H2 Chemistry on Ionic Equilibria (crazy chapter!)
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Hi
I need help. So difficult cant do!!!!
25 cm3 of Phosphoric Acid H3PO3 of initial concentration 0.1M was titrated against NaOH, of concentration 0.1M.
Given the following equations and corresponding Ka values:-
H3PO3 = H2PO3- + H+ Ka1 = 1x 10-2
H2PO3- = HPO32- + H+ Ka2 = 2.5 x10-7
Qn: Sketch the shape of the pH curve during this titration, labelling significant pH values and significant volumes of NaOH on the axes of the curve.
So far I am only understand that
- 50 cm3 of NaOH is required at end point.
- I am only guessing that there is only one acidic buffer region at pKa1 at 10cm3 mark?
- I am not sure the pH at first and second equivalence point. So confusing already.
Thanks for helping me to understand. -
Originally posted by contemporarydancer:
Hi
I need help. So difficult cant do!!!!
25 cm3 of Phosphoric Acid H3PO3 of initial concentration 0.1M was titrated against NaOH, of concentration 0.1M.
Given the following equations and corresponding Ka values:-
H3PO3 = H2PO3- + H+ Ka1 = 1x 10-2
H2PO3- = HPO32- + H+ Ka2 = 2.5 x10-7
Qn: Sketch the shape of the pH curve during this titration, labelling significant pH values and significant volumes of NaOH on the axes of the curve.
So far I am only understand that
- 50 cm3 of NaOH is required at end point.
- I am only guessing that there is only one acidic buffer region at pKa1 at 10cm3 mark?
- I am not sure the pH at first and second equivalence point. So confusing already.
Thanks for helping me to understand.First of all, here's some background info for students :
H3PO3 is more correctly named (for common name; for systematic name, work out the oxidation state of phosphorus in the acids) "phosphorous acid" (rather than "phosphoric acid", H3PO4). Unlike phosphoric acid, a triprotic acid, phosphorous acid is indeed a diprotic acid (hence Ka1 and Ka2) - the proton bonded to the phosphorus atom is non-acidic (ie. cannot be dissociated).
Now, while I'm *not* going to work out for you the exact pH values and significant NaOH volumes, here are some guidelines on what the question requires you to do. Based on what you've written, you've got your fundamentals, and are doing pretty ok. Just keep at it, and you'll do fine.
[quote]50 cm3 of NaOH is required at end point.[/quote]
Firstly, that's "equivalence point", not end point. Equivalence point refers to the exact stoichiometric neutralization point per each acidic proton. "End point" of a titration occurs when the (relevant) indicator used changes colour. The end point approximates the equivalence point.
Secondly, there are two equivalence points, and hence two end points. This is because phosphorous acid is a diprotic acid (ie. one mole of acid gives two moles of protons; aka dibasic acid).
[quote]I am only guessing that there is only one acidic buffer region at pKa1 at 10cm3 mark?[/quote]
Firstly, since phosphorous acid is a diprotic acid, there will hence be 2 buffer regions (ie. regions in which any protons or hydroxide ions added will be immediately removed from solution by the conjugate base and conjugate acid respectively, hence the 'buffering' effect). As to the pH of these buffer regions (or to be precise the pH of the exact points of maximum buffer capacity for the 2 regions), use pKa1 and pKa2 to determine.
Secondly, how did you obtain the value of "10cm3" mark for NaOH? Remember that an acid buffer is most effective (ie. at its maximum buffer capacity) when molarity of acid = molarity of conjugate base / salt. Understand that this also means the pH of maximum buffer capacity occurs midpoint(s) of the neutralization process, before the equivalence point(s) is/are reached.
Hence you should be able to work out the volumes of NaOH for the 2 points of maximum buffer capacity, which given that the question is kind enough to have the molarities of phosphorous acid and sodium hydroxide as being equal, makes your job much easier.
[quote]I am not sure the pH at first and second equivalence point. So confusing already.[/quote]
Ahhh, here's the biggie, the main meat of the question. To determine the pH at 1st and 2nd equivalence point, you need 2 tables :a) ICF (Initial Change Final) table for 1st equivalence point, in moles.
b) ICF (Initial Change Final) table for 2nd equivalence point, in moles.
