Isaac Chemistry Questions
I've been struggling on two Chemistry questions on Isaac Physics for quite a while now, and would appreciate some help.
Here are the links to the questions:
https://isaacphysics.org/questions/chem_16_i3_10?board=chem_16_book_i3&stage=a_level
and
https://isaacphysics.org/questions/chem_16_i4_7?board=chem_16_book_i4&stage=further_a
Here is my working so far for the two questions:
For the first question, I found the square root of the given solubility product to find the concentration of thallium (I) ions left in the solution after the precipitate has been formed, multiplied this by 0.08 dm^3 to find the number of moles, and then added this to the moles of thallium in the precipitate, which i just calculated by doing 0.173 / (204.4 + 35.5). I then divided this final amount by 0.06 dm^3 to find the initial concentration of thallium and got 0.0297 mol dm^-3, which is apparently wrong.
For the second question, I just thought that the question was a bit vague about what evaporates (e.g. does the precipitate also evaporate with the ether?) and some other things. My first instinct was to do ((0.00014 g / 0.125 dm^3) / 1.820*10^-4) * (0.05 - 0.03) dm^3 and got 0.1231 grams, which is also wrong. I'm assuming I have to do something with the fact that some of the ether evaporates while equilibrium is maintained, but I'm not sure what.
Thanks for any help.
Here are the links to the questions:
https://isaacphysics.org/questions/chem_16_i3_10?board=chem_16_book_i3&stage=a_level
and
https://isaacphysics.org/questions/chem_16_i4_7?board=chem_16_book_i4&stage=further_a
Here is my working so far for the two questions:
For the first question, I found the square root of the given solubility product to find the concentration of thallium (I) ions left in the solution after the precipitate has been formed, multiplied this by 0.08 dm^3 to find the number of moles, and then added this to the moles of thallium in the precipitate, which i just calculated by doing 0.173 / (204.4 + 35.5). I then divided this final amount by 0.06 dm^3 to find the initial concentration of thallium and got 0.0297 mol dm^-3, which is apparently wrong.
For the second question, I just thought that the question was a bit vague about what evaporates (e.g. does the precipitate also evaporate with the ether?) and some other things. My first instinct was to do ((0.00014 g / 0.125 dm^3) / 1.820*10^-4) * (0.05 - 0.03) dm^3 and got 0.1231 grams, which is also wrong. I'm assuming I have to do something with the fact that some of the ether evaporates while equilibrium is maintained, but I'm not sure what.
Thanks for any help.
Original post by CelestialMaster7
I've been struggling on two Chemistry questions on Isaac Physics for quite a while now, and would appreciate some help.
Here are the links to the questions:
https://isaacphysics.org/questions/chem_16_i3_10?board=chem_16_book_i3&stage=a_level
and
https://isaacphysics.org/questions/chem_16_i4_7?board=chem_16_book_i4&stage=further_a
Here is my working so far for the two questions:
For the first question, I found the square root of the given solubility product to find the concentration of thallium (I) ions left in the solution after the precipitate has been formed, multiplied this by 0.08 dm^3 to find the number of moles, and then added this to the moles of thallium in the precipitate, which i just calculated by doing 0.173 / (204.4 + 35.5). I then divided this final amount by 0.06 dm^3 to find the initial concentration of thallium and got 0.0297 mol dm^-3, which is apparently wrong.
For the second question, I just thought that the question was a bit vague about what evaporates (e.g. does the precipitate also evaporate with the ether?) and some other things. My first instinct was to do ((0.00014 g / 0.125 dm^3) / 1.820*10^-4) * (0.05 - 0.03) dm^3 and got 0.1231 grams, which is also wrong. I'm assuming I have to do something with the fact that some of the ether evaporates while equilibrium is maintained, but I'm not sure what.
Thanks for any help.
Here are the links to the questions:
https://isaacphysics.org/questions/chem_16_i3_10?board=chem_16_book_i3&stage=a_level
and
https://isaacphysics.org/questions/chem_16_i4_7?board=chem_16_book_i4&stage=further_a
Here is my working so far for the two questions:
For the first question, I found the square root of the given solubility product to find the concentration of thallium (I) ions left in the solution after the precipitate has been formed, multiplied this by 0.08 dm^3 to find the number of moles, and then added this to the moles of thallium in the precipitate, which i just calculated by doing 0.173 / (204.4 + 35.5). I then divided this final amount by 0.06 dm^3 to find the initial concentration of thallium and got 0.0297 mol dm^-3, which is apparently wrong.
