Question

Hydrolysis constant (Kh) of 0.01 M
ammonium chloride solution at 250
C when ionization constant Kb=1.8x
10-5 and ionic product of water (Kw)
at 250 C = 1.0 x 10-14​

Answer 1

Answer:

Hydrolysis

Explanation:

Hydrolysis constant (Kh) of 0.01 M

ammonium chloride solution at 250

C when ionization constant Kb=1.8x

10-5 and ionic product of water (Kw)

at 250 C = 1.0 x 10-14

Answer 2

Given:

The concentration of ammonium chloride (NH₄Cl) = 0.01 M

The ionization constant of ammonia (Kb) = 1.8 × 10⁻⁵

The ionic product of water (Kw) = 1.0 × 10⁻¹⁴

To Find:

The hydrolysis constant (Kh) of ammonium chloride (NH₄Cl) at 25°C.

Solution:

The relation between hydrolysis constant (Kh), ionization constant of a weak base (Kb), and ionic product of water (Kw) for a salt of a weak base and strong acid is

Kh = Kw/ Kb = Ionic product of water / Ionization constant of ammonia

⇒ Kh = 1.0 × 10⁻¹⁴ / 1.8 × 10⁻⁵ = 5.6 × 10⁻¹⁰.

Hence, the hydrolysis constant (Kh) of ammonium chloride (NH₄Cl) salt is  5.6 ×10⁻¹⁰.