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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 pKa.
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Solution
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`
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