Titration Equations
Titration Equations
Hi! In the chemistry review book, titrations are calculated by using N1V1 = N2V2 (1 being acid, 2 being base). However, when I was looking at the Khan Academy video for some extra help, they had used the equation M1V1 = M2V2. Is there any advantage/disadvantage between using either equations? Or would both lead me to the same answer for a question on the MCAT that asks how much of a certain acid/base was needed to neutralize the reaction? Thanks!

 Posts: 469
 Joined: Sat Mar 30, 2019 8:39 pm
Re: Titration Equations
You may want to be a little careful, "N" refers to normality. Technically, the appropriate equation is always and forever N1V1=N2V2 without any exceptions.
But, of course, when dealing with strong monoprotic acids and bases, N = M. That means that the normality is exactly the same as the molarity, since each mole of, for example, HCl dissociates into exactly one mole of H+ and one mole of Cl. In the case of titrations of polyprotic acids and bases, you do want to be very sure to use normality as your measuring stick though  if you wanted to, for example, know how much 1 M NaOH to add to neutralize 1 L of 1M H3PO4 (phosphoric acid), you would definitely want to account for all 3 protons of phosphoric acid. So 3 N x 1 L = 1 N x ? L
And if you solve that, you end up at the answer: 3 liters of 1 M NaOH.
(Real examples are of course vastly more complex, and since H3PO4 is a weak triprotic acid, this 'neutralization' really puts you at the third equivalence point  but everything I said holds up)
But, of course, when dealing with strong monoprotic acids and bases, N = M. That means that the normality is exactly the same as the molarity, since each mole of, for example, HCl dissociates into exactly one mole of H+ and one mole of Cl. In the case of titrations of polyprotic acids and bases, you do want to be very sure to use normality as your measuring stick though  if you wanted to, for example, know how much 1 M NaOH to add to neutralize 1 L of 1M H3PO4 (phosphoric acid), you would definitely want to account for all 3 protons of phosphoric acid. So 3 N x 1 L = 1 N x ? L
And if you solve that, you end up at the answer: 3 liters of 1 M NaOH.
(Real examples are of course vastly more complex, and since H3PO4 is a weak triprotic acid, this 'neutralization' really puts you at the third equivalence point  but everything I said holds up)
Re: Titration Equations
Hi,
I am looking at a titrationbased passage question, #8 on Chemistry Qbank 4.
I don't understand how the protein ligand reaction can be both exothermic and endothermic (in the explanation).
Can someone explain?
Thanks,
S.
I am looking at a titrationbased passage question, #8 on Chemistry Qbank 4.
I don't understand how the protein ligand reaction can be both exothermic and endothermic (in the explanation).
Can someone explain?
Thanks,
S.

 Posts: 469
 Joined: Sat Mar 30, 2019 8:39 pm
Re: Titration Equations
Careful, it is both exergonic (deltaG is negative) and endothermic (deltaH is positive).
And since deltaG = deltaH  TdeltaS, we know that the only way for those two things to be true is that deltaS must be positive. Be sure to pay close attention to the difference in the word exergonic (negative delta G, spontaneous forward reaction under given conditions) and endothermic (positive deltaH, requiring a net input of heat in the system).
And since deltaG = deltaH  TdeltaS, we know that the only way for those two things to be true is that deltaS must be positive. Be sure to pay close attention to the difference in the word exergonic (negative delta G, spontaneous forward reaction under given conditions) and endothermic (positive deltaH, requiring a net input of heat in the system).