Question

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

Answer 1

Correct Answer - A
`CH_(3)COOHhArrCH_(3)COOH^(-)+H^(+)`
`K_(a_(1))=([CH_(3)COO^(-)][H^(+)])/([CH_(3)COOH])` (`i`)
`HCNhArrH^(+)+ CN^(-)`
`K_(a_(2))=4.5xx10^(-10)=([H^(+)][CN^(-)])/([HCN])` (`ii`)
By (`i`)/(`ii`)
`K= (K_(a_(1)))/(K_(a_(2)))=([HCN][CH_(3)COO^(-)])/([CN^(-)][CH_(3)COOH])`
`=(1.5xx10^(-5))/(4.5xx10^(-10))=3xx10^(4)`