Chemistry M09 IB SL/HL
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do you guys remember this question?
900cm3 of water is added to 100cm3 of NaOH,whose pH is 12.
What is the pH of the solution?
I chose 13 after a long calculation involving -log(x) or something
However, my friend easily says that since pH of water is 7, the answer is obviously between 12 and 7 and there was only one option, which was 11.
Who is right? Is the answner 11 or 13?
proudofmyself
do you guys remember this question?
900cm3 of water is added to 100cm3 of NaOH,whose pH is 12.
What is the pH of the solution?
I chose 13 after a long calculation involving -log(x) or something
However, my friend easily says that since pH of water is 7, the answer is obviously between 12 and 7 and there was only one option, which was 11.
Who is right? Is the answner 11 or 13?
900cm3 of water is added to 100cm3 of NaOH,whose pH is 12.
What is the pH of the solution?
I chose 13 after a long calculation involving -log(x) or something
However, my friend easily says that since pH of water is 7, the answer is obviously between 12 and 7 and there was only one option, which was 11.
Who is right? Is the answner 11 or 13?
Well, how would the pH go up? You are diluting the NaOH. Therefore, the pH will decrease, and be closer to neutral.
excarnifico
oxidation: CH3CH2OH --> CH3CHO + 2e-
reduction: (Cr2O7)2- + 14H+ + 6e- --> 2Cr3+ + 7H2O
..i think.
reduction: (Cr2O7)2- + 14H+ + 6e- --> 2Cr3+ + 7H2O
..i think.
Ah i just wrote Cr6+ + 3e- --> Cr3+ for the reduction, i assumed the rest were spectator ions
as for the oxidation; it said the final product was an acid hence it can't be an aldehyde... it would have to be CH3C00H ; which means an additional oxygen will have to be part of the equation... also there is one less hydrogen (or 2 in the aldehyde) so that would also have to feature....
@liverpool: yea..and all for 2 marks
RNQ
The pH would be 11 because the pOH increases by 1 as the concentration of OH- ions decreased 10fold.
If pOH increases by 1, pH decreases by 1.
Basically, if you think about it, if you dilute the base it is going to be a weaker base, so its pH would decrease.
If pOH increases by 1, pH decreases by 1.
Basically, if you think about it, if you dilute the base it is going to be a weaker base, so its pH would decrease.
If you're using the logic common-sense thing that works,
however if you use the -log[H+] rule; you've just decreased the concentration of H+ 10 fold (like if you had diluted an acid) in which case the answer would be 13.... thats the problem with -log thing, since it means a HIGHLY diluted acid could have a pH of lets say, 100
i.e: HCl has a pH of 3, it is diluted by factor 10^9; so pH increases by 9, suddenly it is a strong base of pH 12
Still... i assume IB would expect you to use logic, so 11 must be the correct answer
radam
If you're using the logic common-sense thing that works,
however if you use the -log[H+] rule; you've just decreased the concentration of H+ 10 fold (like if you had diluted an acid) in which case the answer would be 13.... thats the problem with -log thing, since it means a HIGHLY diluted acid could have a pH of lets say, 100
i.e: HCl has a pH of 3, it is diluted by factor 10^9; so pH increases by 9, suddenly it is a strong base of pH 12
Still... i assume IB would expect you to use logic, so 11 must be the correct answer
however if you use the -log[H+] rule; you've just decreased the concentration of H+ 10 fold (like if you had diluted an acid) in which case the answer would be 13.... thats the problem with -log thing, since it means a HIGHLY diluted acid could have a pH of lets say, 100
i.e: HCl has a pH of 3, it is diluted by factor 10^9; so pH increases by 9, suddenly it is a strong base of pH 12
Still... i assume IB would expect you to use logic, so 11 must be the correct answer
No, the concentration of OH- decreased 10fold, so pOH would rise from 2 to 3. So pH decreases from 12 to 11.
Why overcomplicate things?
radam
If you're using the logic common-sense thing that works,
however if you use the -log[H+] rule; you've just decreased the concentration of H+ 10 fold (like if you had diluted an acid) in which case the answer would be 13.... thats the problem with -log thing, since it means a HIGHLY diluted acid could have a pH of lets say, 100
i.e: HCl has a pH of 3, it is diluted by factor 10^9; so pH increases by 9, suddenly it is a strong base of pH 12
Still... i assume IB would expect you to use logic, so 11 must be the correct answer
however if you use the -log[H+] rule; you've just decreased the concentration of H+ 10 fold (like if you had diluted an acid) in which case the answer would be 13.... thats the problem with -log thing, since it means a HIGHLY diluted acid could have a pH of lets say, 100
i.e: HCl has a pH of 3, it is diluted by factor 10^9; so pH increases by 9, suddenly it is a strong base of pH 12
Still... i assume IB would expect you to use logic, so 11 must be the correct answer
Your logic fails because H+ and OH- are always in equilibrium with water. Oh and it's definitely 11. Now tell me, what's the answer to the HIn indicator question?
