The ionization constant of ammonium hydroxide | vedclass

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Question

(D) The ionization constant of the weak base $NH_{4}OH$ is given as $K_{b} = 1.77 	imes 10^{-5}$.Ammonium chloride $(NH_{4}Cl)$ is a salt of a weak b

Vedclass · Accepted Answer

$5.65 	imes 10^{-10}$

Vedclass · Answer

(D) The ionization constant of the weak base $NH_{4}OH$ is given as $K_{b} = 1.77 	imes 10^{-5}$.Ammonium chloride $(NH_{4}Cl)$ is a salt of a weak base $(NH_{4}OH)$ and a strong acid $(HCl)$.The hydrolysis constant $(K_{h})$ for such a salt is given by the formula:$K_{h} = \frac{K_{w}}{K_{b}}$Where $K_{w}$ is the ionic product of water,which is $1.0 	imes 10^{-14}$ at $298 \ K$.Substituting the values:$K_{h} = \frac{1.0 	imes 10^{-14}}{1.77 	imes 10^{-5}}$$K_{h} \approx 0.56497 	imes 10^{-9} = 5.65 	imes 10^{-10}$.

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

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