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Chemistry HL/SL help


Hedron123

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  • 2 weeks later...

Hey Chem Gurus,

Could you help me out with Voltaic Cell calculations? I was absent from the lesson when our class went through theory, and now I'm in trouble with the exercises!

So: I'm supposed to calculate theoretical voltages in a circuit with different half-cells. In my questions I've got Cu, Zn, Fe, Mg and Al mixed with each other in all possible combinations, so I've got 10 different voltaic cells, and I need to figure out the voltage for each one of them. I've got no idea how to even start! :D

Btw, They're all in 0.1 molar solutions: Cu electrode in CuSo4 solution, Zn in ZnSO4, Fe in FeSO4, Mg in Mg(NO3)2 and Al in Al(NO3)3. Do these details have an effect on the calculations?

I think Table 14 in the data booklet has something to do with these calculations. How do I know which one is the cathode or the anode? And in which direction the reaction goes?

I've got time 'till Sunday evening to figure all this out! :)

For example a sample calculation of one situation, with explanations, would be very much appreciated! Or do you know some good internet links that explain the calculations?

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aim3.h2.jpg

Just using it for an example.

If I remember right Pb is a greater reducing agent than Ag, since reducing agents are oxidized Pb is the anode. Oxidation occurs at the anode, always, won't change, remember it. This means the Ag will be reduced. Now when you calculate the voltage most books give a rather complicated way of doing with addition and switch and all this weird stuff but when it comes down to simplicity it is the E value of the reduced thing (Ag) minus the E value of the oxidized thing (Pb). If you do not know which is which because you don't get a diagram, or you simply don't have a reactivity series (paper1, you don't get that handy data book) it will be the greater value minus the lesser value.

In this case the E value of Ag and Pb are 0.80 and -0.13 respectively according to http://hyperphysics.phy-astr.gsu.edu/hbase/tables/electpot.html

So in this case, reduced-oxidized would be 0.80-(-0.13) which is 0.93V :)

Remember redox, reduction minus oxidation, it's in order for you right there :D

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I'm also working on voltaic cells ! To add to what Drake Glau already said, and to answer your question on whether the concentration has an effect or not:

"You are not working under standard conditions, but the important thing is that the ratio of the concentrations is 1:1. Therefore, you would still measure the standard cell potential. Any differences in the readings would come from other sources such as resistance in the salt bridge, lack of calibration of the voltmeter and the concentrations of the two cells not being exactly the same, for whatever reason."

Sorce: http://chemistry.bd.psu.edu/jircitano/electrochem07a.pdf That link also contains some extra info than might or might not be useful to you.

Hope this helps. Best of luck with your calculations! XD

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Thank you guys so much for help!! I actually got it! And well....You showed that its not so complicated after all - once you get the idea. :( And hey, the hyperphysics website was really good, and so was your link too, Hasta!

Thanks once again for saving me. Next time I won't forget the textbook at school for the weekend hahaha :)

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  • 2 weeks later...

Guys, I need help :( this lab report is just too confusing.. I have no idea on how to proceed :( the title is: analysis of aspirin tablet by back titration.

  1. I reacted a known mass of crushed aspirin tablets with a known concentration of NaOH and with H2O.
    Based on the background information that my teacher gave, the reactions that are supposed to take place are:
    CH3COO(C6H4)COOH + OH- --> CH3COO(C6H4)COO- + H2O (fast reaction)
    CH3COO(C6H4)COO- + OH- --> HO(C6H4)COO- + CH3COO- (slow reaction)
    However it’s just reactions of aspirin with NaOH. What about its reaction with water? Or is it just a ‘dilution’?
  2. I warmed the mixture. I did not boil it.. and the manual says “ to hydrolyse the acetylsalicyclic acid”.
    is that the aspirin? So in this step hydrolysis takes place? Through Google, I found this reaction which they call the hydrolysis of acetylsalicyclic acid:
    CH3COO(C6H4)COOH + H2O --> CH3COOH + HO(C6H4)COOH
    If I look at it again, wasn’t that supposed to occur in step 1 too? But then in the first one since I reacted it with NaOH, I got HO(C6H4)COONa + CH3COONa. Then if I also got CH3COOH + HO(C6H4)COOH, I had so many products! :o
    In this part does any other reaction that involves heating occur?
  3. Then I diluted the mixture.
  4. Next I titrated it against a known concentration of HCl using phenolphthalein indicator.
    What is the reaction that takes place in this step?
    FYI in step 1 I had excess NaOH so was it supposed to react with HCl too? Or did something happen to it when I warmed the mixture?
    Theoretically, HCl would react with HO(C6H4)COONa, CH3COONa, CH3COOH and HO(C6H4)COOH, right? And maybe with the excess NaOH too?
  5. Last I standardised NaOH with HCl.
    Not to worry, I know what happened in this step.

