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`.

Answer

A, C, D

Explanation

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)`