A adrakdavra Full Member 10+ Year Member Joined Nov 1, 2012 Messages 179 Reaction score 0 Points 0 Other Health Professions Student May 6, 2013 #1 Advertisement - Members don't see this ad 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?? Last edited: May 7, 2013
Advertisement - Members don't see this ad 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??
A adrakdavra Full Member 10+ Year Member Joined Nov 1, 2012 Messages 179 Reaction score 0 Points 0 Other Health Professions Student May 7, 2013 #2 adrakdavra said: 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?? Click to expand... 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 Upvote 0 Downvote
adrakdavra said: 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?? Click to expand... 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
