Chemistry help, if you're up for it :D
Post Reply
 
Thread Rating:
  • 0 Votes - 0 Average
  • 1
  • 2
  • 3
  • 4
  • 5
20-10-2016, 08:40 AM (This post was last modified: 20-10-2016 08:37 PM by Grimm.)
Chemistry help, if you're up for it :D
Hi everyone. I'm sure you're all glad I'm back with another chemistry problem giving me problems. You can see below is the attached question. Here is the work I have for it.

kb = kw/ka = 1 E-14 / 1.8 E-5 = 5.55 E-10

NaC2H3O2 .151 M HC2H3O2 .082 M

C2H3O2- + H2O <--> HC2H3O2 + OH-
I .151 .082 0
C -x +x +x
E .151-x .082+x x

kb = [HC2H3O2][OH-] / [C2H3O2-]

5.55 E-10 = [.082 + x][x] / [.151 - x]

Since the kb is so low we can make assumptions and use the 5% rule.

5.55 E-10 = [.082][x] / [.151]

x = 1.023 E-9 = [OH-]

-log[OH-] = 8.990 = pOH

14 - pOH = 5.010 = pH

Someone please help me. I already have it marked wrong so right now my main concern is just making sure that I know it for the exam. Please let me know where I went wrong here.

Edit: It messed up my nice ICE table... If you can't decipher it let me know.


Attached File(s) Thumbnail(s)
   
Find all posts by this user
Like Post Quote this message in a reply
20-10-2016, 03:56 PM
RE: Chemistry help, if you're up for it :D
Did the question require you to use Kb? If not, it's the long way around given that you're dealing with a buffered acetic acid solution.

Use Ka and check your 5% assumption.

---
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
20-10-2016, 06:40 PM
RE: Chemistry help, if you're up for it :D
Sounds ideal for the Henderson Hasselbach equation. Are equal volumes of acetic acid and sodium acetate solutions used?
Find all posts by this user
Like Post Quote this message in a reply
20-10-2016, 06:58 PM
RE: Chemistry help, if you're up for it :D
The amounts are:

.151 M Sodium Acetate

.082 M Acetic Acid
Find all posts by this user
Like Post Quote this message in a reply
20-10-2016, 08:52 PM
RE: Chemistry help, if you're up for it :D
I attached a screen capture of the the question itself in my OP. Here is what I get using the ka:

HC2H3O2 = .082 M
NaC2H3O2 = .151 M

ka for Acetic Acid is 1.75 E-5

HC2H3O2 + H2O <---> C2H3O2- + H3O+

ka = [C2H3O2-][H3O+] / [HC2H3O2]

1.75 E-5 = (.082)(x) / (.151)

9.503 E-6 = x = [H3O+]

-log[H3O+] = 5.022 = pH[/size]
Find all posts by this user
Like Post Quote this message in a reply
20-10-2016, 09:12 PM
RE: Chemistry help, if you're up for it :D
We do actually have competent people here who can help you with this stuff if they're available, but a fantastic resource for homework help is Yahoo Answers and Chegg. With Chegg, you'll find tons of previously asked questions, or you can just ask your own question. They also have the most of the text books with the answers worked out. I use their tutoring services and get into sessions online with a whiteboard and a mic.

I love Chegg! it's a great product and its saved my sanity many, many times.
Find all posts by this user
Like Post Quote this message in a reply
20-10-2016, 09:28 PM
RE: Chemistry help, if you're up for it :D
(20-10-2016 09:12 PM)Aliza Wrote:  We do actually have competent people here who can help you with this stuff if they're available, but a fantastic resource for homework help is Yahoo Answers and Chegg. With Chegg, you'll find tons of previously asked questions, or you can just ask your own question. They also have the most of the text books with the answers worked out. I use their tutoring services and get into sessions online with a whiteboard and a mic.

I love Chegg! it's a great product and its saved my sanity many, many times.

Chegg makes me pay money. I'm a student I have no money lol

Paleophyte helped me last time, he was great.
Find all posts by this user
Like Post Quote this message in a reply
21-10-2016, 12:21 AM
RE: Chemistry help, if you're up for it :D
(20-10-2016 08:52 PM)Grimm Wrote:  I attached a screen capture of the the question itself in my OP. Here is what I get using the ka:

HC2H3O2 = .082 M
NaC2H3O2 = .151 M

ka for Acetic Acid is 1.75 E-5

HC2H3O2 + H2O <---> C2H3O2- + H3O+

ka = [C2H3O2-][H3O+] / [HC2H3O2]

1.75 E-5 = (.082)(x) / (.151) <---- Tongue

9.503 E-6 = x = [H3O+]

-log[H3O+] = 5.022 = pH[/size]

As Inigo suggests, this problem is most easily solved by the Henderson-Hasselbalch method. The technique that you're using is the long way around but should also yield the correct answer if you stop transposing your acetate and your acetic acid concentrations. Tongue

Both yield a pH of 4.5ish

---
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, 12:35 AM
RE: Chemistry help, if you're up for it :D
(21-10-2016 12:21 AM)Paleophyte Wrote:  
(20-10-2016 08:52 PM)Grimm Wrote:  I attached a screen capture of the the question itself in my OP. Here is what I get using the ka:

HC2H3O2 = .082 M
NaC2H3O2 = .151 M

ka for Acetic Acid is 1.75 E-5

HC2H3O2 + H2O <---> C2H3O2- + H3O+

ka = [C2H3O2-][H3O+] / [HC2H3O2]

1.75 E-5 = (.082)(x) / (.151) <---- Tongue

9.503 E-6 = x = [H3O+]

-log[H3O+] = 5.022 = pH[/size]

As Inigo suggests, this problem is most easily solved by the Henderson-Hasselbalch method. The technique that you're using is the long way around but should also yield the correct answer if you stop transposing your acetate and your acetic acid concentrations. Tongue

Both yield a pH of 4.5ish

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.
Find all posts by this user
Like Post Quote this message in a reply
21-10-2016, 12:40 AM
RE: Chemistry help, if you're up for it :D
(21-10-2016 12:35 AM)Grimm Wrote:  
(21-10-2016 12:21 AM)Paleophyte Wrote:  As Inigo suggests, this problem is most easily solved by the Henderson-Hasselbalch method. The technique that you're using is the long way around but should also yield the correct answer if you stop transposing your acetate and your acetic acid concentrations. Tongue

Both yield a pH of 4.5ish

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?!?

---
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: