Question

The ionisation constant of ammonium hydroxide is $ 1.77 \times 10^{-5} $ at $298\, K.$ Hydrolysis constant of ammonium chloride is

• A. $ 5.65 \times 10^{-10} $
• B. $ 6.50 \times 10^{-12} $
• C. $ 5.65 \times 10^{-13} $
• D. $ 5.65 \times 10^{-12} $

Answer 1

Answer: (A)

Given that $K _{ b }\left( NH _{4} OH \right)=1.77 \times 10^{-5}$
$K _{ W }=10^{-14}$ at $298\, K$
$K _{ h }\left( NH _{4} Cl \right)=\frac{ K _{ W }}{ K _{ b }}=\frac{10^{-14}}{1.77 \times 10^{-5}}$
$=5.65 \times 10^{-10}$