Acidic solution is defined as a solution whose `[H^(o+)] gt [overset(Theta)OH]`. Base solution has `[overset(Theta)OH] gt [H^(o+)]`. During acid-base titrations, `pH` of the mixture will change depending on the amount base added. This variation is shown in the form of graph by making plot as titration curves `100mL` of `1.0 M H_(3)A (K_(a_(1)) = 10^(-3), K_(a_(2)) = 10^(-5), K_(a_(3)) = 10^(-7))` is titrated against `0.1M NaOh`. The titration curve is as follows.

What is the `pH` at point `A?

Options

A. `3`
B. `4`
C. `5`
D. `6`

Answer

B

Explanation

Since `K_(a_(1))` and `K_(a_(2))` are high values therfore point `A, K_(a_(1))` and `K_(a_(2))` are considered.
`H_(3)A hArr H^(o+) + H_(2)A^(Theta)`
`H_(2)A^(Theta) hArr H^(o+) + HA^(2-)`
`K_(a_(1)) xx K_(a_(2)) = ([H^(o+)][HA^(2-)])/([H_(2)A]) [["At equivalence point",],[[HA^(2-)]=,[H_(3)A]]]`
`-log K_(a_(1)) - log K_(a_(2)) =- 2log [H^(o+)]`
`pK_(a_(1)) + pK_(a_(2)) = 2pH`.
`:. pH = (pK_(a_(1)) + pK_(a_(2)))/(2) = (3+5)/(2) = 4`