The ionization constant of ammonium hydroxide is `1.77xx10^(-5)` at `298 K`. Hydrolysis constant of ammonium chloride is

Options

A. `5.665xx10^(-12)`
B. `5.65xx10^(-10)`
C. `6.50xx10^(-12)`
D. `6.65xx10^(-13)`

Answer

B

Explanation

`NH_(4)Cl` is a salt of strong acid and weak base, so
`K_(h)=(K_(w))/(K_(b))=(10^(-14))/(1.771x10^(-5))=5.65x10^(-19)`