A certain weak acid has a dissocation constant of `1.0 xx 10^(-4)`. The equilibrium constant for its reaction with a strong base is

Options

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

Answer

C

Explanation

`HA: K_(a) = 10^(-4)`
`HA +NaOH hArr NaA +H_(2)O`
Clearly, the reverse reaction is the hydrolysis reaction.
`rArr K_("Required") = (1)/(K_(h)) = (K_(a))/(K_(w)) = (10^(-10))/(10^(-14)) = 10^(10)`