1.

A solution of monobasic acid with molarity 3xx10^(-2)M has a freezing point depression of 0.06^(@)C. Calculate pK_(a) of the acid (Molal depression constant of water is 1.86^(@)C//m)

Answer»

Solution :`DeltaT_(f)=I XX K_(f)xxm`
Taking molality = MOLARITY (as the solution is DILUTE)
`0.06=i xx 1.86xx3xx10^(-2)"or"i=(0.06)/(1.86xx3xx10^(-2))=1.0753`
For the acid HA,
`{:(,HA ,hArr,H^(+),+,A^(-),),("Initial moles",C,,0,,0,),("At. eqm.",C(1-alpha),,Calpha,,Calpha,"Total "=C(1-alpha)+Calpha+Calpha=C(1+alpha)):}`
`therefore"i"=("Total no. moles after dissociation")/("Inital moles")=(C(1+alpha))/(C)=1+alpha`
`"or"alpha=i-1=1.0753-1=0.0753`
`K_(a)=(Calpha.Calpha)/(C(1-alpha))=Calpha^(2)`
`=(3xx10^(-2))(0.0753)^(2)=1.74xx10^(-4)""(therefore alpha lt lt 1)`
`pK_(a)=-logK_(a)=-log(1.74xx10^(-4))=3.769~=3.77`


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