The pH of a monoacidic base is 10. Calculate | vedclass

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(D) For a monoacidic base $BOH$,the $pH = 10$. Since $pH + pOH = 14$,we have $pOH = 14 - 10 = 4$. Therefore,$[OH^-] = 10^{-pOH} = 10^{-4} \ M$. The co

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(D) For a monoacidic base $BOH$,the $pH = 10$. Since $pH + pOH = 14$,we have $pOH = 14 - 10 = 4$. Therefore,$[OH^-] = 10^{-pOH} = 10^{-4} \ M$. The concentration of the base $C = 0.01 \ M = 10^{-2} \ M$. For a weak base,$[OH^-] = C 	imes \alpha$,where $\alpha$ is the degree of dissociation. $10^{-4} = 10^{-2} 	imes \alpha$. $\alpha = \frac{10^{-4}}{10^{-2}} = 10^{-2} = 0.01$. Percentage dissociation $= \alpha 	imes 100 = 0.01 	imes 100 = 1 \%$.

The $pH$ of a monoacidic base is $10$. Calculate its percentage dissociation in $0.01 \ M$ solution at $298 \ K$. (in $\%$)

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