Chemistry help, if you're up for it :D
Post Reply
 
Thread Rating:
  • 0 Votes - 0 Average
  • 1
  • 2
  • 3
  • 4
  • 5
21-10-2016, 01:01 AM (This post was last modified: 21-10-2016 01:06 AM by Paleophyte.)
RE: Chemistry help, if you're up for it :D
The Henderson-Hasselbalch method is fairly simple:

Start with Ka = [H+][A-]/[HA]

Since we want pH, solve for [H+]

[H+] = Ka[HA]/[A-]

Then take the negative log of both sides to give

pH = pKa - log([HA]/[A-])

and since a negative log is just the log of its inverse we can flip the [HA]/[A-] term to get

pH = pKa + log([A-]/[HA]) <--- Compact, tidy and ICE-free

That's it. Same trick for pKb.

For this question:

pKa = -log(1.75e-5) = 4.75
[A-] = 0.151 M
[HA] = 0.082 M

pH = 4.75 + log(0.151M/0.082M)
pH = 4.75 + 0.265 or so
pH = 5.01 give or take rounding

Quick-checking our work, 4.75 is the pH we'd have if we were simply working with pure acetic acid. We expect the buffer to force some of the H3O+ out of solution and into acetic acid courtesy of the excess acetate, so the +0.265 makes sense.

Aword of caution, Henderson-Hasselbalch runs on the same 5% assumption that we use for all weak acid/base calcs that don't involve quadratics. If that 5% assumption fails then it will cause you grief. And quadratics.

---
Flesh and blood of a dead star, slain in the apocalypse of supernova, resurrected by four billion years of continuous autocatalytic reaction and crowned with the emergent property of sentience in the dream that the universe might one day understand itself.
Find all posts by this user
Like Post Quote this message in a reply
21-10-2016, 01:13 AM
RE: Chemistry help, if you're up for it :D
Solving using the quadratic equation yields a pH of 5.02. That's the most precise and least assumption-prone. In this case it's identical to the other methods except that it take a lot longer.

Answer: 5.02 and 4.98 are within error of one another. Throw the damned computer out a window for causing you this much grief.

---
Flesh and blood of a dead star, slain in the apocalypse of supernova, resurrected by four billion years of continuous autocatalytic reaction and crowned with the emergent property of sentience in the dream that the universe might one day understand itself.
Find all posts by this user
Like Post Quote this message in a reply
[+] 1 user Likes Paleophyte's post
21-10-2016, 01:18 AM
RE: Chemistry help, if you're up for it :D
(21-10-2016 12:40 AM)Paleophyte Wrote:  
(21-10-2016 12:35 AM)Grimm Wrote:  I think I see what went wrong here. On paper I have .151(x) / .082 = 1.75 E-5
That gives me a pH of 5.022

I think when I typed it out I did transpose. Facepalm I do that a lot with regular numbers as well. I'll see a 5465 as a 5645 for a second or two. I really have to focus on every little number.

I don't know how to use the Henderson-Hasselbalch method properly, I tried it and got a pH of 1.3. We haven't gone over it in class yet and it hasn't been in my homework.

Ignore me. Mosty. pH 4.5 is wrong. My bad.

5.02 is right, your computer program is telling you that you have the wrong answer because of rounding differences. Depending on the Kb/Ka/H-H route you take you will get very slightly different answers. They're al correct within error but the idiot machine doesn't understand that.

WTF would anybody ever use Kb on an acid acetate buffer?!?

That's what I was leaning towards but I wanted to make extra sure it wasn't just me being stupid. I guess they want me to be able to use multiple routes, idk. I just followed the instructions lol. There are a few problems that I'm 95% sure I calculated correctly but I probably didn't enter properly. If I want the points I have to go to my prof and demonstrate each one of the problems that the software wouldn't accept. Shit, I feel like my college should be paying ME at this point.

Thank you everyone for your help. BowingBowingBowing
I'm very determined. I'd have to be to be up at 3 AM doing chemistry right? Consider Or maybe just a masochist....
Find all posts by this user
Like Post Quote this message in a reply
[+] 1 user Likes Grimm's post
21-10-2016, 01:32 AM
RE: Chemistry help, if you're up for it :D
(21-10-2016 01:01 AM)Paleophyte Wrote:  The Henderson-Hasselbalch method is fairly simple:

Start with Ka = [H+][A-]/[HA]

Since we want pH, solve for [H+]

[H+] = Ka[HA]/[A-]

Then take the negative log of both sides to give

pH = pKa - log([HA]/[A-])

and since a negative log is just the log of its inverse we can flip the [HA]/[A-] term to get

pH = pKa + log([A-]/[HA]) <--- Compact, tidy and ICE-free

That's it. Same trick for pKb.

For this question:

pKa = -log(1.75e-5) = 4.75
[A-] = 0.151 M
[HA] = 0.082 M

pH = 4.75 + log(0.151M/0.082M)
pH = 4.75 + 0.265 or so
pH = 5.01 give or take rounding

Quick-checking our work, 4.75 is the pH we'd have if we were simply working with pure acetic acid. We expect the buffer to force some of the H3O+ out of solution and into acetic acid courtesy of the excess acetate, so the +0.265 makes sense.

Aword of caution, Henderson-Hasselbalch runs on the same 5% assumption that we use for all weak acid/base calcs that don't involve quadratics. If that 5% assumption fails then it will cause you grief. And quadratics.

Oh I like this. Thumbsup I see why everyone has suggested this to me. I get why it's a + log and not - log now and why the anion is in the numerator. I've used
[H+] = Ka[HA]\[A-] several times. This is a nice tie-in to that.
Find all posts by this user
Like Post Quote this message in a reply
21-10-2016, 06:30 PM
RE: Chemistry help, if you're up for it :D
(21-10-2016 01:18 AM)Grimm Wrote:  I'm very determined. I'd have to be to be up at 3 AM doing chemistry right? Consider Or maybe just a masochist....

Why not both? Big Grin

---
Flesh and blood of a dead star, slain in the apocalypse of supernova, resurrected by four billion years of continuous autocatalytic reaction and crowned with the emergent property of sentience in the dream that the universe might one day understand itself.
Find all posts by this user
Like Post Quote this message in a reply
Post Reply
Forum Jump: