The hydrolysis constant (K_h) for a salt of a | vedclass

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Question

(C) For a salt derived from a weak acid $(HA)$ and a weak base $(BOH)$,the hydrolysis reaction is: $B^+ + A^- + H_2O \rightleftharpoons BOH + HA$. The

Vedclass · Accepted Answer

$K_h = \frac{K_w}{K_a 	imes K_b}$

Vedclass · Answer

(C) For a salt derived from a weak acid $(HA)$ and a weak base $(BOH)$,the hydrolysis reaction is: $B^+ + A^- + H_2O ightleftharpoons BOH + HA$. The equilibrium constant for this reaction is the hydrolysis constant $(K_h)$. It is related to the ionic product of water $(K_w)$,the dissociation constant of the acid $(K_a)$,and the dissociation constant of the base $(K_b)$ by the expression: $K_h = \frac{K_w}{K_a 	imes K_b}$.

Column $I$	Column $II$
$A$. $CH_3COONa$	$i$. Almost neutral,$pH > 7$ or $< 7$
$B$. $NH_4Cl$	$ii$. Acidic,$pH < 7$
$C$. $NaNO_3$	$iii$. Alkaline,$pH > 7$
$D$. $CH_3COONH_4$	$iv$. Neutral,$pH = 7$

The hydrolysis constant $(K_h)$ for a salt of a weak acid and a weak base is given by:

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Column $I$ Column $II$

$A$. $CH_3COONa$ $i$. Almost neutral,$pH > 7$ or $< 7$

$B$. $NH_4Cl$ $ii$. Acidic,$pH < 7$

$C$. $NaNO_3$ $iii$. Alkaline,$pH > 7$

$D$. $CH_3COONH_4$ $iv$. Neutral,$pH = 7$

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