`K_(a)` of `HA` is `10^(-4)`. The equilibrium constant for its reaction with a strong base is

Options

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

Answer

C

Explanation

For the weak acid `HA`,
`HA hArr H^(o+) + A^(Theta)`
`K_(a) = ([H^(o+)]xx[A^(Theta)])/([HA])`
Reaction of weak acid with strong base.
`HA + OH^(Theta) hArr A^(Theta) + H_(2)O`
`K = ([A^(Theta)])/([HA]xx[overset(Theta)OH])`
Divident (1) by (2), we get
`(K_(a))/(K) = [H^(o+)] [overset(Theta)OH] = K_(w)`
or `K = (K_(a))/(K_(w)) = (10^(-14))/(10^(-14)) = 10^(10)`