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Chemical Equilibria Problem

  • Chemistry
  • Thread starter Fig Neutron
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  • #1
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Homework Statement:

Recalculate the pH after the addition of 0.1 M HCl. Compare this value with that where the same 0.05 ml of 0.1 M HClis added to the equivalent amount of pH 7 water. You may assume no significant change in volume since you are only adding 1 drop to each.

Relevant Equations:

pH = pKa + [salt]/[acid]
pKa = - log (Ka)
This is for a high school chemistry class. In part a of the question, I calculated the pH of the solution to be 3.38. Part a was the question: Calculate the pH of a solution containing 0.75 M lactic acid (Ka= 1.4 *10^-4) and 0.25 M sodium lactate.
For part b I am having trouble determining how to start. I believe I will be adding 5*10^-6 to the 0.75 and 0.25 in the expression: pH = pKa + log (.25/.75) from the previous question. (5*10^-5 L)(.1 M)= 5*10^-6
Please let me know if I have not proved all of the necessary information. Any help is appreciated, and thank you for your time.
 
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Answers and Replies

  • #2
Borek
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3.38 is OK.

What reaction takes place in the solution after you add strong acid?
 
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  • #3
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Hey, @Fig Neutron, I like your HS chem class for working on that not-too-easy problem.
 
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  • #4
epenguin
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You ask whether information sufficient - I'd say it is set out in confusing manner, one has to go back and forth trying to understand and it would be better to quote the original question verbatim.

I cannot quite believe you are asked to calculate the effect of adding an unstated volume of acid to an unstated volume of (buffer) mixture. I have no idea what an "equivalent amount of water" means.

Not realising you need the unstated information might be why you could not get the answer. (The calculation is much the same as for your simpler part a by the way.)
 
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  • #5
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3.38 is OK.

What reaction takes place in the solution after you add strong acid?
Thank you for your help. I’m not exactly sure what the reaction would be. Question b is a bit confusing. The first addition of HCl I believe is into the lactic acid from part a, but I may be wrong. Since the HCl dissociates completely, I believe the pH would be decreasing. And for the second addition of HCl into water, I believe the pH would also be decreasing because of the additional H+ ions.

You ask whether information sufficient - I'd say it is set out in confusing manner, one has to go back and forth trying to understand and it would be better to quote the original question verbatim.

I cannot quite believe you are asked to calculate the effect of adding an unstated volume of acid to an unstated volume of (buffer) mixture. I have no idea what an "equivalent amount of water" means.

Not realising you need the unstated information might be why you could not get the answer. (The calculation is much the same as for your simpler part a by the way.)
Thank you for your help. I’m sorry I made it confusing. The question I put under the “homework statement” is a direct copy of the the problem given to me.
 
  • #6
Borek
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I’m not exactly sure what the reaction would be.
Hint: you have a base (lactate anion) and a strong acid (H+).

I am afraid @epenguin is right that either the wording is incomplete/confusing, or there is some missing information. Once you will realize what the reaction is, what changes in the solution and how it can be used to calculate the pH change, you will see why.
 
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  • #7
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Hint: you have a base (lactate anion) and a strong acid (H+).

I am afraid @epenguin is right that either the wording is incomplete/confusing, or there is some missing information. Once you will realize what the reaction is, what changes in the solution and how it can be used to calculate the pH change, you will see why.
So, it would be a neutralization reaction.

Also, I apologize you are both correct. The question should be: Recalculate the pH after the addition of 0.05 ml of 0.1 M HCl. Compare this value with that where the same 0.05 ml of 0.1 M HCl is added to the equivalent amount of pH 7 water. You may assume no significant change in the volume since you are only adding 1 drop to each.

“0.05 ml of” is missing from the first sentence in my original post. Again, I apologize for the mistake.
 
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  • #8
epenguin
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If the question is incomplete you need to state in an answer what you understand it to be. That way you get credit if it is marked by a human, though not by a computer.

You seem to have made the reasonable guess that you are adding 0.05 ml of 0.1 M HCl to 1 L. That is adding 5×10-3 mmoles HCl to 1L, i.e yes, 5×10-6 moles HCl. You can calculate the effect, but it will not change the pH of a 1 M buffer very much so maybe your 0.05 ml was supposed to be added to 1 ml which you can also calculate. (That is a 5% change in volume, maybe the exegesis is against it).

In all cases the question is, as Borek asks, what does the added acid do? What really is HCl solution? What is your in buffer mixture, if we call lactic acid HL? What things change when you add HCl to your mixture?
 
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  • #9
epenguin
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I had not seen your post when I posted mine at about the same time. Now we know how much of what we add, but do we know what volume of the buffer we add it to?
 
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  • #10
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I am not completely sure, but I believe it is being added to the value from part a. In my first post I guessed that the moles of HCl would be added to those concentrations (.25 M/.75 M) I used in the Henderson-Hasselbalch equation, but I don’t know the volumes. It is moles per liter, but I don’t think I have enough information to find the volume used.
 
