Respuesta :
Explanation:
It is known that in titration of diprotic acid, pH of the half equivalence points represents [tex]pk_{a}'s[/tex] of acid.
Therefore, [tex]pk_{a_{2}}[/tex] = pH of the second half equivalence point = 6.09
Hence, during the titration of a diiprotic acid ([tex]H_{2}A[/tex]) with strong base, the following dissociation take place.
[tex]H_{2}A = H^{+} + HA^{-}[/tex] ( first equivalence point)
[tex]HA^{-} = H^{+} + A^{2-}[/tex] (second equivalence point)
The [tex]pk_{a_{2}}[/tex] of diprotic acid will be calculated as follows.
pH = [tex]pka_{2} + log \frac{[A^{2-}]}{[HA^{-}]}[/tex]
Now, at half equivalence point,
[tex][HA^{-}] = [A^{2-}][/tex]
Therefore,
pH = [tex]pka_{2} + log 1[/tex]
pH = [tex]pka_{2}[/tex]
Thus, we can conclude that [tex]pka_{2}[/tex] for this unknown acid is equal to its pH.
[tex]pK_a_2[/tex] for this unknown acid is equals to the pH of the solution.
It is known that in titration of diprotic acid, pH of the half equivalence points represents pKa of acid.
Hence, during the titration of a diprotic acid ([tex]H_2A[/tex]) with strong base, the following dissociation take place.
[tex]H_2A[/tex] ⇄ [tex]H^++HA^-[/tex] ( first equivalence point)
[tex]HA^-[/tex] ⇄ [tex]H^++A^{2-}[/tex] (second equivalence point)
Calculation for [tex]pK_a_2[/tex] for diprotic acid:
[tex]pH=pK_a_2+log\frac{[A^{2-}]}{[HA^-]}[/tex]
Now, at half equivalence point,
[tex][HA^-]=[A^{2-}][/tex]
Thus,
[tex]pH=pK_a_2+log1\\\\pH=pK_a_2[/tex]
Thus, we can conclude that [tex]pK_a_2[/tex] for this unknown acid is equal to its pH.
Therefore, [tex]pK_a_2[/tex] = pH of the second half equivalence point = 6.09
Find more information about equivalence point here:
brainly.com/question/24584140