To determine the volume of NaOH required for both equivalence points, use stoichiometry (ie. looking at the balanced chemical equation(s); in this context, for neutralization of the first and second acidic protons).
The important and tricky PitTrap here (which many students fall into and lose their precious marks and hence also their chance for a distinction grade), is when calculating the molarity of the salt (or acid, or base), you have to be very careful to take into consideration and correctly add up the total volumes of both the acid and alkali used (including for 2nd equivalence point).
Calculating the pH of the 2nd equivalence point.
Notice that for Na2HPO3(aq) of which the anion is the HPO3 2-(aq) aka monohydrogen phosphite ion (ie. the twice deprotonated phosphorous acid), because Na+ has no affinity to covalent bond with (or remove from solution) OH- ions, while HPO3 2- has significant affinity to covalent bond with (or remove from solution) H+ ions aka protons, the salts Na2HPO3 is basic/alkaline upon hydrolysis. (Most students find it easier to think of it summarily and remember it as "strong base + weak acid = basic salt").
Hence, using Kb2 expression and value (you are given Ka2 value; hence using the relationship between Ka, Kb and Kw, you can determine the Kb2 value), work out the molarity of hydroxide ions, hence pOH, and therefore finally pH.
Calculating the pH of the 1st equivalence point.
Now, the first deprotonated phosphorous acid, NaH2PO3(aq) of which the anion is H2PO3-(aq) aka dihydrogen phosphite ion, is a little trickier. One the one hand, it may protonate itself (ie. remove protons from solution to become phosphorous acid again) and hence be basic. On the other hand, it may serve as a source of protons itself (since its second acidic proton may dissociate, to become monohydrogen phosphite ion) and hence be acidic.
To determine which, compare the relevant Ka value against the relevant Kb value. You should be able to figure out it's the Ka2 value against the Kb1 value (tip : focus on the protonation and deprotonation of the ion being discussed, the dihydrogen phosphite ion).
Because the Ka2 (deprotonation of ion as an acid) value is significantly greater than the Kb1 (protonation of ion as a base) value, you can conclude that the dihydrogen phosphite ion is acidic rather than basic.
Hence, to determine pH at the 1st equivalence point, after working out the ICF table and upon having determined the molarity of the salt NaH2PO3(aq), we hence use the Ka2 expression and value to work out the molarity of protons, and hence pH.
To refresh your memory,
Ka = ( [molarity of protons] x [molarity of conjugate base] ) / [molarity of conjugate acid]
Kb = ( [molarity of hydroxide ions] x [molarity of conjugate acid] ) / [molarity of conjugate base]
So to summarize, this is a good discriminating (distinction) question, in which you have to decide whether the 1st-deprotonated phosphorous acid aka dihydrogen phosphite ion, is acidic or basic (upon hydrolysis), based on the relevant Ka and Kb values. Knowing which, will enable you to (decide whether to use Ka or Kb expression to) calculate the pH at the 1st equivalence point. The twice-deprotonated phosphorous acid aka monohydrogen phosphite ion, on the other hand, is very obviously basic (upon hydrolysis) (since the proton bonded to phosphorus cannot dissociate and hence isn't acidic), and hence you must use the Kb expression to calculate OH, then pOH and finally pH, at the 2nd equivalence point.
[quote]I need help. So difficult cant do!!!! So confusing already. Thanks for helping me to understand.[/quote]
You're welcome. And you're doing pretty ok, actually. And remember, the more you've the passion to understand, enjoy and love Chemistry (for what it *truly* is, and not just memorizing blindly as instructed dogmatically by your JC lecturers/tutors, or for the sake of paper chase exams), the more you will inevitably excel in it!
We All Love Chemistry! -
Thanks....I think I got it. I tried the question again. and I got the following answers.
