Hi UltimaOnline,
The previous thread on equilibria question has been locked due to inactivity (!) so I am creating a new thread here! I have a question on ionic equilibria from RI Prelim 2013.
Q40:
Ans: A
Remarks: For 1, how do I establish the PH of the second equivalence point is 6.84? Secondly, during the first equivalence point, is it a COOH that is deprotonated or the H3N+ that is deprotonated first?
Thank you! :)
Originally posted by gohby:Hi UltimaOnline,
The previous thread on equilibria question has been locked due to inactivity (!) so I am creating a new thread here! I have a question on ionic equilibria from RI Prelim 2013.
Q40:
Ans: A
Remarks: For 1, how do I establish the PH of the second equivalence point is 6.84? Secondly, during the first equivalence point, is it a COOH that is deprotonated or the H3N+ that is deprotonated first?
Thank you! :)
Hi UltimaOnline,
For this question, I am more interested in the mechanics of the experiment rather than the answer.
If I had understood this procedure correctly, it is the part where “saturating the solution with hydrogen sulphide” that is providing the S2- ions which is used to precipitate ZnS, followed by MnS.
A: Now, what is the purpose of passing HCl into the solution prior to the saturation of H2S? My thinking is that it suppresses the dissociation of H2S (by LCP), thereby ensuring that only a very minute of S2- is present in the solution when it is added to prevent both precipitates from forming at the same time, given their very small Ksp values?
B: Next, how is the pH raised during the experiment to allow for the maximum effective precipitation (just right before MnS starts to precipitate)? My understanding for such reactions is that OH- ions will be supplied to decrease [H+] but this is not stated in the question. Is my understanding correct?
C: As I add more H2S into the solution, does the pH increase? I think it does because, the dissociation of H+ ions from H2S is too insignificant to counteract the decrease of [H+] from the dilution, hence resulting to a lower [H+] and thereby a higher pH.
Originally posted by gohby:
Hi UltimaOnline,
For this question, I am more interested in the mechanics of the experiment rather than the answer.
If I had understood this procedure correctly, it is the part where â€œsaturating the solution with hydrogen sulphideâ€� that is providing the S2- ions which is used to precipitate ZnS, followed by MnS.
A: Now, what is the purpose of passing HCl into the solution prior to the saturation of H2S? My thinking is that it suppresses the dissociation of H2S (by LCP), thereby ensuring that only a very minute of S2- is present in the solution when it is added to prevent both precipitates from forming at the same time, given their very small Ksp values?
B: Next, how is the pH raised during the experiment to allow for the maximum effective precipitation (just right before MnS starts to precipitate)? My understanding for such reactions is that OH- ions will be supplied to decrease [H+] but this is not stated in the question. Is my understanding correct?
C: As I add more H2S into the solution, does the pH increase? I think it does because, the dissociation of H+ ions from H2S is too insignificant to counteract the decrease of [H+] from the dilution, hence resulting to a lower [H+] and thereby a higher pH.
Happy New Year, UltimaOnline! :)
Why does the pH decrease with the addition of H2S? Wouldn’t the mixing of a strong acid (HCl) with a weak acid (H2S) result to a lower [H+], thereby increasing the pH of the solution?
Now if the pH does decrease with the addition of H2S, and given that we need to increase the pH of the solution to point when MnS starts to precipitate by adding OH-, procedurally how does the experiment work – I assume that I add the H2S and OH- simultaneously? The question doesn’t offer a very clear picture tbh.
(P.S: I am deviating so much from the original question and I feel that I am being too scholastic… but I think a sound understanding of how this experiment works will aid one in arriving at the solution..)
Hi UltimaOnline,
In a mixture of 0.1M of ethanoic acid (pKa 4.8 ) and 0.2M of bromic acid (pKa 8.7), the initial pH is 2.90. Why is it that the initial pH of the mixture is solely based on the [H+] from the ethanoic acid, and not from the bromic acid? Wouldn't the different [H+] from the bromic acid affect the overall [H+] in the mixture? (The question does not state the proportions of the acids in the mixture.)
Next, the 2 acids are titrated with NaOH. In drawing the titration curve, I understand that the base will always react with the stronger acid first, then the weaker acid after the neutralisation with the stronger acid. However, what is stopping the base from reacting with the weaker acid at the same time when it was reacting with the stronger acid?
Thank you! :)
Originally posted by gohby:Hi UltimaOnline,
In a mixture of 0.1M of ethanoic acid (pKa 4.8 ) and 0.2M of bromic acid (pKa 8.7), the initial pH is 2.90. Why is it that the initial pH of the mixture is solely based on the [H+] from the ethanoic acid, and not from the bromic acid? Wouldn't the different [H+] from the bromic acid affect the overall [H+] in the mixture? (The question does not state the proportions of the acids in the mixture.)
