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The freezing point of 0.20 M solution of...

The freezing point of `0.20 M` solution of week acid `HA` is `272.5 K`. The molality of the solution is `0.263 "mol" kg^(-1)`. Find the pH of the solution on adding `0.25 M` sodium acetate solution.
`K_(f)` of water =`1.86 K m^(-1)`

Text Solution

Verified by Experts

`DeltaT_(f)=(273-272.5)=0.5 K`
`DeltaT_(f)=iK_(f)m`
`:.i=DeltaT_(f)//K_(f)m=0.5//1.86xx0.263=1.022`

`i=1+alpha`
`1.022=1+alpha rArr alpha = 0.022`
`K_(a)=C alpha^(2)=0.2xx(0.022)^(2)=9.6xx10^(-5)`
`rArr pK_(a)=4.0177`
On adding `0.25 M NaA`, buffer is formed
`:.`[Salt]=0.25 M[Acid]=0.2 M
`pH=pK_(a)+log["Salt"//"Acid"]=4.0177+log(0.25//0.2)`
`=4.0177+0.0969=4.1146`
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