A weak base is 5 % dissociated in its 0.01 M | vedclass

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(D) For a weak base,the dissociation constant $K_b$ is given by the formula $K_b = \frac{c \alpha^2}{1 - \alpha}$.Given: Concentration $c = 0.01 \ M$,

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$2.5 	imes 10^{-5}$

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(D) For a weak base,the dissociation constant $K_b$ is given by the formula $K_b = \frac{c \alpha^2}{1 - \alpha}$.Given: Concentration $c = 0.01 \ M$,degree of dissociation $\alpha = 5 \% = 0.05$.Since $\alpha$ is very small,we can approximate $1 - \alpha \approx 1$.Thus,$K_b \approx c \alpha^2$.Substituting the values: $K_b = (0.01) 	imes (0.05)^2$.$K_b = 10^{-2} 	imes (2.5 	imes 10^{-3}) = 2.5 	imes 10^{-5}$.

$A$ weak base is $5 \%$ dissociated in its $0.01 \ M$ solution. Calculate the dissociation constant.

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