  • #11
epenguin
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OK but I see no value in part a. Rereading the question I find even
. Part a was the question: Calculate the pH of a solution containing 0.75 M lactic acid (Ka= 1.4 *10^-4) and 0.25 M sodium lactate.
somewhat objectionable just as English. It is psyching slightly to guess it means a solution that is 0.75 M in lactic acid and 0.25 M in sodium lactate. I don't know whether others at this site, not everyone a native English speaker, agrees with me.

In the light of that I now think that they must be asking in their dialect, calculate the pH after adding 0.1 mole of HCl to 1L of the previous buffer (whose pH was 3.38 OK) = making 0.1 M in Cl-. That would be a reasonable question.

So can you solve it in that case. What will this addition change? It is easiest to think of this in terms of molaritities of the chemical components, molecules and ions.

If this is marked work make your understanding of what the question is clear in your answer, Then you have some comeback if marked down for not answering the question if it was different from that.
 
  • #12
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@epenguin Thank you very much for your help. I appreciate your time. Your clarifications made it much easier to understand and solve the problem. I did manage to get the question right. :smile: And thank you to everyone else for helping.
 
  • #13
epenguin
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Please tell us the answers you got.

Did you choose a major since writing about that a couple of years ago?
 
  • #14
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I calculated the pH with the addition of the HCl to the solution from part a to be 3.38. It is the same answer that I got for part a, but I wasn’t marked off. Perhaps because there wasn’t a significant enough amount of HCl being added to make a noticeable difference in the pH?

Then I calculated the [H+] of water with a pH of 7 and add the additional H+ from the HCl. The new pH I got was 5.02.

I mostly just noted the change in the pH for each as the comparison it asked for, but again that seemed to be enough for credit. Thank you again for your help.

Please tell us the answers you got.

Did you choose a major since writing about that a couple of years ago?
Honestly, I am still a bit uncertain. I have broad interests and it has made it difficult to narrow down my decision. This is my last year dual enrolled in high school and community college. For now I am in the Biotechnology program but still open to other options.
 
  • #15
epenguin
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Sorry, I don't know what "wasn't marked off" means. You don't say what volume you have added your acid to. The effect could well be negligible, but "negligible" is not very discriminating.

For the dilution into water, this training in quantitative science you are getting will maybe have been successful the day when the answer 5.02 rings an alarm that tells you a whole number or near to cannot possibly be the answer with any volumes mentioned so far. :oldsmile:
 
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  • #16
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@epenguin Sorry, by saying “wasn’t marked off” I meant I did not receive any deductions.

Unfortunately, as you noticed, I still do not understand the question as well as I would like. But, current events have made it difficult to communicate with my instructor since all classes are now in an online format.
 
  • #17
epenguin
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It is not a problem that you can't understand an unsolvable, because incomplete, question.

But it is important you be able to recognise information is missing, that is, what information you need in order to solve, find a pH for example.

A shame you cannot contact your prof to say you think you have been given insufficient information. Bad questions and textbook mistakes are unfortunate for students as the standing of Profs and books makes students think themselves inadequate for not understanding them!

That said, one of the most frequent causes of inability to solve problems that we see here is students not using all the information they are given. This happens because they haven't understood the importance of this information and its role in obtaining a solution. And because of this they don't give the information when they come here asking for help! So then a lot of the work of the homework helpers is trying to find out what the question really is! That is the reason for the questions I was asking you insistently above.

Purpose of this exercise (b) is to practice and test your ability to work out the pH changes on adding acid to buffer or to water.

You have not yet demonstrated that ability. Your answer for water appears to be wrong as already mentioned, but then it has the same defect as we started with - that you have not stated the problem that you are solving. Your answer for the buffer was right, but scarcely significant, as mentioned. To show ability to solve these problems I suggest you answer the case of if you add 0.05 ml 0.1 M HCl to 1 ml of the buffer (you get a pH change that is significant even if only just) and to 1 ml water. If you can do that it will stand you in good stead for biotech or other likely studies - where you may come across other students still rather floundering in this area.

Finally I am not happy with the phrase in the question "You may assume no significant change in volume since you are only adding 1 drop to each." The implication most people would read from that would be that the dilution does change the pH, but this change is small enough we shall ignore it. That's not quite right – the point is that the dilution, be it small or large, will make no difference at all to the pH! For if you look at the concentration terms in your Henderson-Hasselbalch equation you will see that if you corrected concentration for the change in volume in the numerator you would also do the same correction in the denominator, so you needn't do it! To say it another way, the ratio of moles is and remains the same as the ratio of molarities on dilution. (The equation is treating the solution as ideal, which it isn't, but that's what we do and where we should leave it.)
 
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