If the base ionization constant (K_b) of hydr | vedclass

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(D) The salt $N_2H_5Cl$ is a salt of a weak base $(N_2H_4)$ and a strong acid $(HCl)$.For such salts,the hydrolysis constant $(K_h)$ is given by $K_h

Vedclass · Accepted Answer

$0.003$

Vedclass · Answer

(D) The salt $N_2H_5Cl$ is a salt of a weak base $(N_2H_4)$ and a strong acid $(HCl)$.For such salts,the hydrolysis constant $(K_h)$ is given by $K_h = \frac{K_w}{K_b}$.Given $K_w = 1.0 	imes 10^{-14}$ and $K_b = 9.6 	imes 10^{-9}$.$K_h = \frac{1.0 	imes 10^{-14}}{9.6 	imes 10^{-9}} \approx 1.04 	imes 10^{-6}$.The degree of hydrolysis $(h)$ is given by $h = \sqrt{\frac{K_h}{C}}$,where $C = 0.1 \ M$.$h = \sqrt{\frac{1.04 	imes 10^{-6}}{0.1}} = \sqrt{1.04 	imes 10^{-5}} \approx 3.22 	imes 10^{-3}$.Percentage of hydrolysis = $h 	imes 100 = 3.22 	imes 10^{-3} 	imes 100 = 0.322 \%$.Rounding to the nearest option provided,the value is approximately $0.3 \%$ or $0.003$ as a fraction.

If the base ionization constant $(K_b)$ of hydrazine $(N_2H_4)$ is $9.6 \times 10^{-9}$,what is the percentage of hydrolysis of a $0.1 \ M$ solution of the salt $N_2H_5Cl$?

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