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Is it right that Exothermic deltaHsoln (delHsoln <0 means soluble in water? ) means water gets more hot(gain energy hence temp rise) , as ions release energy(which outweighs LEdissociation) on hydration?
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Originally posted by Flying grenade:
Is it right that Exothermic deltaHsoln (delHsoln <0 means soluble in water? ) means water gets more hot(gain energy hence temp rise) , as ions release energy(which outweighs LEdissociation) on hydration?
Solution Enthalpy = (endothermic) Lattice Dissociation Enthalpy + (exothermic) Hydration EnthalpySolution Entropy = (positive) Lattice Dissociation Entropy + (negative) Hydration Entropy
Gibbs Free Energy of Solution = Solution Enthalpy - Temperature x Solution Entropy
A species is soluble in water (ie. solution process is thermodynamically feasible, ie. when energetically and entropically considered at a given temperature) only if Gibbs Free Energy change is negative.
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pH + pOH = 14 only for 25 degrees celcius is it?
Above equation does not hold true for other temperature?
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Why cs toh book, didnt show reaction of phenol with ROCl , RCOX , RX ? Only show reaction with base,metal, no2+, and Cl-
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In chemistry, whats the difference between state and phase?
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Originally posted by Flying grenade:
In chemistry, whats the difference between state and phase?
You can easily google this out for yourself. -
Originally posted by Flying grenade:
pH + pOH = 14 only for 25 degrees celcius is it?
Above equation does not hold true for other temperature?
Obviously. -
Originally posted by Flying grenade:
Why cs toh book, didnt show reaction of phenol with ROCl , ROCX , RX ? Only show reaction with base,metal, no2+, and Cl-
Because CS Toh expects you to be able to work out such reactions yourself. And for A levels, students are expected to know that, phenol being a weaker nucleophile than alcohols, will need to first be deprotonated before it can react with the less reactive electrophiles. -
Advanced cs toh guide page 330, hydroxyl compounds
Near btm of the page,
What does "lower aryl alcohol" means?
Does it mean aryl alcohols with less c atoms, around 6? Just now didn't think of this cos haven't read page 331,and most aryl alcohols we encounter only with 6 c atoms i.e. phenol?
My guess from page 331, term "lower alcohol" is used, it means lower number of c atoms?
Thanks
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Page 334 advanced cstoh guide
What does 'esterification can be used as a test for alcohol - the ester formed floats on water and has a sweet smell' means?
Isnt this a test of presence of ester instead?as we deduce ester present from sweet smell?
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Originally posted by Flying grenade:
Advanced cs toh guide page 330, hydroxyl compounds
Near btm of the page,
What does "lower aryl alcohol" means?
Thanks
Lower = not too many C atoms present. Aryl = benzene ring present. Idea is simply : most alcohols are liquid at rtp (due to both van der Waals interactions and hydrogen bonding present), unless it's a large molecule with many C atoms, which would be solid at rtp. Cambridge question will give guidance or hints, don't worry about details like "which alcohol liquid? which solid?". -
Originally posted by Flying grenade:
Page 334 advanced cstoh guide
What does 'esterification can be used as a test for alcohol - the ester formed floats on water and has a sweet smell' means?
Isnt this a test of presence of ester instead?as we deduce ester present from sweet smell?
You can test for the presence of alcohols by adding an acyl halide. If a sweet smell is observed, meaning that an ester is generated, which in turn means the original analyte must have contained an alcohol. -
Hi Ultima,
https://www.dropbox.com/s/2i79w0ow9hq7bsp/1111.PNG?dl=0
for part ii) , why is this thought process wrong : i want find [H+] . i use ka formula. tried, but didnt get the correct answer.
(p.s. , i've internalised the 2 methods/approach that was taught to me , but want to know why above concept wrong. )
is it because, inside the solution, its not just the weak acid inside, there's the cj base too? Help my concept please T-T
thanks ultima
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Originally posted by Flying grenade:
Hi Ultima,
https://www.dropbox.com/s/2i79w0ow9hq7bsp/1111.PNG?dl=0
for part ii) , why is this thought process wrong : i want find [H+] . i use ka formula. tried, but didnt get the correct answer.
(p.s. , i've internalised the 2 methods/approach that was taught to me , but want to know why above concept wrong. )
is it because, inside the solution, its not just the weak acid inside, there's the cj base too? Help my concept please T-T
thanks ultima
Being a buffer soln, you *can* indeed use Ka, or even Kb. It's just that the Henderson-Hasselbalch equation saves you time as you can work in moles instead of molarities. If you didn't get the correct answer, it means your molarities (ie. moles / new total volume of soln) of both HA and A- at that point (double-equivalence) are wrong. Obviously you need to plug in the molarities of both conjugate acid and conjugate base into the Ka (or Kb) equation, and not all questions will be as kind as this question to let you have maximum buffer capacity, so you better start practicing tougher questions in which the moles or molarities of both conjugate acid and conjugate base are different. Then try using all 3 formulae : Ka, Kb, and Henderson-Hasselbalch equation, to prove *to yourself* that all 3 formulae will give you the same final answer of the correct pH. -
Is it okay/fine/chemically correct to write
E.g.
