`Hin` is an acidic indicator `(K_(Ind) =10^(-7))` which dissociates into aqueous acidic solution of `30mL` of `0.05M H_(3)PO_(4) (K_(1) = 10^(-3), K_(2) = 10^(-7), K_(3) = 10^(-13))`
If `Hin` and `Ind^(Theta)` posses colour `P` and `Q`, respectively, and concentration of `HIn` is `120` times than that of `Ind^(Theta)`. colour `Q` predominates over `P` when concnetration of `Ind^(Theta)` is `127` times of `HIn`.
What is the `pH` range of the indicator.

Options

A. `4.896 to 9.0792`
B. `4.896 to 8.0792`
C. `4.896 to 7.0792`
D. `4.896 to 6.0792`

Answer

A

Explanation

When colour `P` redominates over `Q`, then,
`pH = pK_(Ind) + log 120`
`pH = 7+ log 120 = 9.0792`
When colour `Q` predominates over `P`, then
`pH = pK_(Ind) +"log"(1)/(127)`
`pH = 7 - log 127 = 4.896`
`pH` range of indicator is `4.896` to `9.0792`