Degree of dissociation of NH4OH in water is 1.8× 10-5, then hydrolysis constant of NH4Cl is:

Question

Degree of dissociation of NH4OH in water is 1.8× 10-5, then hydrolysis constant of NH4Cl is: (A)  1.8× 10-5  (B)  1.8× 10-10  (C)  5.55× 10-5  (D)  5.55× 10-10

Tardigrade Admin · Accepted Answer

Kh=(Kw/Kb) where Kw= ionic product of water =1 × 10-14 Kb= degree of dissociation of NH 4 OH =1.8 × 10-5 Kh=(1 × 10-14/1.8 × 10-5) =0.555 × 10-9 =5.55 × 10-10

Q. Degree of dissociation of $N H_{4} O H$ in water is $1.8 \times 10^{- 5},$ then hydrolysis constant of $N H_{4} Cl$ is:

$1.8 \times 10^{- 5}$

$1.8 \times 10^{- 10}$

$5.55 \times 10^{- 5}$

$5.55 \times 10^{- 10}$

$K_{h} = \frac{K _{w}}{K _{b}}$
where $K_{w} =$ ionic product of water $= 1 \times 1 0^{- 14}$
$K_{b} =$ degree of dissociation of $N H_{4} O H$ $= 1.8 \times 1 0^{- 5}$
$K_{h} = \frac{1 \times 1 0 ^{- 14}}{1.8 \times 1 0 ^{- 5}}$
$= 0.555 \times 1 0^{- 9}$
$= 5.55 \times 1 0^{- 10}$

Q. Degree of dissociation of NH4​OH in water is 1.8×10−5, then hydrolysis constant of NH4​Cl is:

1.8×10−5

1.8×10−10

5.55×10−5

5.55×10−10