Next, the 2 acids are titrated with NaOH. In drawing the titration curve, I understand that the base will always react with the stronger acid first, then the weaker acid after the neutralisation with the stronger acid. However, what is stopping the base from reacting with the weaker acid at the same time when it was reacting with the stronger acid?
Thank you! :)
Q1
As a follow up to the previous question, the question was correct as I omitted the (I) in the bromic (I) acid.
The answer indicates the precise volumes of sodium hydroxide required for during the 2 equivalence points as 10cm³ and 30cm³. However, wouldn’t that mean we have to assume that the mixture contains an equal volume of ethanoic acid and bromic (I) acid? Is this a reasonable assumption to make in light of the information provided?
Q2
For part (ii) my pH at equivalence point is 8.61 but the answer suggests 8.58. Did I overlook something? As for (iii), I am only able to get a pH of 11.946 (5sf).
I carried out my calculations using the exact values from the calculator. Alternatively, I would write 5sf for my intermediate workings and round it up to 3sf for the final answer. Both of them do not yield a pH of 12.0. If my calculations are correct, should I reduce the number of significant figures just to “confirm” the value of 12.0?
Q3
(Ka of acid = 5.9x10^-4M)
I reckon this is a blind spot on my part, but using the formula Ka ≈ [H+]²/[acid] yields an answer of 0.169M instead of 0.179M as suggested by the answer. As for (ii) the answer suggests 0.0937M - I drew the ICE table but was unable to obtain 0.0937M exact - similar to the problem above. Is there something wrong with my calculation method and my degree of accuracy?
Thank you once again, UltimaOnline :)
Originally posted by gohby:Q1
As a follow up to the previous question, the question was correct as I omitted the (I) in the bromic (I) acid.
The answer indicates the precise volumes of sodium hydroxide required for during the 2 equivalence points as 10cmÂ³ and 30cmÂ³. However, wouldnâ€™t that mean we have to assume that the mixture contains an equal volume of ethanoic acid and bromic (I) acid? Is this a reasonable assumption to make in light of the information provided?
Q2
For part (ii) my pH at equivalence point is 8.61 but the answer suggests 8.58. Did I overlook something? As for (iii), I am only able to get a pH of 11.946 (5sf).
I carried out my calculations using the exact values from the calculator. Alternatively, I would write 5sf for my intermediate workings and round it up to 3sf for the final answer. Both of them do not yield a pH of 12.0. If my calculations are correct, should I reduce the number of significant figures just to â€œconfirmâ€� the value of 12.0?
Q3
(Ka of acid = 5.9x10^-4M)
I reckon this is a blind spot on my part, but using the formula Ka â‰ˆ [H+]Â²/[acid] yields an answer of 0.169M instead of 0.179M as suggested by the answer. As for (ii) the answer suggests 0.0937M - I drew the ICE table but was unable to obtain 0.0937M exact - similar to the problem above. Is there something wrong with my calculation method and my degree of accuracy?
Thank you once again, UltimaOnline :)
Hi UltimaOnline,
Many thanks for your reply.
Q2 (YJC Prelims 04 P3): Urgh my bad. The Ka of ethanoic acid is given as 1.8x10^-5M. It was embedded in part (a) of the same question which was unrelated to part (b), save for this Ka value, so I extracted part (b) in order not to clutter my question, but I forgot to include the Ka value from embedded in part (a).
I think these are standard questions; normally I am not very concerned about slight deviations with the answers. However this question specifically required the confirmation that the pH is 12.0, and I could only get 11.946 (and 8.61 for part (ii)).
For Q3, this is a HCJC Prelims 02/P3 question. My workings are as follows:
Given that the initial pH of the acid is 2.0, so [H+] = 10^-2. I used the formula Ka ≈ [H+]²/[initial acid], i.e. 5.9x10^-4 = [10^-2]²/ [initial acid]. So I found [initial acid] to be 0.16949M (5sf).
My detailed workings for part (b) are as follows:
Kb = 10^-14/5.9x10^-4 ≈ [OH-]²/[sodium benzenesulfonate]
Since pH at equivalence point is 8.1, [OH-]² = 1.5849x10^-12. Hence [sodium benzenesulfonate] = 0.093509M. However the answer provided was 0.0937M.
Originally posted by gohby:Hi UltimaOnline,
Many thanks for your reply.
Q2 (YJC Prelims 04 P3): Urgh my bad. The Ka of ethanoic acid is given as 1.8x10^-5M. It was embedded in part (a) of the same question which was unrelated to part (b), save for this Ka value, so I extracted part (b) in order not to clutter my question, but I forgot to include the Ka value from embedded in part (a).
