1.

The correct relation between hydrolysis constant (K_(b)) and degree of hydrolysis (alpha) for the following equilibrium is

Answer»

`alpha = sqrt((K_(W).C)/(K_(a)))`
`alpha= sqrt((K_(w))/(K_(a).C))`
`alpha = sqrt((K_(a).C)/(K_(w))`
`alpha = sqrt((K_(a))/(K_(w).C))`

SOLUTION :Salt of weak ACID and strong BASE.
`K_(h) = (K_(w))/(K_(a))`
`h = sqrt((K_(h))/(c))`
`h = sqrt((K_(w))/(K_(a)c))`


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