A student is having issues with determining the pH at the equivalence point in a titration of a weak acid with a strong base. How would you explain the process of making this determination?
Answer this question and then write an issue that someone would have with the material in Chapter 16 so that the next student can explain that issue.
The pH of a weak acid solution prior to any added base is at a greater pH then a strong acid would be because it's not 100% ionized. After strong base is added, the reaction forms a conjugate base of the weak acid resulting in a buffer region. The vertical portion of the titration curve goes over a smaller pH change with the equivalence point at a pH greater then 7.
ReplyDeleteI'm actually having issues with the same type of titration problem, I understand it conceptually but the math problems confuse me. The ones that ask pH initially, at half equivalence point, at equivalence point, and beyond equivalence point.
Well I may have missed some points on the last lab summary for that one.
ReplyDeleteFor your issue, I believe it usually relates to diprotic acids or bases that undergo a second ionization (or even a third but we never discussed that) and you'll need to refer to the Henderson-Hasselbach equation.
pH = pKa + log([A-]/[HA])
When you titrate a weak acid/base with a strong base/acid, a conjugate weak base/acid is formed as you stated. Since it is a diprotic acid undergoing it's first ionization the initial H-H equation would be written as following:
pH = pKa1 + log([HA-]/[H2A])
At the 1/2 equivalence point only half as many moles of strong base/acid are added to moles of weak acid/base, leading to the formation of an equimolar amount of the conjugate base/acid. This would cause ([HA-]/[H2A]) = 1 and in turn cause log([HA-]/[H2A]) = 0. As a result, you are left with pH = pKa1.
At the first equivalence point, there are equal amounts of weak acid/base and strong base/acid, which will completely react with each other leaving you with only the amphoteric conjugate base/acid. Since there is no more weak acid/base (H2A), this renders the H-H equation useless and a new equation must be derived from the information that you do have ( Ka1 and Ka2). It's a long process to derive the equation needed but it is listed on the amphoteric salt handout from class.The equation for the first equivalence point comes out to be the following:
pH = 1/2(pKa1 + pKa2)
The second 1/2 equivalence point is a repeat process for the first 1/2 equivalence except the original H-H equation has change since the diprotic acid/base is undergoing it's second ionization.
pH = pKa2 + log ([A2-]/[HA-])
Once again at this point there will be equimolar amount of the weak acid/base and conjugate base/acid leaving you with the following:
pH = pKa2
Honestly, for the second equivalence point and beyond completely went over my head in lecture and that would probably be my issue.
All I can infer from the graphs seen is that after the second equivalence point, the solution just becomes more and more concentrated in what is being added whether acid or base. I guess to me I imagine adding a bucket of a strong base or acid to a drop of weak base or acid.I feel that peers may struggle with when and when not to use the Henderson-Hasselbach equation if anyone could clear this up.
ReplyDeleteThanks for your responses!
ReplyDelete