For the second question, I just thought that the question was a bit vague about what evaporates (e.g. does the precipitate also evaporate with the ether?) and some other things. My first instinct was to do ((0.00014 g / 0.125 dm^3) / 1.820*10^-4) * (0.05 - 0.03) dm^3 and got 0.1231 grams, which is also wrong. I'm assuming I have to do something with the fact that some of the ether evaporates while equilibrium is maintained, but I'm not sure what.
Thanks for any help.
In the first question the solubility product must take into account the concentration of the chloride ions that have been added. Therein lies your error.
ksp = [Tl+][Cl-] = 1.75×10−4
First determine the concentration of the chloride ions after addition of the sodium chloride solution.
Then determine the concentration of thallium remaining in the solution and hence the mol.
Can you take it from there?
Original post by charco
In the first question the solubility product must take into account the concentration of the chloride ions that have been added. Therein lies your error.
ksp = [Tl+][Cl-] = 1.75×10−4
First determine the concentration of the chloride ions after addition of the sodium chloride solution.
Then determine the concentration of thallium remaining in the solution and hence the mol.
Can you take it from there?
ksp = [Tl+][Cl-] = 1.75×10−4
First determine the concentration of the chloride ions after addition of the sodium chloride solution.
Then determine the concentration of thallium remaining in the solution and hence the mol.
Can you take it from there?
Thanks! I managed to do the question now. I did actually calculate the concentration of Cl in a previous attempt at the question, but I think the steps I did after that were incorrect. Your guidance helped me clear up the mistakes I made - many thanks.
Original post by CelestialMaster7
I've been struggling on two Chemistry questions on Isaac Physics for quite a while now, and would appreciate some help.
Here are the links to the questions:
https://isaacphysics.org/questions/chem_16_i3_10?board=chem_16_book_i3&stage=a_level
and
https://isaacphysics.org/questions/chem_16_i4_7?board=chem_16_book_i4&stage=further_a
Here is my working so far for the two questions:
For the first question, I found the square root of the given solubility product to find the concentration of thallium (I) ions left in the solution after the precipitate has been formed, multiplied this by 0.08 dm^3 to find the number of moles, and then added this to the moles of thallium in the precipitate, which i just calculated by doing 0.173 / (204.4 + 35.5). I then divided this final amount by 0.06 dm^3 to find the initial concentration of thallium and got 0.0297 mol dm^-3, which is apparently wrong.
For the second question, I just thought that the question was a bit vague about what evaporates (e.g. does the precipitate also evaporate with the ether?) and some other things. My first instinct was to do ((0.00014 g / 0.125 dm^3) / 1.820*10^-4) * (0.05 - 0.03) dm^3 and got 0.1231 grams, which is also wrong. I'm assuming I have to do something with the fact that some of the ether evaporates while equilibrium is maintained, but I'm not sure what.
Thanks for any help.
Here are the links to the questions:
https://isaacphysics.org/questions/chem_16_i3_10?board=chem_16_book_i3&stage=a_level
and
https://isaacphysics.org/questions/chem_16_i4_7?board=chem_16_book_i4&stage=further_a
Here is my working so far for the two questions:
For the first question, I found the square root of the given solubility product to find the concentration of thallium (I) ions left in the solution after the precipitate has been formed, multiplied this by 0.08 dm^3 to find the number of moles, and then added this to the moles of thallium in the precipitate, which i just calculated by doing 0.173 / (204.4 + 35.5). I then divided this final amount by 0.06 dm^3 to find the initial concentration of thallium and got 0.0297 mol dm^-3, which is apparently wrong.
For the second question, I just thought that the question was a bit vague about what evaporates (e.g. does the precipitate also evaporate with the ether?) and some other things. My first instinct was to do ((0.00014 g / 0.125 dm^3) / 1.820*10^-4) * (0.05 - 0.03) dm^3 and got 0.1231 grams, which is also wrong. I'm assuming I have to do something with the fact that some of the ether evaporates while equilibrium is maintained, but I'm not sure what.
Thanks for any help.