For Mg(OH)2 to be precipitated, Qsp (Mg(OH)2) >Ksp (Mg(OH)2)
Looking at 2015 jc prelims solutions, all wrote I.P. or just Solubility = ____
None wrote Qsp
My school uses I.P. , but sometimes i forgot, i wrote Qsp a number of times already ,but now trying to get the habit back of writing I.P. But is it wrong to write 'Qsp' ?
Btw, SAJC uses ICF table, lol, cool. that makes BJC students, and them ,that know about it
Update : error above. Solubility not equals Qsp. Always specify molar/mass solubility
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Originally posted by Flying grenade:
Is it okay/fine/chemically correct to write
E.g.
For Mg(OH)2 to be precipitated, Qsp (Mg(OH)2) >Ksp (Mg(OH)2)
Looking at 2015 jc prelims solutions, all wrote I.P. or just Solubility = ____
None wrote Qsp
My school uses I.P. , but sometimes i forgot, i wrote Qsp a number of times already ,but now trying to get the habit back of writing I.P. But is it wrong to write 'Qsp' ?
Btw, SAJC uses ICF table, lol, cool. that makes BJC students, and them ,that know about it
Both are acceptable by Cambridge, use whichever you prefer, IP or Qsp, to be compared with Ksp. Personally, I'll recommend Qsp (since it's closer in form to Ksp). For solubility (which is not the same as Ksp. ie. solubility product), always specify "molar solubility" (mol/dm3) or "mass solubility" (g/dm3). -
Originally posted by UltimaOnline:
Both are acceptable by Cambridge, use whichever you prefer, IP or Qsp, to be compared with Ksp. Personally, I'll recommend Qsp (since it's closer in form to Ksp). For solubility (which is not the same as Ksp. ie. solubility product), always specify "molar solubility" (mol/dm3) or "mass solubility" (g/dm3).Oh yeah , right ! Thank you!
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DHS/2015/P2/Q3 part iv )
https://www.dropbox.com/s/x2gzb3vp1ekxl3c/20160502_012359-1.jpg?dl=0
I understand the solution.
But i dun understand the qn, Why to determine solubility of CaC2O4 at high pH, use NaOH? What are the alternative method?
I was thinking, CaC2O4 》《Ca2+ + C2O42- , let solubility of CaC2O4 be x, hence solubility is just root(Ksp)
Thanks ultima
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Originally posted by Flying grenade:
DHS/2015/P2/Q3 part iv )
https://www.dropbox.com/s/x2gzb3vp1ekxl3c/20160502_012359-1.jpg?dl=0
I understand the solution.
But i dun understand the qn, Why to determine solubility of CaC2O4 at high pH, use NaOH? What are the alternative method?
I was thinking, CaC2O4 》《Ca2+ + C2O42- , let solubility of CaC2O4 be x, hence solubility is just root(Ksp)
Thanks ultima
Your molar solubility based strictly on Ksp ignores the effect of pH, which is the usual case within the H2 syllabus. This DHS question goes beyond the H2 syllabus (which Cambridge often does) to include the effect on pH, which is why your answer is inadequate and thus wrong. If you do not add NaOH, how are you going to make the pH high (ie. alkaline)? That's why you add NaOH, which increases solubility by precipitating out the cation (in contrast, adding acid increases solubility by protonating the anion). There's no alternative method ; Wilbur's equation only covers acidic to neutral pH. -
Thanks for ur explanation !!!
I'm not fully understanding the problem yet
Is it :
I want to study the solubility of CaC2O4.
In this case, the solubility of CaC2O4 in high pH(alkaline conditions)
In a way, we want to study the effect of [H+] in the solubility of CaC2O4. (Can be seen from Wilber's equation)
In part iii) , the [H+] plays a role, in protonating anion, in turn having an effect in the solubility of CaC2O4.
However in part iv) , we actually want to study the effect of [H+] in high pH, but however, the NaOH actually causes increased solubility of CaC2O4, as it precipitates out the cation, instead of due to the effect of [H+] , which we want to study. Hence the problem.
Is it?
For part iii) , the solution elucidates about the effect of LCP for the explanation. But can i also explain , based on the equation given in part iii, solubility increases, as total [H+] in numerator increases faster than total [H+] in denominator, hence solubility increases
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Originally posted by Flying grenade:
Thanks for ur explanation !!!
