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MoeDent44

Titration calculation help, don't know what to do next

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Hey!

So the task is: you have 25mL 0,10 mol/L CH3CH2COOH and 0,20 mol/L 12,5mL NaOH. What is the pH at the equivalence point.

So I did as following:

nCH3CH2COOH = c * V = 0,10 mol/L * 0,0250 L = 2,5 * 10^-3 mol

nNaOH = c * V = 0,20 mol/l * 0,0125L = 2,5*10^-3 mol

I don't know what to next, since this is a weak acid - strong base titration. If it were strong acid strong base and you had the same amount of mol, you'd get a pH of 7. I don't know what to do here.

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What's the pka of propionic acid? You must know how weak it is, right?

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25 minutes ago, joigus said:

What's the pka of propionic acid? You must know how weak it is, right?

The Ka of C2H5COOH is 1,3*10^-5 mol/L. pKa is 3,8. How do I interpret this in the task? Is this my final answer?

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OK. I'm a bit hazy on this right now. But I think the key words you're looking for are Henderson-Hasselbach. Try to work it out, and if I'm not helping, maybe someone can provide better help. You must picture the tritation curve in your mind. It's what biologists call a sigmoid curve.

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2 minutes ago, joigus said:

OK. I'm a bit hazy on this right now. But I think the key words you're looking for are Henderson-Hasselbach. Try to work it out, and if I'm not helping, maybe someone can provide better help. You must picture the tritation curve in your mind. It's what biologists call a sigmoid curve.

Thank you. The thing is that I've never encountered a task like this. I would know how to solve it if it were a strong base - strong acid titration, but this is a strong base - weak acid titration. I have the same amount of each solutions and I need to find the pH in the equivalence point. 

13 minutes ago, joigus said:

OK. I'm a bit hazy on this right now. But I think the key words you're looking for are Henderson-Hasselbach. Try to work it out, and if I'm not helping, maybe someone can provide better help. You must picture the tritation curve in your mind. It's what biologists call a sigmoid curve.

I also only have the concentration and the volume of C2H5COOH and the same of NaOH, which makes it difficult to use the Henderson Hasselbalch equation. 

 

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27 minutes ago, MoeDent44 said:

which makes it difficult to use the Henderson Hasselbalch equation. 

Why? Something wrong with the logarithm?

I may be missing something. Please, wait for the experts, I'm not one. In the meantime, maybe we both get lucky and you can find the key to your problem with my help, by discussing the basic concepts. If not, I'm sorry if I lead you into more confusion. I'm answering this because nobody else is reacting to your post so far and you seem to be in a bit of a hurry.

Good luck!

26 minutes ago, joigus said:

tritation

I meant "titration."

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22 minutes ago, joigus said:

Why? Something wrong with the logarithm?

I may be missing something. Please, wait for the experts, I'm not one. In the meantime, maybe we both get lucky and you can find the key to your problem with my help, by discussing the basic concepts. If not, I'm sorry if I lead you into more confusion. I'm answering this because nobody else is reacting to your post so far and you seem to be in a bit of a hurry.

Good luck!

I meant "titration."

Your help has necessary for me to understand more of the consept, but I'm still stuck, but thank you.

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Posted (edited)
1 hour ago, joigus said:
1 hour ago, MoeDent44 said:

which makes it difficult to use the Henderson Hasselbalch equation. 

Why? Something wrong with the logarithm?

I may be getting this totally wrong, but what's the problem with [acid]=[base] and Henderson-Hasselbach?

Most people are concerned with the murder of George Floyd now, which is only fair. So it seems that you're stuck with me, the chemical idiot of the lot. :-) :-) 

So, what's wrong with it?

Edited by joigus
added smiles

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So I did this:

nC2H5COOH = c * V = 0,10 mol/L * 0,025 L = 2,5 * 10^-3 mol

nNaOH = c * V = 0,20 mol/L * 0,0125 L = 2,5 * 10 ^-3 mol

When these two components react, we get:

C2H5COOH + NaOH = C2H5COONa + H2O

C2H5COONa = C2H5COO- + Na+

C2H5COO- + H20 = C2H5COOH + OH-

Kb= 1,0*10^-14 (mol/l)^2 / Ka = 7,7*10^-10 mol/L

[C2H5COOH] =c/Vtotal = 2,5*10^-3 mol / (0,0125 + 0,025) L = 0,0667 mol/L

x^2=7,7*10^-10 * 0,0667 

x^2 = 5,14 * 10^-11

x = 7,2 * 10 ^-6 mol/L OH-

pOH = -log[OH-] = -log(7,2*10^-6) = 5,145

pH=14 - pOH = 14 - 5,145 = 8,86

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OK. It took me a while to realize what you were trying to do (plus I've been busy.) I was missing some words and/or some data in the statement of your problem. I should have noticed that there must be a reason that you were given the volumes. For some reason I thought you were given the concentrations at equilibrium.

It was a neutralization of two solutions with hydrolysis of the salt with initial concentrations and volumes specified, and you were given the k_a of the weak acid, right? I didn't know that.

It seems you've solved it. I haven't checked the numbers in the quadratic eq. for the moles in equilibrium, but the method is correct. Plus it should give you a pH slightly above 7, so everything checks AFAICT.

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13 hours ago, joigus said:

OK. It took me a while to realize what you were trying to do (plus I've been busy.) I was missing some words and/or some data in the statement of your problem. I should have noticed that there must be a reason that you were given the volumes. For some reason I thought you were given the concentrations at equilibrium.

It was a neutralization of two solutions with hydrolysis of the salt with initial concentrations and volumes specified, and you were given the k_a of the weak acid, right? I didn't know that.

It seems you've solved it. I haven't checked the numbers in the quadratic eq. for the moles in equilibrium, but the method is correct. Plus it should give you a pH slightly above 7, so everything checks AFAICT.

My teacher said that my answer was correct. Thank you for helping out, you made me understand what the pKa and pKb does. 

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