1.

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 will be the change in `pH` from point `B` to point `C?`A. `2.8`B. `3.2`C. `4.6`D. `0.94`

Answer» Correct Answer - D
At point `B,Na_(2)HA` is formed, hydrolysis of `HA^(2-)` is considered and hydrolysis of `NaH_(2)A `is supressed `(K_(a_(2))` is considered) due to common ion `[overset(Theta)OH]` effect.
`HA^(2-) + H_(2)O hArr H_(2)A^(Theta) + overset(Theta)OH`
`H_(2)A^(Theta) + H_(2)O hArr H_(3)A + overset(Theta)OH`
At point `B`, salt `Na_(2)HA` (salt of `W_(A)//S_(B))` is formed. So the volume of `NaOH` used at point `B` is `2` times the volume of `H_(3)A` used.
Total volume of solution
`= (100 mL + 200 mL) = 300 mL`
`M_(1)V_(1) (Beofre) = M_(2)V_(2)(After)`
`0.1 xx 100 = M_(2) xx 300`
`M_(2) = (0.1)/(3) = (1)/(30) = 0.033`
`:. [Na_(2)HA] = 0.033 M`
`:. pH = (1)/(2) (pK_(w) + pK_(a_(3)) + log C)`
`= (1)/(2) (14 + 5+ "log"(1)/(30)) = (1)/(2) (19 - 1.48) = 8.76`
change is `pH = 9.7 - 8.76 = 0.94`


Discussion

No Comment Found

Related InterviewSolutions