For the indicator 'Hin' the ratio `(Ind^(Θ))//(HIn)` is `7.0` at `pH` of `4.3`. What is `K_(eq)` for the indicator.

Options

A. `3.5 xx 10^(-4)`
B. `3.5 xx 10^(-5)`
C. `3.5 xx 10^(-2)`
D. `3.5 xx 10^(-3)`

Answer

A

Explanation

`pH = pK_(a) + log [("Salt")/("Acid")]`
`pH = pK_(a) _ log [(In^(Theta))/(HIn)]`
`4.3 = pK_(a) + log 7`
`pK_(a) = 4.3 - log 7`
`pK_(a) = 4.3 - 0.845 = 3.455`
`K_(a) = "Antilog" (-3.455) = (-3-4.55 +1-1)`
`bar(4).545 = 3.5 xx 10^(-4)`