The ionization constant of ammonium hydroxide is 1-77-10-5 at 298 K- Hydrolysis constant of ammonium chloride is-

Given that K_b (NH_4OH)= 1.77 × 10^ 5 We know, nbsp; K_w = 10^ 14 at 298 K From salt hydrolysis, nbsp; K_h (NH_4Cl) = K_wK_b= 10^ 141.77 × 10^ 5= 5.649 × ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Salt of Weak Acid and Weak Base

The ionizatio...

Question

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

5.56 \times 10^{- 12}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5.65 \times 10^{- 10}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

6.50 \times 10^{- 12}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5.65 \times 10^{- 13}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $5.65 \times 10^{- 10}$
Given that $K_{b} (N H_{4} O H) = 1.77 \times 10^{- 5}$

We know,
$K_{w} = 10^{- 14}$ at $298 K$

From salt hydrolysis,
$K_{h} (N H_{4} C l) = \frac{K_{w}}{K_{b}} = \frac{10^{- 14}}{1.77 \times 10^{- 5}} = 5.649 \times 10^{- 10}$

Hence, option (b) is correct.

Suggest Corrections

Similar questions

Q. The ionization constant of Ammonium hydroxide is

1.77 \times 10^{- 5}

at 298 K. It is hydrolysis constant is

Q. Calculate

[O H^{-}]

of 0.01 M solution of ammonium hydroxide solution. The ionization constant (

K_{b}

) for

N H_{4} O H = 1.8 \times 10^{- 5}

Q. Calculate the degree of hydrolysis and

p H

0.02 M

ammonium cyanide

(N H_{4} C N)

at 298 K.

[K_{1}

H C N = 4.99 \times 10^{- 9}, K_{b}

for

N H_{4} O H = 1.77 \times 10^{- 5}]

Q. Calculate

[O H^{-}]

of 0.01 M ammonium hydroxide solution. The ionization constant for

N H_{4} O H

1.8 \times 10^{- 5} M

Q. Calculate

[O H^{-}]

and % age dissociation of

0.01 M

solution of ammonium hydroxide solution. The ionization constant for

N H_{3} O H

K_{b} = 1.8 \times 10^{- 5}

Join BYJU'S Learning Program

The ionization constant of ammonium hydroxide is 1.77×10−5 at 298 K. Hydrolysis constant of ammonium chloride is:

The correct option is B 5.65×10−10Given that Kb(NH4OH)=1.77×10−5 We know, Kw=10−14 at 298 K From salt hydrolysis, Kh(NH4Cl)=KwKb=10−141.77×10−5=5.649×10−10 Hence, option (b) is correct.

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