JEE Main Chemistry question
The ionization constant of ammonium hydroxide is 1.77×10⁻⁵ at 298 K. Hydrolysis constant of ammonium chloride is:
Answer: (B)
Solution
Kh=Kw/Kb=10⁻¹⁴/1.77×10⁻⁵=5.65×10⁻¹⁰.
The ionization constant of ammonium hydroxide is 1.77×10⁻⁵ at 298 K. Hydrolysis constant of ammonium chloride is:
Kh=Kw/Kb=10⁻¹⁴/1.77×10⁻⁵=5.65×10⁻¹⁰.