For the second question, I just thought that the question was a bit vague about what evaporates (e.g. does the precipitate also evaporate with the ether?) and some other things. My first instinct was to do ((0.00014 g / 0.125 dm^3) / 1.82010^-4) (0.05 - 0.03) dm^3 and got 0.1231 grams, which is also wrong. I'm assuming I have to do something with the fact that some of the ether evaporates while equilibrium is maintained, but I'm not sure what.
No, the ether evaporates leaving the other component behind.
The idea is that you determine the concentration of the solute in the two phases using the partition coefficient.
This gives you the solute in ether.
Then you use the new volumes and concentrations to find the new mol of solute in the ether, which is then separated off.
Original post by charco
For the second question, I just thought that the question was a bit vague about what evaporates (e.g. does the precipitate also evaporate with the ether?) and some other things. My first instinct was to do ((0.00014 g / 0.125 dm^3) / 1.82010^-4) (0.05 - 0.03) dm^3 and got 0.1231 grams, which is also wrong. I'm assuming I have to do something with the fact that some of the ether evaporates while equilibrium is maintained, but I'm not sure what.
No, the ether evaporates leaving the other component behind.
The idea is that you determine the concentration of the solute in the two phases using the partition coefficient.
This gives you the solute in ether.
Then you use the new volumes and concentrations to find the new mol of solute in the ether, which is then separated off.
No, the ether evaporates leaving the other component behind.
The idea is that you determine the concentration of the solute in the two phases using the partition coefficient.
This gives you the solute in ether.
Then you use the new volumes and concentrations to find the new mol of solute in the ether, which is then separated off.
I'm still not quite sure about what I'm meant to do. The concentration of the solute in water is 0.00014g/0.125dm^3 = 1.12x10-3 gdm^-3. Using this, the concentration of the solute in the ether is 1.12x10^-3 / 1.82x10^-4 = 6.15 gdm^-3. The new volume of ether is then 0.05 - 0.03 = 0.02 dm^3, so the moles of solute in the ether at the end is 0.1231 grams, which is what I got before. Am I misunderstanding a part of the question? Thanks
Original post by CelestialMaster7
I'm still not quite sure about what I'm meant to do. The concentration of the solute in water is 0.00014g/0.125dm^3 = 1.12x10-3 gdm^-3. Using this, the concentration of the solute in the ether is 1.12x10^-3 / 1.82x10^-4 = 6.15 gdm^-3. The new volume of ether is then 0.05 - 0.03 = 0.02 dm^3, so the moles of solute in the ether at the end is 0.1231 grams, which is what I got before. Am I misunderstanding a part of the question? Thanks
OK - I think I managed to get it, but would like to make sure that my working is valid, and that I didn't get the answer due to sheer luck. Here is what I did:
I calculated the mass of solute in ether before some of the ether evaporated by doing 1.12x10^-3/1.82x10^-4 x 0.05 = 4/13 grams. The mass of solute in ether then changes by y amount after some of the ether evaporates, and I know that this solute goes from the ether to the water, so I said that 4/13 - y = ((0.00014 + y)/0.125)/1.8210^-4 x 0.02, and solved for y to get y (the change in mass of solute) = 0.2098mg. This means that the final amount of solute in ether is 4000/13 - y = 307.5mg (4s.f.). Is this logic correct?
There is also one last question I was hoping to get help on if you don't mind:
https://isaacphysics.org/questions/chem_16_i3_7?board=chem_16_book_i3&stage=a_level
I think this question has something to do with the common ion effect, but I'm not sure how to do the calculations. As there are two of the same ion, the regular ways of calculating concentration don't seem to work when I try them. Any advice on what to do? Many thanks.
Original post by CelestialMaster7
OK - I think I managed to get it, but would like to make sure that my working is valid, and that I didn't get the answer due to sheer luck. Here is what I did:
I calculated the mass of solute in ether before some of the ether evaporated by doing 1.12x10^-3/1.82x10^-4 x 0.05 = 4/13 grams. The mass of solute in ether then changes by y amount after some of the ether evaporates, and I know that this solute goes from the ether to the water, so I said that 4/13 - y = ((0.00014 + y)/0.125)/1.8210^-4 x 0.02, and solved for y to get y (the change in mass of solute) = 0.2098mg. This means that the final amount of solute in ether is 4000/13 - y = 307.5mg (4s.f.). Is this logic correct?
