Question

The ionization constant of Ammonium hydroxide is $1.77\times {10}^{-5}$ at 298 K. It is hydrolysis constant is

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

$K_h=\frac{k_w}{k_b}=\frac{1\times {10}^{-14}}{1.77\times {10}^{-5}}=5.65\times {10}^{-10}$