InterviewSolution
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InterviewSolution
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The dissociation constant of a weak acid `HA si 4.9 xx 10^(-8)`. After making the necessary approximations, calculate i. Percentage ionisation ii. `pH` iii. `overset(Theta)OH` concentration in a decimolar solution of the acid. Water has a `pH of 7`. |
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Answer» Correct Answer - A::C::D The dissociation constant of a weak acid `HA` is `4.9 xx 10^(-8)`. i. For the weak acid, `alpha = sqrt(((K_(a))/(C))) =sqrt(((4.9xx10^(-8))/(1//10)))` `= 7 xx 10^(-4) = 0.07 %` ii. `pH = - log [H^(o+)]` `=- log [C alpha] = [(1)/(10) xx7 xx 10^(-4)]` `=- log [7 xx 10^(-5)] = 4.15` iii. `[overset(Theta)OH] [H^(o+)] = 10^(-14)` or `[overset(Theta)OH] = (10^(-14))/([H^(o+)])` `= (10^(-14))/([Calpha]) = (10^(-14))/(7xx10^(-5)) =1.43 xx 10^(-10)` |
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