There is also one last question I was hoping to get help on if you don't mind:
https://isaacphysics.org/questions/chem_16_i3_7?board=chem_16_book_i3&stage=a_level
I think this question has something to do with the common ion effect, but I'm not sure how to do the calculations. As there are two of the same ion, the regular ways of calculating concentration don't seem to work when I try them. Any advice on what to do? Many thanks.
I calculated the mass of solute in ether before some of the ether evaporated by doing 1.12x10^-3/1.82x10^-4 x 0.05 = 4/13 grams. The mass of solute in ether then changes by y amount after some of the ether evaporates, and I know that this solute goes from the ether to the water, so I said that 4/13 - y = ((0.00014 + y)/0.125)/1.8210^-4 x 0.02, and solved for y to get y (the change in mass of solute) = 0.2098mg. This means that the final amount of solute in ether is 4000/13 - y = 307.5mg (4s.f.). Is this logic correct?
There is also one last question I was hoping to get help on if you don't mind:
https://isaacphysics.org/questions/chem_16_i3_7?board=chem_16_book_i3&stage=a_level
I think this question has something to do with the common ion effect, but I'm not sure how to do the calculations. As there are two of the same ion, the regular ways of calculating concentration don't seem to work when I try them. Any advice on what to do? Many thanks.
Yes, it's perfectly sound logic.
Original post by CelestialMaster7
OK - I think I managed to get it, but would like to make sure that my working is valid, and that I didn't get the answer due to sheer luck. Here is what I did:
I calculated the mass of solute in ether before some of the ether evaporated by doing 1.12x10^-3/1.82x10^-4 x 0.05 = 4/13 grams. The mass of solute in ether then changes by y amount after some of the ether evaporates, and I know that this solute goes from the ether to the water, so I said that 4/13 - y = ((0.00014 + y)/0.125)/1.8210^-4 x 0.02, and solved for y to get y (the change in mass of solute) = 0.2098mg. This means that the final amount of solute in ether is 4000/13 - y = 307.5mg (4s.f.). Is this logic correct?
There is also one last question I was hoping to get help on if you don't mind:
https://isaacphysics.org/questions/chem_16_i3_7?board=chem_16_book_i3&stage=a_level
I think this question has something to do with the common ion effect, but I'm not sure how to do the calculations. As there are two of the same ion, the regular ways of calculating concentration don't seem to work when I try them. Any advice on what to do? Many thanks.
I calculated the mass of solute in ether before some of the ether evaporated by doing 1.12x10^-3/1.82x10^-4 x 0.05 = 4/13 grams. The mass of solute in ether then changes by y amount after some of the ether evaporates, and I know that this solute goes from the ether to the water, so I said that 4/13 - y = ((0.00014 + y)/0.125)/1.8210^-4 x 0.02, and solved for y to get y (the change in mass of solute) = 0.2098mg. This means that the final amount of solute in ether is 4000/13 - y = 307.5mg (4s.f.). Is this logic correct?
There is also one last question I was hoping to get help on if you don't mind:
https://isaacphysics.org/questions/chem_16_i3_7?board=chem_16_book_i3&stage=a_level
I think this question has something to do with the common ion effect, but I'm not sure how to do the calculations. As there are two of the same ion, the regular ways of calculating concentration don't seem to work when I try them. Any advice on what to do? Many thanks.
Yes, it's perfectly sound logic.
