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Author: Subject: Weak Acid/Base Titration Questions
smaerd
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[*] posted on 31-3-2011 at 17:22
Weak Acid/Base Titration Questions


I'm having a hard time figure this all out, I've been at it for maybe collectively 10 hours so far. I understand equilibrium of weak acids/bases and buffer solutions no problem at all. I've read the chapter in the text book, watched a few videos online, been to each class, but I still can't manage to solve these problems? Something is missing here, does anyone have any good resources for this?

For example a problem in the textbook says : 500mL of 0.167M NaOH solution is combined with 500mL of 0.1M CH3COOH. What are the concentrations of Na+, OH-, Ch3COO-, H+, and CH3COOH?

So it's obvious to see that the 0.167M NaOH solution is going to consume all of the H+ from the CH3COOH, shifting the equilibrium to completely consume the CH3COOH itself because it is in a greater concentration. So you're left with a concentration of 0.0835M Na+, and 0.05M Ch3COO-. Though I'm not sure what to do, I tried setting up an ice table, using the henderson-hesselbalch(spelling?), and I'm not getting the right answers. How would you go about solving this?
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smaerd
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[*] posted on 31-3-2011 at 18:09


Nevermind I figured it out, I see now...

NaOH = H2O -> Na + OH
CH3COOH <-> CH3COO- + H+
CH3COOH + OH- <-> H+ + CH3COO-

So:
OH+ = 0.0835 - 0.05 = 0.0335M
CH3COO- = 0 + 0.05M = 0.05M
H+ = 10^(14 - (-log([OH-]))) = 2.985*10^(-13)M
Na = 0.0835M(it doesn't do anything here)

Therefore, CH3COOH = X
-log(2.985*10^(-13)M) = -log(1.8*10^(-5)) + log(0.05M/x)
12.52 = 4.744 + log(0.05M/x)
7.775 = log(0.05M/x)
0.05/10^(7.775) = x = 8.39*10^(-10) = [CH3COOH]

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[*] posted on 1-4-2011 at 18:23


Yep, that looks right.
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smaerd
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[*] posted on 3-4-2011 at 11:11


Have a test tomorrow and I'm stuck again on a problem :(.

25ml of HF of unknown M is titrated with 0.2M NaOH. After 20mL of the base solution has been added the ph was 3.00. What was the concentration of the original solution? Ka of HF = 7.1E-4

So here's what I did first...(Henderson Hesselbalch)

3 = -log(7.1E-4) + log(X)
3 = 3.148 + log(X)
-0.148 = X
10^(-0.148) = 0.7112135 = X(X = [conjugate base]/[acid])

So now I also know that the pH is 3.
So, 10^(-3) = 0.001M (H+)
0.001M / 1000mL * 45ml(total soln.) = 4.5E-5 Mols [H+]

And 0.2M NaOH/1000mL * 20mL = 0.002Mols NaOH in 20mL

So then,
4.5E-5mols(H+) after NaOH addition + 0.002Mols OH- = 0.002045mols H+

H+ = F- = 0.002045mols / 45mL * 1000mL = 0.045444M F-

Therefore,
X = [F-]/[HF]
X = 0.7112135 = 0.045444M/[Acid]
So,
0.04544M/0.7112135 = [Acid] = 0.0638M HF

Which is incorrect. The answer is 0.39M. Can anyone lend a hand I can't find any examples in the book like this, or online examples, or videos on youtube explaining this....

[Edited on 3-4-2011 by smaerd]
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smaerd
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[*] posted on 3-4-2011 at 16:54


Nevermind again, I figured out a system to do it! It only took me like... 10 hours hehe. If anyone wants to know it I will post it up here on request otherwise this thread can hopefully go to detritus after tomorrow.
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[*] posted on 3-4-2011 at 16:56


why do you assume [H+]=[F-]?

If you assume that all the OH- from the NaOH will absorb the proton from the HF, you can assume that [F-]=[Na]
so [F-]=0.888888889
using your ratio of [HF]/[F], [HF]=0.21387
since the initial [HF] is equal to to the [HF] at equilibrium + [F-] at equilibrium, initial [HF] at the new volume is 0.045 (truncated)
adjust for the initial volume, you get [HF]=0.3849 ( I assume that if you round differently youll get 0.385)
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smaerd
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[*] posted on 3-4-2011 at 17:11


Thanks for that spirocycle, here's how I figured out how to do them. Which is pretty much what you said.

HA(initial) = Volume * C Mol/1000mL (C being a variable representing mols of acid)
OH-(initial) = Volume * Moles/1000mL

HAeq = HA(initial) - OH-(initial)
A- = OH-(initial) (what your essentially saying here)

pH = -log(Ka) + log( [A-] / [HAeq] )

I ended up with 0.385 which is close enough to 0.39 using this.

[Edited on 4-4-2011 by smaerd]
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entropy51
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[*] posted on 4-4-2011 at 16:48


Quote: Originally posted by smaerd  
And 0.2M NaOH/1000mL * 20mL = 0.002Mols NaOH in 20mL
For starters, you might start checking your arithmetic and learn to keep your units straight.
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