I'm not fully understanding the problem yet
Is it :
I want to study the solubility of CaC2O4.
In this case, the solubility of CaC2O4 in high pH(alkaline conditions)
In a way, we want to study the effect of [H+] in the solubility of CaC2O4. (Can be seen from Wilber's equation)
In part iii) , the [H+] plays a role, in protonating anion, in turn having an effect in the solubility of CaC2O4.
However in part iv) , we actually want to study the effect of [H+] in high pH, but however, the NaOH actually causes increased solubility of CaC2O4, as it precipitates out the cation, instead of due to the effect of [H+] , which we want to study.
Is it?
For part iii) , the solution elucidates about the effect of LCP for the explanation. But can i also explain , based on the equation given in part iii, solubility increases, as total [H+] in numerator increases faster than total [H+] in denominator, hence solubility increases
Wrong, in high pH, it's OH- that's relevant, not H+. "We" do *not* want to study the effect of [H+] in high pH, "We" only want to study the effect of pH on solubility, so Wilbur's equation (involving H+) is obsolete and can be retrenched.Your mathematics-based answer will not be acceptable, because firstly, the H2 Chem syllabus is always testing students about shifting of position of equilibrium (use of acronyms like LCP not allowed in A level exams); and secondly, the actual question (on Wilbur's work, or similar) in the A levels might not give you an mathematical equation or formula at all, so if all you can give are mathematical reasonings (as opposed to chemical reasoning that's required by the mark scheme), you'll be screwed in the A levels.
BedokFunland JC exam tip : If you're exam-smart, you'll give a variety of reasons, whatever you can think of. As long as they don't contradict each other, you'll be given credit for whichever answer Cambridge wanted.
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I understand the discussion about OH- , with the cation, can precipitate out an ionic solid, and , in acidic pH, anion is protonated. But i dont see how these is relevant for the explanation in iv) .
I dont see how "NaOH might precipitate out Ca(OH)2 , which would decrease [Ca2+] , therefore potentially increasing solubility of CaC2O4 at high pH" (given answer solution) addresses the question of iv) , which is to study the effect of (high) pH on solubility.
We want to study the effect of pH on solubility.
For part iv) , understood, Wilber's equation is obsolete for alkaline pH.For part iv) , Then, why is adding NaOH a problem? it should be Wilber's equation that's the problem.
i dont see how adding NaOH is a problem studying effect of pH on solubility.
Adding NaOH IS A Problem if we want to study effect of pH on solubility, USING Wilber's equation.
I dont see a link here. maybe i dun understand/disagree with the phrasing of the qn, and the ans. might be my command/understanding of english is inadequate.
Ok real sorry, just move on, i dont mean to annoy you, sorry, its ok if u dont reply to this qn
Just move on -
Originally posted by Flying grenade:
I understand the discussion about OH- , with the cation, can precipitate out an ionic solid, and , in acidic pH, anion is protonated. But i dont see how these is relevant for the explanation in iv) .
I dont see how "NaOH might precipitate out Ca(OH)2 , which would decrease [Ca2+] , therefore potentially increasing solubility of CaC2O4 at high pH" (given answer solution) addresses the question of iv) , which is to study the effect of (high) pH on solubility.
We want to study the effect of pH on solubility.
For part iv) , understood, Wilber's equation is obsolete for alkaline pH.For part iv) , Then, why is adding NaOH a problem? it should be Wilber's equation that's the problem.
i dont see how adding NaOH is a problem studying effect of pH on solubility.
Adding NaOH IS A Problem if we want to study effect of pH on solubility, USING Wilber's equation.
I dont see a link here. maybe i dun understand/disagree with the phrasing of the qn, and the ans. might be my command/understanding of english is inadequate.
Ok real sorry, just move on, i dont mean to annoy you, sorry, its ok if u dont reply to this qn
Just move on
sourceJust skip the damn question in the A levels if you don't like Cambridge's phrasing. Oh wait, only I can do that and still get an A grade. *You* can't.
Yeah, go chat with (and annoy) your school teacher or private tutor instead.
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I understand this :
By Hess law,
🔺Hsoln = 🔺Hhyd + 🔺LEdissociation
Can help me check where i've gone wrong thinking in this way?? :
🔺Hsoln = energy released on hydration by hydrated ions Minus Energy absorbed by Ionic solid to dissociate
So 🔺Hsoln = (-ve enthalpy change) minus (positive enthalpy change )
If i put modulus on the RHS terms, then it becomes all +ve
Help please thanks!
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HCO3- is both amphoteric and amphiprotic right?