H2 chemistry buffer solutions
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This a pretty good practice question for 'A' level H2/H3/H1 students. It's not overly difficult, complicated or tricky; but it does a good job of testing the student's conceptual understanding of this topic (of acid-base ionic equilibria). It's somewhat tedious for an actual 'A' level exam question, but quite typical for a prelim exam question.
(Students be reminded/warned : JC prelim papers are often deliberately designed as speed exam papers - even if you actually know how to do every single question in the paper, you might not have enough time to do so!)
Instead of revealing the solution, it's more beneficial for the students here on this forum (and more entertaining for us moderators *gRiN*), to attempt this question yourself and share your solution here in this thread, for peer discussions' sake.
I'll post my comments on your attempted solution submitted here on this thread.
Not only Bigcable22, but all 'A' level H2/H3/H1 students here are invited to send in your attempted solutions. It's no problem if you're unsure of your answers.
Remember, the student who asks questions may appear to be a fool (among his classmates) for that moment; but the student who does not ask, remains a fool forever.
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(a) Assumption made would be that ammonia is a weak base thus the dissociation to form hydroxide ion is negligible, with that, you should be able to form an equation using initial and final concentration of ammonia with ammonium ion and hydroxide ion.
You will need to find the basicity constant first, using the pKb. Then equating it, you should get something like:
Let [NH4+] = x
Kb = [NH4+][OH-] / [NH3] (because H2O here is a constant)
Kb = x^2 / 0.20 (assuming 0.20 - x = 0.20)
Then solve for x and you'll get the pOH then subsequently the pH.
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Nice of d3sT1nY to kickstart the sharing, but he's an ORD personnel going onto University. How about you current JC students? Attempt and post your solutions leh! Come come, no need to ke4 qi4! Dare to "let the horse come over!"
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im stucked at the mass of nh4cl added part.. any hint?
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Originally posted by Bigcable22:
im stucked at the mass of nh4cl added part.. any hint?
Ok, here's a hint :Use Kb expression to find the no. of moles of NH4+ at indicated pH.
(Again of course, assumptions regarding approximations necessarily have to be made).
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is it something like the Ph = pkb + log 10 [salt]/[base]??
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Originally posted by Bigcable22:
is it something like the Ph = pkb + log 10 [salt]/[base]??
Instead of blindly memorizing the Henderson–Hasselbalch equation without understanding (an all-too common practice among JC students which I discourage; because I want students to allow themselves to enjoy Chemistry with understanding, and memorizing stuff blindly isn't all that much fun, is it?), realize this formula is simply a mathematical derivation from the fundamental Kb (or in the case of acids, Ka) expression.Bottomline : you DON'T have to memorize or use the Henderson–Hasselbalch equation at all.
Just apply the fundamental Kb (or Ka) expression (which is so basic I'd expect almost all students can understand and appreciate the underlying meaningful concept behind it), and from there you can calculate [OH-] (or [H+]) and hence determine pOH (or pH).
Wikipedia on Henderson–Hasselbalch equation :
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(Other than Bigcable22 who posted this qn), have all the other current 'A' level students on this forum given up trying to solve this problem already? Can no current 'A' level student solve this problem?
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How many A's students lurking here anyway?
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Originally posted by dkcx:
How many A's students lurking here anyway?
The fact that many of them are lurkers, means they won't own up (their presence) anyway. -
Alright, I was trying this question. It got confusing with the 50cm3 of acid added to (I suppose) 1 dm3 of the base.
Anyway, any answer? I got my mass of ammonium chloride as 25.2g. (I suppose it's wrong.)
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basmoth messaged me his solution (my comments in blue) :
4a)
Find pOH
Kb = 10^-4.76 = [OH-]^2/0.20
[OH-] = 0.001864 (4 s.f)
pOH = 2.729
pH = 11.27b) (i)
Kb = [OH-][NH4+]/[NH3]
-lgKb = -lg[OH-] + (-lg[NH4+]/[NH3])
pKb - pOH = (-lg[NH4+]/[NH3])
lg[NH4+]/[NH3] = 0.34
[NH4+]/[NH3] = 10^0.34
[NH4+] = 0.4375 = concentration of NH4Cl added (assuming NH4+ from NH3 is negligible)In 1 dm3, mass of NH4Cl added = 23.4g (3 s.f)
b) (ii) I would assume that a large amount of NH3 would act as a buffer? Change in pH should be almost negligible. I don't know if I should use Le Chatelier's Principle in saying that as the [NH4+] increases, equilibrium shifts to the right to reduce it and hence, the [OH-] drops too and pOH increases a little. Thus, pH should be slightly lower.
While this is not strictly required by the question here, (I always advise students to) go ahead and write down whatever you suspect may be relevant, as long as you know it's correct (in terms of chemistry) on its own, you won't be penalized. Often, questions are perceived (fairly or unfairly) as being ambiguous, and every mark is precious, so the exam-smart students would go ahead to write down everything possibly relevant, as long as it's correct. And also, qualify your answers whenever you are compelled to give alternative answers. See the 1st post of my thread here for examples on question ambiguity and alternative answers.