I think these are standard questions; normally I am not very concerned about slight deviations with the answers. However this question specifically required the confirmation that the pH is 12.0, and I could only get 11.946 (and 8.61 for part (ii)).
For Q3, this is a HCJC Prelims 02/P3 question. My workings are as follows:
Given that the initial pH of the acid is 2.0, so [H+] = 10^-2. I used the formula Ka â‰ˆ [H+]Â²/[initial acid], i.e. 5.9x10^-4 = [10^-2]Â²/ [initial acid]. So I found [initial acid] to be 0.16949M (5sf).
My detailed workings for part (b) are as follows:
Kb = 10^-14/5.9x10^-4 â‰ˆ [OH-]Â²/[sodium benzenesulfonate]
Since pH at equivalence point is 8.1, [OH-]Â² = 1.5849x10^-12. Hence [sodium benzenesulfonate] = 0.093509M. However the answer provided was 0.0937M.
Originally posted by UltimaOnline:
Ah ok Gohby, no worries then, your working is fine. It's likely that the question setter used a slightly different (eg. 3sf instead of the given 2sf) value for Ka, in his/her own calculations. Another possibility could be related to my next point.
What I've pointed out earlier, still holds. Benzenesulfonic acid should have a much larger Ka (and Cambridge can give a few Ka values of other acids and ask the student to deduce, with reasons, which is the correct Ka of benzenesulfonic acid). And if the initial [H+] is so high, the approximation is no longer valid. In other words, you must use the formula Ka = ([H+] x [A-]) / (initial [HA] - [H+]).
Perhaps this is what the question setter did. You can try it out to see if the answer is closer to the question setter's answer (and explain to your students why this is a more correct working ; ie. because the proton dissociation is no longer negligible compared to the initial molarity of the acid, hence the approximation is no longer mathematically valid to 3 sf).
No prob! ;)
Hi UltimaOnline,
Your explanation was spot on! For 3(i) when I calculate the pH without the approximation I got the answer to be 0.179M. :)
However, how do we ascertain if the [H+] is sufficiently high for which the approximation would not be valid? Separately, when we calculate the pH at the equivalence point are there circumstances where we cannot approximate the [salt] as the intial concentration of the salt as the dissociation is deemed to be significant?
Originally posted by gohby:Hi UltimaOnline,
Your explanation was spot on! For 3(i) when I calculate the pH without the approximation I got the answer to be 0.179M. :)
However, how do we ascertain if the [H+] is sufficiently high for which the approximation would not be valid? Separately, when we calculate the pH at the equivalence point are there circumstances where we cannot approximate the [salt] as the intial concentration of the salt as the dissociation is deemed to be significant?
Originally posted by gohby:Q3
(Ka of acid = 5.9x10^-4M)
I reckon this is a blind spot on my part, but using the formula Ka ≈ [H+]²/[acid] yields an answer of 0.169M instead of 0.179M as suggested by the answer. As for (ii) the answer suggests 0.0937M - I drew the ICE table but was unable to obtain 0.0937M exact - similar to the problem above. Is there something wrong with my calculation method and my degree of accuracy?
Thank you once again, UltimaOnline :)
Hi UltimaOnline,
Apologies for digging out this post but I realised some of the discrepancies with sf arose because I did the approximation method to negate the dissociation of the acid when I shouldn't have done so. Hence I am relooking at other questions where there are discrepancies wrt the accuracy of the answers.
For 3(ii) (Ka of acid = 5.9x10^-4M), here are my workings but I think I have missed out on an aspect which would explain the disparity between my answer and the suggested answer of 0.0937M. Could you see if my workings/assumptions are in order?
For 4a would it be wrong if I suggest litmus as a suitable indicator? The answer suggested methyl orange but I doubt the working range of methyl orange coincides with the rapid pH change over the equivalence point.
As for 4b, my workings are as follows:
The answer suggested that the pH is 9.24 - is there something wrong with my workings?
Thank you :)
Originally posted by gohby:Hi UltimaOnline,
Apologies for digging out this post but I realised some of the discrepancies with sf arose because I did the approximation method to negate the dissociation of the acid when I shouldn't have done so. Hence I am relooking at other questions where there are discrepancies wrt the accuracy of the answers.
For 3(ii) (Ka of acid = 5.9x10^-4M), here are my workings but I think I have missed out on an aspect which would explain the disparity between my answer and the suggested answer of 0.0937M. Could you see if my workings/assumptions are in order?
For 4a would it be wrong if I suggest litmus as a suitable indicator? The answer suggested methyl orange but I doubt the working range of methyl orange coincides with the rapid pH change over the equivalence point.
As for 4b, my workings are as follows:
The answer suggested that the pH is 9.24 - is there something wrong with my workings?
Thank you :)