InterviewSolution
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InterviewSolution
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The percentage ionization of a weak base is given byA. `(sqrt((K_(a))/c))xx 100`B. `((1)/(1+10^(pK_(b)-pOH))) xx 100`C. `(sqrt((K_(b))/c))xx 100`D. `(sqrt((K_(w))/(c xx K_(a) "of conjugate acid")))100` |
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Answer» Correct Answer - C::D Let weak base be `= BOH hArr B^(+) + OH^(-)` `{:("Initial conc.",c,0,0,),("Final conc.",c-c alpha,c alpha,c alpha,):}` Let bolume of container be 1 litre `rArr` % dissociation `= (c alpha)/(c) xx 100 = 100 alpha` Also `(c alpha^(2))/(1-alpha) = K_(b)` As `alpha` is negligible `rArr K_(b) = c alpha^(2) , alpha = sqrt((K_(b))/(c))` `rArr` % ionization `= 100 alpha = (sqrt((K_(b))/(c))) 100` `rArr 100alpha = (sqrt((K_(b))/(c))) 100 = (sqrt((K_(w))/(K_(a)c))) 100` `[ because K_(b) = (K_(w))/(K_(a))]` (where `c alpha = [OH^(-)]`) `= log_(10) c alpha = pOH = -log_(10) (K_(b)c)^(1//2) rArr pOH = -(1)/(2)log K_(b)c` `pOH = -(1)/(2)log K_(b)c = -log c alpha` `rArr pOH = (1)/(2)[pK_(b) - log c] = -log c alpha` `2pOH = pK_(b) - logc rArr = pK_(b) - 2pOH` `rArr c = 10^(pK_(b) - 2pOH)` Hence choices (c) and are correct while (a) and (b) are not correct. |
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