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 be

Options

A. `9900 K_(a)`
B. `10000K_(a)`
C. `99K_(a)`
D. `K_(a)`

Answer

A

Explanation

`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)`