1.

The measured freezing point depression for a 0.1 m aqueous acetic acid solution is 0.19^(@)C. The acid dissociation constant K_(a) at this concentration will be (K_(f)= 1.86 "Km"^(-1))

Answer»

`4.76xx10^(-5)`
`4xx10^(-5)`
`8xx10^(-5)`
`2xx10^(-5)`

Solution :`DeltaT_(f)=0.19, m=0.1, K_(f)=1.86"KM"^(-1)`
`DeltaT_(f)=iK_(f)xxm`
or `i=(DeltaT_(f))/(K_(f)xxm)=(0.19)/(1.86xx0.1)=1.02`
If `ALPHA` is the degree of dissociation
`{:(,CH_(3)COOH,hArr,CH_(3)COO^(-),+,H^(+)),("Initial moles","1",,"0",,"0"),("After dissociation",""1-alpha,,""alpha,,alpha):}`
Total number of moles after dissociation `=1-alpha+alpha+alpha=1+alpha`
`i=(1+alpha)/(1)=1.02` or `alpha=0.02`
Now `K_(alpha)=C alpha^(2)=0.1xx(0.02)^(2)=4xx10^(-5)`


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