1.

The dissociation constants for acetifc acid and HCN at `25^(@)C` are `1.5xx10^(-5) and 4.5xx10^(-10),` respectively. The equilibrium constant for the equilibrium `CN^(-)+CH_(3)COOHhArrHCN+CH_(3)COO^(-)` would beA. `3.0xx10^(5)`B. `3.0xx10^(-5)`C. `3.0xx10^(-4)`D. `3.0xx10^(4)`

Answer» Correct Answer - D
`CH_(3)COOHhArrCH_(3)COO^(-)+H,K_(a)=1.5xx10^(-5)`
`H^(+)+CN^(-)hArrHCN,(1)/(K_(a))=(1)/(4.5xx10^(-10))`
`thereforeK_(a)"for"CN^(-)+CH_(3)COOHhArrCH_(3)COO^(-)+HCN` is
`(1.5xx10^(-5))/(4.5xx10^(-10))=1/3xx10^(5)=3.33xx10^(4).`


Discussion

No Comment Found

Related InterviewSolutions