Lastly, I have no idea how to account for the hydrolysis of NH4+Cl- (NH4+Cl- + H2O -> NH3 + H3O+ + Cl-)
Consider #1
Using Kb and Ka values, compare the strength of NH3 as a base, versus NH4+ as an acid. (Also notice that the factor difference in Kb/Ka values outweighs the factor in molarity difference).
Consider #2
The majority basic species present is NH3 (rather than OH- from water). For every proton (H+) you transfer from NH4+ to NH3, you end up back with the same no. of moles hence molarity of NH3 and NH4+.
Consider #3
A pH of 8.9 (at 25 deg C) is stated by the question. If the solution is acidic, the Ka expression is more relevant; if it's alkaline, the Kb expression is more relevant.
I assumed that NH3 reacted with HCl to give 0.01 mols of NH4+
New concentration of NH4+ = 0.447/1.05 = 0.426 mol/dm3
New concentration of NH3 = 0.19/1.05 = 0.180 mol/dm3[OH-] = Kb ([NH3])/[NH4+] = 7.343 x 10^-6 mol/dm3
pOH = 5.134
pH = 8.67 (3 s.f)Change is -0.23.
You made a careless mistake here? At 25 deg C, 14 - 5.134 = 8.87 (3 sig fig). My figures are slightly different from yours, but no difference at 3 sig fig. To answer the question, you need to comment about "the change in pH of the solution is small (pH lowered by only approximately 0.032), due to the buffer action of the weak base NH3 and its conjuagte acid salt NH4Cl".
c) My graph would start at 0.699 (-lg(0.2)) and end at 11.3 (from Kb expression). There would be a region of rapid change around pH values 3.5-6.5 as the solution is acidic due to the hydrolysis of the salt. The end-point should have a pH of 5.5.
As stated by the question, the x-axis shouuld be volume of HCl(aq) added. Hence, your graph should start with pH 11.27, maximum buffer capacity at 9.24 for 25cm3 HCl(aq), equivalence point* at 5.12 for 50cm3 HCl(aq) due to hydrolysis of acidic NH4Cl(aq) salt, and the graph levels off approaching (but never reaching) pH 0.699 at excess HCl(aq) added.
(Note that stoichiometric "equivalence point" is approximated by "end point" via use of suitable indicator during practical. Some students confuse the two terms.)
*Sorry if I got everything wrong. We've just started on physical chemistry and I'm still horrible at it. I'm trying to get better, though. :p
You got virtually everything correct. You're currently an undergrad like dkcx and d3sT1nY eh? Enjoy your days in the University while you can! Every phase in life brings to the table different joys, pain, issues, priorities, purpose and meaning. Make the most of life (Carpe Diem) and enjoy each and all of these to the max, the best you can. Live in The Now. And live it well.
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Originally posted by UltimaOnline:
*Sorry if I got everything wrong. We've just started on physical chemistry and I'm still horrible at it. I'm trying to get better, though. :p
Just a correction, this is an analytical chemistry question and not physical chemistry.
Physical chemistry deals with thermodyanamics, kinetics, quantum etc etc which sucks
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Hey, thanks for taking the time to go through my solutions. :) And nah, I'm sitting for my A'Levels this year.
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Good luck. Any plans to do chemistry at uni? UltimaOnline seems to think highly of you :p
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Originally posted by dkcx:
Good luck. Any plans to do chemistry at uni? UltimaOnline seems to think highly of you :p
Just so you know, my comment "You're currently an undergrad like dkcx and d3sT1nY eh?" has naught to do with "thinking highly" of anyone (I tell my students*, "It doesn't matter what others think of you, it only matters what you think of yourself").
It's simply that its unusual for a JC student to "just beginning on physical chemistry only now", because physical chemistry is always the 1st part of the JC syllabus taught (before inorganic or organic chem), and moreover acid-base equilibria is usually only taught in JC2, so it appears illogical for a JC2 student to say "just beginning on physical chemistry only now" when phy chem is the 1st part of JC1 syllabus. So based on his statement, it didn't seem likely that he was a JC student.
Just so you know.
PS.
A corollary or extension of *, that everyone would do well to take under advisement :
When someone says something nasty about you, it doesn't mean you're nasty, it means he's nasty.
When someone says something nice about you, it doesn't mean you're nice, it means he's nice.
I remember when sharing such wisdom when I was teaching in school, it is often 'bo hiu' by the majority of students (who go "simi lai eh?"), but for the minority of students who get a glimpse of understanding and are possibly enriched by such sharing, it makes it worthwhile.
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Do you need to get this defensive and bomb such a long paragraph just because of 1 sentence :p
Most students i would say regardless of age hates to be preached with logic... If my tuition boy can wakeup his idea, i think his parents would be the happiest people in this world and not feel sad why their son is behaving like that...