Welcome Guest Search | Active Topics |

Tag as favorite
AAMC chemistry question pack Q # 24
Options
Previous Topic Next Topic
Sophia_6535
#1 Posted : Tuesday, May 18, 2021 10:02:49 PM
Rank: Member

Groups: Registered
Joined: 4/25/2021
Posts: 10

Thanks: 0 times
Was thanked: 0 time(s) in 0 post(s)
In the last question of passage 4, I don't understand how they got the ratio 100:1 from
10^2 = [HPO4^-2]/[H2PO4^-]

Can someone please help me out? Thanks!
Back to top
Report?
INSTR_Katrina_128
#2 Posted : Tuesday, May 18, 2021 10:56:24 PM
Rank: Advanced Member

Groups: Registered
Joined: 5/18/2021
Posts: 73

Thanks: 0 times
Was thanked: 0 time(s) in 0 post(s)
Hi Sophia!

Great question.

The ions HPO4^-2 and H2PO4^- differ from each other by ONE H+ ion, so they are acid/base conjugate pairs!
Since H2PO4^- HAS the additional H+, it is the acid, and HPO4^-2 is MISSING the H+, it is the conjugate base. So, [HPO4^-2]/[H2PO4^-] is [base]/[acid].

This question is a perfect example of the Henderson-Hasselbalch equation:

pH = pKa + log([base] / [acid])

In the passage, we are told pH = 8.7 and we are given pKa = 6.7 in the problem statement. We are being asked to solve for [base]/[acid], so I will replace this term with x in the equation to make it easier to write out the equation.

8.7 = 6.7 + log x

Then, solve for x.

log x = 2

To "get out of" a log, we need to do the opposite action, which is "10 to the power of" on both sides.

10^(log x) = 10^2

Therefore, x = 10^2 = 100.

Let me know if this helps. :)
Back to top
Report?
Users browsing this topic
Guest (3)
Tag as favorite
You cannot post new topics in this forum.
You cannot reply to topics in this forum.
You cannot delete your posts in this forum.
You cannot edit your posts in this forum.
You cannot create polls in this forum.
You cannot vote in polls in this forum.

Clean Slate theme by Jaben Cargman (Tiny Gecko)
Powered by YAF | YAF © 2003-2009, Yet Another Forum.NET
This page was generated in 0.072 seconds.