- Nov 12, 2008
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I am weak with genchem (I simply don't look at it much) and I am sooooo busy with school myself right now. If I find free time I will look into this, but I can't make any garuntees.
F**k Buffer Solutions
- Thread starter bickelz
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[MENTION=387]IndigoSensor[/MENTION] don't bother with it. I wasn't actually asking for help. Unless you have to do a bunch of calculations to prepare weird ass buffer solutions by titrating acetic acid and NaOH when you TA, you should concentrate on more important things.
Actually, you probably wouldn't use NaOH all that much because it's an inorganic strong base. I'd imagine you would use weak acids and bases that are organic such as ammonia.
- Nov 12, 2008
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Actually, I use NaOH a lot. More so KOH, at very high concentrations.
You woudn't think it, but ochem uses some pretty nasty chemicals. Muuuuuch more so then genchem. Hell I'll use bases that are 10x stronger then sat. NaOH all the time! :tongue1:
You woudn't think it, but ochem uses some pretty nasty chemicals. Muuuuuch more so then genchem. Hell I'll use bases that are 10x stronger then sat. NaOH all the time! :tongue1:
I am kind of rusty with chem as I haven't had it for 2 years, but here is my attempt:
The HH Equation:
4.65 = 4.75 + log [A-/HA]
Solving for the ratio of Base/Acid I got:
0.7943
So from here, I did the equivalance point:
0.471 L of 0.0791 M of NaOH --> Gives 0.0372561 mol of NaOH
It's a 1:1:1:1 ratio. This means 0.0372561 mol of Acetate ions are formed. And 0.0372561 mol of Acetic acid were required.
Set a proportion equal to the [Base/Acid] so that you can solve for how much acid you require:
0.7943 = [Acetate/Acetic Acid]
0.7943 = 0.0372561/Acetic Acid
Solving for Acetic Acid gets:
0.0469 mol of Acetic Acid
So in the end, you used a total of:
0.0372561 mol + 0.0469 mol of acetic acid.
If you convert this to mL using molecular weight and density:
4.82 mL is the total I got.
I feel like that is way too little though... Idk. I wouldn't trust me. lol.
I just did this for the hell of it. I am too lazy to try to do anything involving more math at the moment. lol.
The HH Equation:
4.65 = 4.75 + log [A-/HA]
Solving for the ratio of Base/Acid I got:
0.7943
So from here, I did the equivalance point:
0.471 L of 0.0791 M of NaOH --> Gives 0.0372561 mol of NaOH
It's a 1:1:1:1 ratio. This means 0.0372561 mol of Acetate ions are formed. And 0.0372561 mol of Acetic acid were required.
Set a proportion equal to the [Base/Acid] so that you can solve for how much acid you require:
0.7943 = [Acetate/Acetic Acid]
0.7943 = 0.0372561/Acetic Acid
Solving for Acetic Acid gets:
0.0469 mol of Acetic Acid
So in the end, you used a total of:
0.0372561 mol + 0.0469 mol of acetic acid.
If you convert this to mL using molecular weight and density:
4.82 mL is the total I got.
I feel like that is way too little though... Idk. I wouldn't trust me. lol.
I just did this for the hell of it. I am too lazy to try to do anything involving more math at the moment. lol.
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That looks like it could be it [MENTION=2670]Phage[/MENTION]. The problem I have with this is that I don't think it takes into account what the produced acetate ion will do. Idk though, maybe it does. I had this thought process I guess but I was trying to do it with ice tables instead of HH because I didn't know the HH equation. You should still be able to do it with ice tables though because HH is derived from them and I was struggling with finding an answer through ice tables. Damn, I was pretty close though. I just didn't know how to fit it all together.
LOL. Not me! Remember, I took Chem for Non Technical Majors--we learned how come conditioner works on hair and other crap like that. About the only thing I really remember is that explosions are caused when the molecules recombine, not when they separate.... f**k science! (Sonyab does not like science!!!!) Now Econ, Sociology, History and such....
I tried out the ice chart for the dissociation of acetate ions and the amount of acetic acid form was of the 10^-6 degrees. I thought I could just get away with approximating that this dissociation was negligible. But I couild have done it wrong. hehe.
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I got the test back today. 55/100 but the median was ~50 and he said the mean was "<50" and the high score was 75. There were a ton of 35's so I actually did good.
So I think I know how to do that problem.
1.Acetic Acid + NaOH --> Water + Sodium Acetate
Since the M and volume of NaOH was given, we can calculate for the Sodium Acetate produced when you titrate NaOH to the equivalence point.
2. C2H3O2- + H20 <--> Acetic Acid + OH-
Now the sodium acetate dissolved into acetate ion (the conjugate base of acetic acid). We know the pKa, target pH and moles of Acetate ion so we can use the HH equation...
pH = pKa + log([A-]/[HA])
using the moles of A- (acetate), the given pKa and target pH, we can figure out the moles of HA (plug and chug). We're given the density of HA so we can go from moles to grams and then grams to mL!
So simple.
So I think I know how to do that problem.
1.Acetic Acid + NaOH --> Water + Sodium Acetate
Since the M and volume of NaOH was given, we can calculate for the Sodium Acetate produced when you titrate NaOH to the equivalence point.
2. C2H3O2- + H20 <--> Acetic Acid + OH-
Now the sodium acetate dissolved into acetate ion (the conjugate base of acetic acid). We know the pKa, target pH and moles of Acetate ion so we can use the HH equation...
pH = pKa + log([A-]/[HA])
using the moles of A- (acetate), the given pKa and target pH, we can figure out the moles of HA (plug and chug). We're given the density of HA so we can go from moles to grams and then grams to mL!
So simple.