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

Options

A. `1.0xx10^(-4)`
B. `1.0xx10^(-10)`
C. `1xx10^(10)`
D. `1.0xx10^(-14)`

Answer

C

Explanation

`HA hArr H^(+)+A^(-)`
`K_(a)= ([H^(+)][A^(-)])/([HA])` …..(i)
Also `HA+B^(+)+OH^(-)rarr B^(+)+A^(-)+H_(2)O`
`K_(eq.)=([H_(2)O][A^(-)])/([HA][OH])` ….(ii)
By eqs. (i) and (ii),
`(K_(eq.))/(K_(a))= (1)/([H^(+)][OH^(-)])=(1)/(K_(w))`
`:. K_(eq.)=(K_(a))/(K_(w))=(10^(-4))/(10_(-14))=10^(10)`