Aamc 10 q 27

[adrakdavra]

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27. Which of the following expressions gives an approximate value of the equilibrium constant (Ka) of the acid in Titration 5?

A) log([H3O+]2 ´ [HOAc])
B) [H3O+]2 ´ [HOAc]
C) log([H3O+]
D) * (H3O+)2 / (HoAC)

Why D is the right answer??

 

[adrakdavra]

27. Which of the following expressions gives an approximate value of the equilibrium constant (Ka) of the acid in Titration 5?

A) log([H3O+]2 ´ [HOAc])
B) [H3O+]2 ´ [HOAc]
C) log([H3O+]
D) * (H3O+)2 / (HoAC)

Why D is the right answer??

They said that Ka = product / reactants
and that product is H+ and HoAC ( acetic acid) and because it is a buffer HoAC = H+
Ka = X2 / X

They put the log at the answer choices to make you think of PH = Pka + log A-/AH