At 25,^o C, the dissociation const | vedclass

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Question

(d) ${K_b} = \frac{{{C^2}{\alpha ^2}}}{{C(1 - \alpha )}} = C{\alpha ^2}$ assuming $\alpha < < 1$;$1 - \alpha 	ilde -1$

${10^{ - 12}} = {10^{

Vedclass · Accepted Answer

$1.0 	imes {10^{ - 7}}\,mol\;{L^{ - 1}}$

Vedclass · Answer

(d) ${K_b} = \frac{{{C^2}{\alpha ^2}}}{{C(1 - \alpha )}} = C{\alpha ^2}$ assuming $\alpha < < 1$;$1 - \alpha 	ilde -1$

${10^{ - 12}} = {10^{ - 2}} 	imes {\alpha ^2}$; ${\alpha ^2} = {10^{ - 10}}$; $\alpha = {10^{ - 5}}$

$[O{H^ - }] = C\alpha = .01 	imes {10^{ - 5}} = {10^{ - 7}}$

Acid	$K_a$
$HCN$	$6.2\times 10^{-10}$
$HF$	$7.2\times 10^{-4}$
$HNO_2$	$4.0\times 10^{-4}$

At $25\,^o C$, the dissociation constant of a base $BOH$ is $1.0 \times {10^{ - 12}}$. The concentration of Hydroxyl ions in $0.01\, M$ aqueous solution of the base would be

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Values of dissociation constant, $K_a$ are given as follows

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$HCN$ $6.2\times 10^{-10}$

$HF$ $7.2\times 10^{-4}$

$HNO_2$ $4.0\times 10^{-4}$

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