`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 this solution is treated with `30mL`of `NaOH` solution, then what molarity of `NaOH` is needed to reach the equivalence point with indicator?

Options

A. `0.1M`
B. `0.2M`
C. `0.3M`
D. `0.4M`

Answer

A

Explanation

At `1st` neutralisation point, `H_(3)PO_(4)` is converted to `KH_(2)PO_(4)`, here `pH` of the solution
`(pK_(1)+pK_(2))/(2) = (3+7)/(2) = 5`.
At `2nd` neutralisation point, `H_(3)PO_(4)` is converted to `KH_(2)PO_(4)`,
`:. pH = (pK_(2)+pK_(3))/(2)=(7+13)/(2) = 10`
The equivalence point will be obtained when acid `H_(3)PO_(4)` will change into `K_(2)HPO_(4)`.
('n' factor for acid =2) `(2H^(o+)` ions are replaced)
`mEq of H_(3)PO_(4) = mEq` of `NaOH`
`0.05 xx 30 xx2 = 30 xx M`
`M_(NaOH) = 0.1M`