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. `11.0`
B. `10.2`
C. `9.7`
D. `7.7`

Answer

C

Explanation

At point `C,Na_(3)A` is formed, hydrolysis of `A^(3-)` occurs and hydrolysis of `Na_(2)HA` is supressed, `(K_(a_(3))` is considered) due to common ion `[overset(Theta)OH]` effect
`A^(3+) + H_(2)O hArr HA^(2-) + overset(Theta)OH`
`HA^(2-)+H_(2)O hArr H_(2)A^(Theta) + overset(Theta)OH`
At point `C` salt `Na_(3)A`(salt of `W_(A)//S_(B))` is formed. since the concentration of `NaOH` and `H_(3)A (0.1)` are equal. so the volume of `NaOH` used at point `C` is `3` times the volume of `H_(3)A` used.
Total volume of solution `= 100 mL (H_(3)A)+ 300 mL (NaOH) = 400 mL`.
`M_(1)V_(1) (Before) = M_(2)V_(2) (After)`
`0.1 xx 100 = M_(2) xx 400`
`M_(2) =(0.1)/(4) = 0.025`
Hence the concentration of salt `(Na_(3)A)` formed at point `C =0.025`.
Using hydrolysis equation for `N_(3)A` (using `K_(a_(3))` value).
`pH = (1)/(2) (pK_(w) + pK_(a_(2)) + log C)`
`= (1)/(2) (14 + 7 + log 25 xx 10^(-3)) = (1)/(2) (21 - 1.6) = 9.7`
`:. pH = 6.7`.