I'm gonna be lazy and represent it like this:
Ksp1 = [Ba][ox] = 2.1×10−7
Ksp2 = [Fe][ox] = 1.7×10−7
Divide Ksp1 by Ksp2
[Ba]/[Fe] = 2.1/1.7 = 1.24
[Ba] = 1.24[Fe]
and as:
Baox <==> Ba + ox
Feox <==> Fe + ox
then
[Ba] + [Fe] = [ox]
Now substitute for [Ba]
1.24[Fe] + [Fe] = [ox]
[ox] = 2.24[Fe]
Ksp2 = [Fe][ox] = 1.7×10−7
2.24 [Fe][Fe] = 1.7×10−7
[Fe]2 = 7.59 x 10-8
[Fe] = 2.75 x 10-4
Hence
[ox] = 6.17 x 10-4
6.2 x 10-4 mol dm-3
Ksp1 = [Ba][ox] = 2.1×10−7
Ksp2 = [Fe][ox] = 1.7×10−7
Divide Ksp1 by Ksp2
[Ba]/[Fe] = 2.1/1.7 = 1.24
[Ba] = 1.24[Fe]
and as:
Baox <==> Ba + ox
Feox <==> Fe + ox
then
[Ba] + [Fe] = [ox]
Now substitute for [Ba]
1.24[Fe] + [Fe] = [ox]
[ox] = 2.24[Fe]
Ksp2 = [Fe][ox] = 1.7×10−7
2.24 [Fe][Fe] = 1.7×10−7
[Fe]2 = 7.59 x 10-8
[Fe] = 2.75 x 10-4
Hence
[ox] = 6.17 x 10-4
6.2 x 10-4 mol dm-3
Original post by charco
I'm gonna be lazy and represent it like this:
Ksp1 = [Ba][ox] = 2.1×10−7
Ksp2 = [Fe][ox] = 1.7×10−7
Divide Ksp1 by Ksp2
[Ba]/[Fe] = 2.1/1.7 = 1.24
[Ba] = 1.24[Fe]
and as:
Baox <==> Ba + ox
Feox <==> Fe + ox
then
[Ba] + [Fe] = [ox]
Now substitute for [Ba]
1.24[Fe] + [Fe] = [ox]
[ox] = 2.24[Fe]
Ksp2 = [Fe][ox] = 1.7×10−7
2.24 [Fe][Fe] = 1.7×10−7
[Fe]2 = 7.59 x 10-8
[Fe] = 2.75 x 10-4
Hence
[ox] = 6.17 x 10-4
6.2 x 10-4 mol dm-3
Ksp1 = [Ba][ox] = 2.1×10−7
Ksp2 = [Fe][ox] = 1.7×10−7
Divide Ksp1 by Ksp2
[Ba]/[Fe] = 2.1/1.7 = 1.24
[Ba] = 1.24[Fe]
and as:
Baox <==> Ba + ox
Feox <==> Fe + ox
then
[Ba] + [Fe] = [ox]
Now substitute for [Ba]
1.24[Fe] + [Fe] = [ox]
[ox] = 2.24[Fe]
Ksp2 = [Fe][ox] = 1.7×10−7
2.24 [Fe][Fe] = 1.7×10−7
[Fe]2 = 7.59 x 10-8
[Fe] = 2.75 x 10-4
Hence
[ox] = 6.17 x 10-4
6.2 x 10-4 mol dm-3
Thanks - I managed to follow all of that except for one bit:
'And as:
Baox <==> Ba + ox
Feox <==> Fe + ox
Then [Ba] + [Fe] = [ox]'
Forgive me if I'm missing something obvious, but why is the bottom line true?
Original post by CelestialMaster7
Thanks - I managed to follow all of that except for one bit:
'And as:
Baox <==> Ba + ox
Feox <==> Fe + ox
Then [Ba] + [Fe] = [ox]'
Forgive me if I'm missing something obvious, but why is the bottom line true?
'And as:
Baox <==> Ba + ox
Feox <==> Fe + ox
Then [Ba] + [Fe] = [ox]'
Forgive me if I'm missing something obvious, but why is the bottom line true?
If the total oxalate that gets into solution does so from iron oxalate and barium oxalate then:
nBaOx ==> nBa2+ + nOxalate from BaOx (in a 1:1 ratio)
mFeOx ==> mFe2+ + mOxalate from FeOx (in a 1:1 ratio)
total oxalate mol = (m+n), which is also equal to mol Ba2+ + mol Fe2+
Then mol Ba2+ + mol Fe2+ must equal the mol of oxalate summed from the two equations.
Original post by charco
If the total oxalate that gets into solution does so from iron oxalate and barium oxalate then:
nBaOx ==> nBa2+ + nOxalate from BaOx (in a 1:1 ratio)
mFeOx ==> mFe2+ + mOxalate from FeOx (in a 1:1 ratio)
total oxalate mol = (m+n), which is also equal to mol Ba2+ + mol Fe2+
Then mol Ba2+ + mol Fe2+ must equal the mol of oxalate summed from the two equations.
nBaOx ==> nBa2+ + nOxalate from BaOx (in a 1:1 ratio)
mFeOx ==> mFe2+ + mOxalate from FeOx (in a 1:1 ratio)
total oxalate mol = (m+n), which is also equal to mol Ba2+ + mol Fe2+
Then mol Ba2+ + mol Fe2+ must equal the mol of oxalate summed from the two equations.
Ah, I see now. Thank you so much for all your help!
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