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 xx 10^(-1)`
B. `1 xx 10^(-10)`
C. `1 xx 10^(10)`
D. `1 xx 10^(14)`

Explanation

When a weak acid combines with a strong base, there salt is formed which undergoes hydrolysis may be considered as the reverse of neutralisation, the equilibrium constant will thus be equal to `1//K_(h).` Therefore . `(1)/(K_(h))=1/(K_(w)//K_(a))=K_(a)/K_(w)=(1 xx 10^(-4))/(1 xx 10^(-14))=1 xx 10^(-10)`