Jaden's Workshop Week 10 - Thermal Equilibrium and Acid/Bases (Key Posted) Topic is endorsed

[Chem_Mod]

Hi everyone,

Thank you for the wonderful two quarters that I've spent with you or one if you didn't have Dr. Lavelle for 14A. I hope I was able to make you laugh a little, study more effectively, and most importantly, understand chemistry more. Some of y'all I will see around campus, some of you I won't. There are too many names in my sections to remember but I will remember faces so please say hi if you see me in passing and officially introduce yourself! My emails will always be open if you have questions about UCLA, how to find resources, etc.

Keep studying hard and good luck with ALL of your finals. Hopefully CHEM 14B is important to y'all, but remember it's one class :).

-Jaden

[Rithwik Narendra 2H]

Hi Jaden!

Will there be an answer key posted after the session? Thanks!

[Student 1H]

I have a question about question 4:

From my understanding, knowing that all the acids have very large Ka and are strong, we only need to calculate the equilibrium constant of acid dissociation Ka = -log(pKa).
But I was just wondering why how the pKa can help rank the pH?

By the way I got:
a. pKa = -9 approx
b. pKa = -3
c. pKa = really negative from complete dissociation
d. pKb = 16, pKa = 14 - pKb = 14 - (16) = -2 approx

Hence my the ranking is c, a, b, d like we got in the workshop.

[Chem_Mod]

I have a question about question 4:
Screenshot 2024-03-14 085842.png

From my understanding, knowing that all the acids have very large Ka and are strong, we only need to calculate the equilibrium constant of acid dissociation Ka = -log(pKa).
But I was just wondering why how the pKa can help rank the pH?

By the way I got:
a. pKa = -9 approx
b. pKa = -3
c. pKa = really negative from complete dissociation
d. pKb = 16, pKa = 14 - pKb = 14 - (16) = -2 approx

Hence my the ranking is c, a, b, d like we got in the workshop.

Sarala that's a fantastic question and unfortunately it's not really a concept that you touch on until 14CL. There are two different ways to solve this problem with the first being the pKa way and the other using Ka values. You will get the exact same answer no matter which way you solve, but the reason why we can use pKa to understand relative acid strength is because it's another "scale" that we can use to see how ionization changes with different pHs. Depending on whether an acid with a pKa of 4 is in a pH solution of less than 4 or greater than 4 will tell us whether the acid deprotonates or is protonated. I hope this answers your question!