1+x) * (0.01/55.5) {No of moles of water in 1L=1000/18=55.55 and approximate 55.55+0.01 ~ 55.55 }
From this we get x=0.2668
Now Kb = Cx^2/(1-x)
substitute values and you will get Kb = 9.7 * 10^-4.
Please approve the answer!!!!!
How do u get Kb=Cx^2/(1-x)???
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