1.

A solution of a weak monoprotic acid has dissociation constant `K_(a)`. The minimum initial concentration `C` such that the concentration of the undissociated acid can be equated to `C` within an error of `1%` would beA. `9900 K_(a)`B. `10000K_(a)`C. `99K_(a)`D. `K_(a)`

Answer» Correct Answer - A
`K_(a) = (C alpha^(2))/((1-alpha))`
For `%` error: `alpha = 0.01`
`K_(a) = (C(0.01)^(2))/((1-0.01)) = (C xx 10^(-4))/(0.99)`
`C rArr (K_(a)xx 0.99)/(10^(-4))`
`rArr 9900 K_(a)`


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