The dissocication constant of a weak acid is `1.0xx10^(-5)`, the equilibrium constant for the reaction with strong base is

Options

A. `1.0xx10^(-5)`
B. `1.0xx10^(-9)`
C. `1.0xx10^(9)`
D. `1.0xx10^(14)`

Answer

C

Explanation

`HAhArrH^(+)+A^(-)` , `K_(a)=([H^(+)][A^(-)])/([HA])` (`i`)
neutralization of the weak acid with strong base is
`HA+OH^(-)hArrA^(-)+H_(2)O`
`K=([A^(-)])/([HA][OH^(-)])` (`ii`)
Dividing (`i`) by (`ii`), `(K_(a))/(K)=[H^(+)][OH^(-)]=K_(w)=10^(-14)`
`K=(K_(a))/(K_(w))=(10^(-5))/(10^(-14))=10^(9)`