A certain weak acid has a dissociation constant `1.0xx10^(-4)`. The equilibrium constant for its reaction with a strong base is :

Options

A. `1xx10^(-4)`
B. `1xx10^(-10)`
C. `oo`
D. `1xx10^(10)`

Answer

D

Explanation

`HA hArr H^(+)+A^(-)` or `K_(a)=([H^(+)][A^(-)])/([HA])=10^(-4)" "....(i)`
The reaction with strong base BOH can be expressed as
`HA+OH^(-) hArr H_(2)O+A^(-)`
`K=([A^(-)][H_(2)O])/([OH^(-)][HA])" "....(ii)`
Also `K_(w)=[H^(+)][OH^(-)]=10^(-14)`
From eqns. (i) , (ii) and (iii)
`K=K_(a)//K_(w)=(10^(-4))/(10^(-14))=1010`