The dissociation constant of NH_4OH is 1.8 ti | vedclass

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Question

(B) The hydrolysis constant $(K_h)$ for a salt of a weak base and a strong acid $(NH_4Cl)$ is given by the formula: $K_h = \frac{K_w}{K_b}$.Given:$K_w

Vedclass · Accepted Answer

$5.55 	imes 10^{-10}$

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(B) The hydrolysis constant $(K_h)$ for a salt of a weak base and a strong acid $(NH_4Cl)$ is given by the formula: $K_h = \frac{K_w}{K_b}$.Given:$K_w = 1.0 	imes 10^{-14}$ (ionic product of water at $25^{\circ}C$)$K_b = 1.8 	imes 10^{-5}$ (dissociation constant of $NH_4OH$)Substituting the values:$K_h = \frac{1.0 	imes 10^{-14}}{1.8 	imes 10^{-5}}$$K_h = 0.555 	imes 10^{-9} = 5.55 	imes 10^{-10}$.

The dissociation constant of $NH_4OH$ is $1.8 \times 10^{-5}$. Calculate the hydrolysis constant of $NH_4Cl$.

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Which of the following compounds give a basic solution on hydrolysis?
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The hydrolysis constant $(K_h)$ for a salt of a weak acid and a weak base is given by:

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