Dan_G
SL Paper 2 Timezone 2 Question 7 (Organic long question)
The mechanism... was it SN1, or SN2.... coz from this thread i've got a horrible feeling i showed the SN2 mechanism and misread the question which said SN1.... please tell me i didn't.... it was a perfect SN2 mechanism as well...
The mechanism... was it SN1, or SN2.... coz from this thread i've got a horrible feeling i showed the SN2 mechanism and misread the question which said SN1.... please tell me i didn't.... it was a perfect SN2 mechanism as well...
While i didn't answer Q7 (i answered the Kc one) , i seem to remember a question being there on SN2 ; besides most of the posts so far have been on the HL paper, so dw and second guess yourself ... its over, be happy
AnonymousPenguin
Your logic fails because H+ and OH- are always in equilibrium with water. Oh and it's definitely 11. Now tell me, what's the answer to the HIn indicator question?
In which case diluting would have no effect at all since its already in equilibrium, and you're adding the same amount of H+ as OH- .... unless ... ah i'm too sleepy to think at the moment
Sadly I do SL so can't help you with the HIn one; as though much of the questions were the same we didn't get that one.
RNQ
No, the concentration of OH- decreased 10fold, so pOH would rise from 2 to 3. So pH decreases from 12 to 11.
Why overcomplicate things?
Why overcomplicate things?
Hmm that's an interesting way to look at it... and would save loads of time in the exam room! Haha i think i'll just accept that its 11.
radam
If you're using the logic common-sense thing that works,
however if you use the -log[H+] rule; you've just decreased the concentration of H+ 10 fold (like if you had diluted an acid) in which case the answer would be 13.... thats the problem with -log thing, since it means a HIGHLY diluted acid could have a pH of lets say, 100
i.e: HCl has a pH of 3, it is diluted by factor 10^9; so pH increases by 9, suddenly it is a strong base of pH 12
Still... i assume IB would expect you to use logic, so 11 must be the correct answer
however if you use the -log[H+] rule; you've just decreased the concentration of H+ 10 fold (like if you had diluted an acid) in which case the answer would be 13.... thats the problem with -log thing, since it means a HIGHLY diluted acid could have a pH of lets say, 100
i.e: HCl has a pH of 3, it is diluted by factor 10^9; so pH increases by 9, suddenly it is a strong base of pH 12
Still... i assume IB would expect you to use logic, so 11 must be the correct answer
Yes, I had the same thoughts at first. But I think that the point is that water isn't just a solvent, its also adding H+ and OH- in 1:1 ratio. Therefore the amount of H+ doesn't stay the same and pH shifts towards neutral.
For the indicator I think it went blue on adding base. You could directly exclude the acid, because 2 of the options were adding strong/weak acid having the same effect, so if one of the options was true, the other would have to be true too. And the 3rd option seemed to be completely irrelevant.
EDIT: I also thought about it in terms of LeChatelier's, so the answer should be correct from the point of chemistry too.
EDIT2: to the pH: although that should mean that [OH] also shouldn't decrease exactly 10 times IMO, but anyway I think this explanation seems to be quite acceptable.
EDIT3: well, now I realise difference between [OH] in water and NaOH is in orders of magnitude, so yes, [OH] will decrease 10x pretty exactly, I guess.
proudofmyself
do you guys remember this question?
900cm3 of water is added to 100cm3 of NaOH,whose pH is 12.
What is the pH of the solution?
I chose 13 after a long calculation involving -log(x) or something
However, my friend easily says that since pH of water is 7, the answer is obviously between 12 and 7 and there was only one option, which was 11.
Who is right? Is the answner 11 or 13?
900cm3 of water is added to 100cm3 of NaOH,whose pH is 12.
What is the pH of the solution?
I chose 13 after a long calculation involving -log(x) or something
However, my friend easily says that since pH of water is 7, the answer is obviously between 12 and 7 and there was only one option, which was 11.
Who is right? Is the answner 11 or 13?
You're diluting by tenfold, hence 11 by inspection.
Chazaa
Thoughts on P3 TZ1 meds and biochem?
Didn't do biochem but heard it was easy. The first two questions weren't really even "chem"... super easy for those who did bio, maybe not so easy for others?
Dan_G
SL Paper 2 Timezone 2 Question 7 (Organic long question)
The mechanism... was it SN1, or SN2.... coz from this thread i've got a horrible feeling i showed the SN2 mechanism and misread the question which said SN1.... please tell me i didn't.... it was a perfect SN2 mechanism as well...
The mechanism... was it SN1, or SN2.... coz from this thread i've got a horrible feeling i showed the SN2 mechanism and misread the question which said SN1.... please tell me i didn't.... it was a perfect SN2 mechanism as well...
it was sn2. and the question was in section b number 6.
Hello! my name is Ana and I take IB HL chemistry. I moved in the middle of the IB which messed with all my subjects but in specific with chemistry. If anyone out there is willing to help me out, i would honestly appreciate if they contacted me! My exams are in a little under two weeks and I know i came very late but I just found this site and i could REALLY use the help!
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