8 Titration of a Diprotic Acid

[@phoenixhollenbeck3318]

This video was super helpful and you are an excellent instructor! Thank you for all that you do!

[@agentsmith7947]

Thank you

[@aKadoki]

Amazing video, just one thing: in 5:33 is that the volume and concentration of the acid or the base used in the titration. Because in that example theyre both the same number and i cant tell. Thanks in advance!

[@lesliehiatt4665]

Good question!! I didn't even realize they were the same number. 😊 in this example, the initial pH is based only on the concentration of acid present at the beginning. At the beginning of the titration, no base has yet been added! Hope this helps!

[@aKadoki]

@@lesliehiatt4665 ive replayed the video a bunch of times bcs our analytics prof makes tests more or less just like what you explained just formulated differently. And after searching youtube for two weeks i can say that your video is extremely helpful, but the title does it injustice. Yes you are showing how the titration would go, but maybe it would get the video more views if you switched the title to something like: calculations of titrations: equivalence points, starting ph, etc, etc. Like i accidentally clicked on the video and it turned out amazing, so just a tip on how to help more students like me, is maybe changind the title and maybe adding points in the vid like at "13:25 second buffer region calculations". Also sorry its 10days late but my tests tmrrw and i came back here to rewatch it.

[@lesliehiatt4665]

Those are really helpful suggestions! Once I have a few free hours I will go in and add those time points and change some titles for all my videos. Great ideas! Thanks!@@aKadoki

[@juludyehya4209]

Why you didn’t divide the concentrations in the first buffer part with the new total volume?

[@lesliehiatt4665]

You absolutely can do this, but since the volume in the numerator is the same as the volume in the denominator, you don't have to do so. The volumes will cancel each other out!

[@the_crackin3610]

18:56 0.1 x 0.1 = 0.01