Initial pH = 1.501st max. buffering capacity occurs when 12.50cm3 NaOH is added. pH = 2.0
1st equivalence point occurs when 25cm3 NaOH is added, pH = 3.95
2nd max. buffering capacity occurs when 37.5cm3 NaOH is added, pH =7.4
2nd equivalence point occurs when 50cm3 NaOH is added, pH =9.71.
Are my values correct?
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There is a part b to this question on Ksp.
Ksp of PbSO4 and SrSO4 are 6.3 x 10^-7 and 3.2x10^-7 respectively.
What are the values of [SO4^2-], [Pb^2+] and [Sr^2+] in a solution at equilibrium with both PbSO4 and SrSO4?
I tried doing them. and my answers are [Pb2+] 0.0116mol/dm3, [Sr2+] 2.94 x 10^(-4) mol/dm3. I cant find [SO4^2-] already. I understand the common ion, SO4^ 2- and the effect it has on the solubility of both sulphates. I got alittle stucked with the calculations part. -
>>> Initial pH = 1.50 <<<
Correct.
>>> 1st max. buffering capacity occurs when 12.50cm3 NaOH is added. pH = 2.0 <<<
Correct.
>>> 1st equivalence point occurs when 25cm3 NaOH is added, pH = 3.95 <<<
Correct.
>>> 2nd max. buffering capacity occurs when 37.5cm3 NaOH is added, pH =7.4 <<<
Wrong. (vol correct, pH wrong).
Correct Ans is pH 6.60
>>> 2nd equivalence point occurs when 50cm3 NaOH is added, pH =9.71. <<<
Wrong. (vol correct, pH wrong).
Correct Ans is 9.56
I won't go into further detail, because if you've read my previous post carefully, the correct method is detailed within.
One possible pitfall you have to check yourself for, is in calculations in determining the MOLARITIES of the acid/base/salts. What are the volumes involved? 25+12.5? 25+25? 25+37.5? 25+50?
Another possible pitfall is in applying the Ka or Kb expression. Did you select the *correct* Ka or Kb expression to use? (there are 4 such expressions in total, phosphorous acid being a diprotic acid.). Did you derive the values of the 2 Kb's correctly from the 2 Ka's? Did you substitute correct molarities into the expressions?
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Solubility Equilibria
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First of all, let me state for the record that the technically correct method to solve this problem is beyond the means (and syllabus requirements) of an 'A' level candidate to handle. It is almost certain that the solution your JC lecturer/tutor has in mind for this question, is oversimplified and technically inaccurate.
(I might be wrong about the abovestated opinion, and indeed I might be wrong about the solution offered below, but this is as per my understanding and interpretation of the problem. Do your best to follow and adopt your JC lecturer/tutor's method for solving this problem).
To begin to attempt to solve this problem, conceptualize or imagine you first begin with two saturated solutions (including solid precipitates), each 1dm3, of lead(II) sulphate and strontium sulphate. You can find the molarities of all the ions present.
Notice that based on the Ksp values, lead(II) sulphate is slightly more soluble than strontium sulphate. This will have an effect on the different relative molarities of the 2 aqueous cations at saturation (equilibrium).
Pour the two solutions (including solid precipitates) into a 2dm3 beaker, to obtain a new solution, which will thereafter adjust itself to a new equilibrium.
You can calculate the initial molarities of all 3 ions in the 2dm3 beaker. What do you expect will happen to the molarities of the 3 ions? (And does it matter if solid precipitates of both sulphates are present initially from the 2 previous separate solutions? Think carefully.)
By Le Chatelier's principle, the effect of increasing the aqueous molarity of an ion (which is the common ion), would be to decrease the solubility of a compound (by shifting position of equilibrium from aqueous towards solid).
In contrast, again by Le Chatelier's principle, the effect of decreasing the aqueous molarity of an ion (which is the common ion), would be to increase the solubility of a compound (by shifting position of equilibrium from solid towards aqueous).
Notice that based on the new initial molarity of the sulphate ions in the 2dm3 container (which is the average of the 2 previous molarities in the 2 separate 1dm3 solutions), one of the sulphates will hence become more soluble, and the other sulphate will hence become less soluble.
