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)`)

Options

A. `4.76xx10^(-5)`
B. `4xx10^(-5)`
C. `8xx10^(-5)`
D. `2xx10^(-5)`

Answer

B

Explanation

`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)`