Question

It has been found that the pH of a 0.01M solution of an organic acid is 4.15. Calculate the concentration of the anion, the ionization constant of the acid and its pK_a.

Answer 1

Let the organic acid be HA.

⇒ \[\ce{HA ↔ H+ + A}\]

Concentration of HA = 0.01 M

pH = 4.15

`-log["H"^+] = 4.15`

`["H"^+] = 7.08 xx 10^(-5)`

Now `"K"_"a" = (["H"^+]["A"^(-)])/(["HA"])`

`["H"^+] = ["A"^(-)] = 7.08 xx 10^(-5)`

[HA] = 0.01

Then

`"K"_"a" = ((7.08 xx 10^(-5))(7.08 xx 10^(-5)))/0.01`

`"K"_"a" = 5.01 xx 10^(-7)`

`"pK"_"a" = -log "K"_"a"`

`= - log (5.01 xx  10^(-7))`

`"pK"_"a" = 6.3001`