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

  • Option 1)

    6.50\times 10^{-12}

  • Option 2)

    5.65\times 10^{-13}

  • Option 3)

    5.65\times 10^{-12}

  • Option 4)

    5.65\times 10^{-10}

 

Answers (1)

best_answer

As learnt

Ionic product of water ( Kw ) -

K_{w}=[H_{3}O^{+}]\:[\bar{O}H]

- wherein

At\:T-298K
 

[H^{+}]=[\bar{O}H^{-}]=10^{-7}M
 

Thus\:the\:value\:of\:K_{w}\:at\:298K=10^{-14}M^{2}
 

Thus\:the\:value\:of\:K_{w}\:is\:temperature\:dependent 

 

 

NH4Cl is a salt of strong acid and weak base, so hydrolysis constant is

K_{h}=\frac{K_{w}}{K_{b}}

Given, K_{b}[NH_{4}OH] \: = \: 1.77\times 10^{-5}

         K_{w}=10^{-14}

\therefore K_{h}=\frac{10^{-14}}{1.77\times 10^{-15}} \: \: = 0.565 \times 10^{-9}

K_{h}= 5.65 \times 10^{-10}


Option 1)

6.50\times 10^{-12}

This option is incorrect

Option 2)

5.65\times 10^{-13}

This option is incorrect

Option 3)

5.65\times 10^{-12}

This option is incorrect

Option 4)

5.65\times 10^{-10}

This option is correct

Posted by

prateek

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