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Topic: Acid/Base equilibria  (Read 11597 times)

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altinure

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Acid/Base equilibria
« on: November 12, 2005, 04:41:25 PM »
I could use help again. Our book is basically designed for high schoolers, I think, so it doesn't cover examples like this, and we did not cover any examples like this in class, so I really have no clue how to do them. I try, but do not come up with the correct pH. I'm starting to understand why this test averages a 33%, with 90% of the class failing it.

I need to calculate the pH from each of these titrations:

4. 20mL of 0.10M NaOH with 20mL of 0.10M HNO2.
5. 20mL of 0.10M HCl with 30mL of 0.10M NH3
7. 20mL of 0.10M NH3 with 20mL of 0.10M NaOH
9. 20mL of 0.10M NaOH with 20mL of 0.10M NaNO2

Ka HNO2 = 4.5x10-4
KbNH3 = 1.8x10-5

If any of you could explain these, I'd really appreciate it.

We basically never covered what to do when adding:
Strong base to weak base
Strong base to basic salt.
Strong acid to weak base when their concentrations differ.

but yet we're still tested on them, so I don't know how our professor expects us to know how to do them.

Thanks.

Offline Borek

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Re:Acid/Base equilibria
« Reply #1 on: November 12, 2005, 06:25:03 PM »
Please check these titration lectures.

Then try to write reaction equation for every case, to find out which ions are present in the solution. Note, that in some cases there can be no reaction at all!  Then you should be able to find out answers.
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altinure

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Re:Acid/Base equilibria
« Reply #2 on: November 12, 2005, 07:00:57 PM »
thanks for the lectures, but they really made no sense at all.  what is Ca?

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Re:Acid/Base equilibria
« Reply #3 on: November 12, 2005, 07:28:18 PM »
Ca is for the concentration (mol/L) of the acid.
My advice is to calculate the moles of acid and base which are involved in each reaction. Then identify what neutralizes what: you will find something (OH- or H+) is in excess.
At that point you will just have to use the henderson and hasselbalch equation.

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Re:Acid/Base equilibria
« Reply #4 on: November 12, 2005, 08:03:26 PM »
thanks for the lectures, but they really made no sense at all.  what is Ca?

Link to symbols used is on the left menu:

http://www.chembuddy.com/?left=pH-calculation&right=symbols

If you can't find the link, you may have troubles finding sense in the lectures  ;)
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altinure

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Re:Acid/Base equilibria
« Reply #5 on: November 12, 2005, 08:26:08 PM »
oh i was assuming that since the lectures increased in difficulty as you went down the column that i didn't need to keep reading down for the symbols.

i figured out 7 and 9 on my own.  i cannot get the correct answer for 4 and 5 for the life of me.  my friend in analytical chemistry can't figure it out either.

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Re:Acid/Base equilibria
« Reply #6 on: November 12, 2005, 08:45:18 PM »
oh i was assuming that since the lectures increased in difficulty as you went down the column that i didn't need to keep reading down for the symbols.

Symbols are explained earlier, once they are introduced. They are also summarized on the page I have posted the link to.

Quote
i figured out 7 and 9 on my own.  i cannot get the correct answer for 4 and 5 for the life of me.  my friend in analytical chemistry can't figure it out either.

4. Endpoint of weak acid dissociation. Explained in the lecture (although you will have to read other lecture as well).
5. No reaction at all. Solution contains two bases - weak and strong. Effect of the weak one can be (probably - it is up to you to check it) ignored.
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Re:Acid/Base equilibria
« Reply #7 on: November 13, 2005, 06:33:48 AM »
I infer you didn't pay any attention to my post. Thank you. >:(

For problem 4: you have a titration with a weak acid and a strong base. You have reached the equivalent point, so:

Kb = [HNO2][OH-] / [NO2-] = [OH-]2 / [NO2-]

[NO2-] = 0.05 M   Kb = 2.2*10-11

pH = 8.02

Now, problem 5. You are doing a titration with a weak base (NH3) and a strong acid (HCl). You haven't reached the equivalent point yet: what you have is a buffer solution.
HCl + NH3 -> NH4+Cl- + NH3 ...because of the NH3 surplus.

After a bit of calculations you get: [NH4+] = 0.04 M and [NH3] = 0.02 M
There is also pKa = 9.26
So: pH = pKa + log [Base] / [Acid] -> pH = 8.95

P.S.: I think this topic should be moved to the high school forum. :coffee1:

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Re:Acid/Base equilibria
« Reply #8 on: November 13, 2005, 06:49:47 AM »
5. No reaction at all. Solution contains two bases - weak and strong. Effect of the weak one can be (probably - it is up to you to check it) ignored.

Sorry, this is answer to question 7, my mistake. Albert gave full answer to the question 5.
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Re:Acid/Base equilibria
« Reply #9 on: November 13, 2005, 06:51:12 AM »
Kb = [HNO2][OH-] / [NO2-] = [OH-]2 / [NO2-]

Have you checked if your assumption ([HNO2]=[OH-]) holds? And have you remembered about dilution?

Quote
P.S.: I think this topic should be moved to the high school forum. :coffee1:

I think it fits General Chemistry, High School Chemistry and Analytical Chemistry as well. If I were to move the thread I will move it to the Analytical Chemistry ;)
« Last Edit: November 13, 2005, 07:00:56 AM by Borek »
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Re:Acid/Base equilibria
« Reply #10 on: November 13, 2005, 07:08:41 AM »
Have you checked if your assumption ([HNO2]=[OH-]) holds? And have you remembered about dilution?
Both checked.

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Re:Acid/Base equilibria
« Reply #11 on: November 13, 2005, 07:42:37 AM »
Both checked.

So what are values for [HNO2 ] and [OH- ]? And what is allowed percent difference between?
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Re:Acid/Base equilibria
« Reply #12 on: November 13, 2005, 08:56:15 AM »
I am not sure I understood what you mean.

Cb / Kb > 100

This gives me the possibility to say that [HNO2] = [OH-] at the equivalent point and [OH] = (Cb * Kb)1/2

[NO2-] = 0.10 M * 20 mL / 40 mL = 0.05 M...because all the NaOH reacts with the acid

However, if you found something wrong in my calculations, would you be so kind as to reveal it?

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Re:Acid/Base equilibria
« Reply #13 on: November 13, 2005, 09:45:15 AM »
Generally speaking your approach is correct and you are right. If you are not interested in pH subtleties don't read any further ;)

You are just out of luck that I have checked the calculations using BATE :)

The problem is, calculated concentration of OH- is not very far from 10-7, thus it is worth checking if autodissociation of water doesn't play its role in the final equilibrium.

Precise calculation (done using BATE, you may read in the lectures cited earlier in the thread how they are performed, Kb=2.2*10-11, Cb=0.05M) shows that

[OH- ] = 1.014*10-6

and

[HNO2 ] = 1.085*10-6

thus the difference is about 7% - slightly more than 5% usually perceived as acceptable. Luckily this difference appears not in a sum or difference, but in the product, thus it doesn't change final result by more than 0.015 unit.
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