Keeping this in mind, you can begin to construct the 2 ICE (Initial Change Equilibrium) tables, for the solution (ie. dissolving) of one salt, and the precipitation of the other salt.
You will soon notice that you end up with rather tedious quadratic equations, which (as stated by the Cambridge-SEAB syllabus specifications) will not be required of the candidate in the actual 'A' level examinations.
The final step is to apply the Ksp values to determine the value of the algebraic variables, and therefore the molarities of the ions at equilibrium.
At this point, you may recognize yet another problem - what if the 2 calculated molarities of the sulphate ion at equilibrium (one based on calculations involving lead(II) sulphate, the other based on calculations involving strontium sulphate), are inconsistent (ie. different?).
At this point, it should be clear enough for everyone to see why this question is not within the scope of the 'A' level H2 syllabus; or to put it in another way, this isn't a valid 'A' level examination question.
There is a much simpler, but technically inaccurate and invalid, method of (apparently or ostensibly) determining the molarities of the 3 ions. And this method, is what I suspect your JC lecturer/tutor has in mind when he/she set came up with this problem.
Because it is technically inaccurate and invalid, and because I do not wish to give wrong ideas in regards to Chemistry, I will not elaborate on the method.
After your JC lecturer/tutor has gone through the answer with you, if you're willing to do so, please share your JC lecturer/tutor's solution or method with us here on this thread, and I may comment further, if appropriate.
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Thanks for your help.
This is actually taken from past year prelim paper from one of the top JC. I didnt get it from the lecturer but from those bookshop. and apparently this particular paper doesnt come with answers.
So I suppose I can skip the second part on Ksp! YIPPIE!! haha...thanks. -
Sorry sir, but I still dont get the part on the 2nd equivalence point and 2nd 1/2 equivalence point.
My understanding is this:
at the 2nd max. buffering capacity, there is only the salt Na2HPO3 (acting as the salt) and NaH2PO3(acting as acid) in equal amounts. So am I right to say it is an acidic buffer?
Thus, from the equation of the acidic buffer
pH = pKa + lg [salt]/[acid]
pH = pKa2 = 6.6
Now at the 2nd equivalence point, this is alittle tricky for me to understand.
There is only the salt Na2HPO3. so this salt will undergo hydrolysis to give OH-.
Hence, I see HPO3- as the conjugate base, and thus have to find the concentration of OH-. And so I will find Kb2, which is 3.98 x 10^(-8) moldm-3
Moles of HPO3- = 0.005
Volume = 75cm3
Therefore, [HPO3-] = 0.0667 moldm-3
pH = 14 - pOH
= 9.71
I dont get the answer of 9.86.
Where did I go wrong? concept?
Thanks. -
>>> at the 2nd max. buffering capacity, there is only the salt Na2HPO3 (acting as the salt) and NaH2PO3(acting as acid) in equal amounts. So am I right to say it is an acidic buffer? <<<
Yes.
>>> Thus, from the equation of the acidic buffer
pH = pKa + lg [salt]/[acid]
pH = pKa2 = 6.6 <<<Right.
>>> Now at the 2nd equivalence point, this is alittle tricky for me to understand. There is only the salt Na2HPO3. so this salt will undergo hydrolysis to give OH-. Hence, I see HPO3- as the conjugate base, and thus have to find the concentration of OH-. And so I will find Kb2, which is 3.98 x 10^(-8) moldm-3
Moles of HPO3- = 0.005
Volume = 75cm3
Therefore, [HPO3-] = 0.0667 moldm-3
pH = 14 - pOH
= 9.71
I dont get the answer of 9.86.
Where did I go wrong? concept? <<<Kb2 = ([OH-][H2PO3 -])/[HPO3 2-]
((1x10^-14)/(2.5x10^-7)) = ([OH-]^2)/((25/1000)x0.1)/(75/1000))
(4x10^-8) = ([OH-]^2) / (0.03333)
[OH-]^2 = 1.333 x 10^-9
[OH-] = 3.651x10^-5
pOH = 4.438
pH = 9.562 = 9.56 (to 3 sig fig)
>>> Thanks. <<<Welcome.