The aim of this experiment is to calculate the mass of aspirin in each aspirin tablet. Don’t worry about the calculations – how I go about finding the mass – I am pretty sure I know what to do. But the only thing I always do not know is the reactions that take place. I cannot proceed now because I do not know what reactions took place :S

So if anybody could help me I would highly appreciate it!! Thank you!!

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I reacted a known mass of crushed aspirin tablets with a known concentration of NaOH and with H2O.

Based on the background information that my teacher gave, the reactions that are supposed to take place are:

CH3COO(C6H4)COOH + OH- --> CH3COO(C6H4)COO- + H2O (fast reaction)

CH3COO(C6H4)COO- + OH- --> HO(C6H4)COO- + CH3COO- (slow reaction)

However it’s just reactions of aspirin with NaOH. What about its reaction with water? Or is it just a ‘dilution’?

Yes the water is just a porduct and does not take part in the reaction. The water will only partially disociate whereas the NaOH will fully disociate. Therefore NaOH will be more reactive and take priority in the reaction.

I warmed the mixture. I did not boil it.. and the manual says “ to hydrolyse the acetylsalicyclic acid”.

is that the aspirin? So in this step hydrolysis takes place? Through Google, I found this reaction which they call the hydrolysis of acetylsalicyclic acid:

CH3COO(C6H4)COOH + H2O --> CH3COOH + HO(C6H4)COOH

If I look at it again, wasn’t that supposed to occur in step 1 too? But then in the first one since I reacted it with NaOH, I got HO(C6H4)COONa + CH3COONa. Then if I also got CH3COOH + HO(C6H4)COOH, I had so many products! :o

In this part does any other reaction that involves heating occur?

Yes the acetylsalicyclic acid is asprin. Yes that is water hydrolysing the asprin, however, as I stated above, the NaOH will react and not the water as it is more reactive. So HO(C6H4)COONa + CH3COONa is your only product.

Then I diluted the mixture.

Next I titrated it against a known concentration of HCl using phenolphthalein indicator.

What is the reaction that takes place in this step?

FYI in step 1 I had excess NaOH so was it supposed to react with HCl too? Or did something happen to it when I warmed the mixture?

Theoretically, HCl would react with HO(C6H4)COONa, CH3COONa, CH3COOH and HO(C6H4)COOH, right? And maybe with the excess NaOH too?

The reaction would be between the Na+ ions and the Cl- and the H+ ions and the COO- part of your products. Yes the HCl would indeed react with excess NaOH, I'm not quite sure what happened. Maybe its the heating or maybe its somthing to do with the use of phenolphthalein indicator. I'm unsure

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Guys, I need help :( this lab report is just too confusing.. I have no idea on how to proceed :( the title is: analysis of aspirin tablet by back titration.

  1. I reacted a known mass of crushed aspirin tablets with a known concentration of NaOH and with H2O.
    Based on the background information that my teacher gave, the reactions that are supposed to take place are:
    CH3COO(C6H4)COOH + OH- --> CH3COO(C6H4)COO- + H2O (fast reaction)
    CH3COO(C6H4)COO- + OH- --> HO(C6H4)COO- + CH3COO- (slow reaction)
    However it’s just reactions of aspirin with NaOH. What about its reaction with water? Or is it just a ‘dilution’?
  2. I warmed the mixture. I did not boil it.. and the manual says “ to hydrolyse the acetylsalicyclic acid”.
    is that the aspirin? So in this step hydrolysis takes place? Through Google, I found this reaction which they call the hydrolysis of acetylsalicyclic acid:
    CH3COO(C6H4)COOH + H2O --> CH3COOH + HO(C6H4)COOH
    If I look at it again, wasn’t that supposed to occur in step 1 too? But then in the first one since I reacted it with NaOH, I got HO(C6H4)COONa + CH3COONa. Then if I also got CH3COOH + HO(C6H4)COOH, I had so many products! :o
    In this part does any other reaction that involves heating occur?
  3. Then I diluted the mixture.
  4. Next I titrated it against a known concentration of HCl using phenolphthalein indicator.
    What is the reaction that takes place in this step?
    FYI in step 1 I had excess NaOH so was it supposed to react with HCl too? Or did something happen to it when I warmed the mixture?
    Theoretically, HCl would react with HO(C6H4)COONa, CH3COONa, CH3COOH and HO(C6H4)COOH, right? And maybe with the excess NaOH too?
  5. Last I standardised NaOH with HCl.
    Not to worry, I know what happened in this step.

The aim of this experiment is to calculate the mass of aspirin in each aspirin tablet. Don’t worry about the calculations – how I go about finding the mass – I am pretty sure I know what to do. But the only thing I always do not know is the reactions that take place. I cannot proceed now because I do not know what reactions took place :S

So if anybody could help me I would highly appreciate it!! Thank you!!

CH3COO(C6H4)COOH + NaOH --> CH3COO(C6H4)COONa + H2O

Next warming it hydrolyzed it: CH3COO(C6H4)COO- + NaOH --> HO(C6H4)COO- + CH3COONa

Titration (I hate these at the moment, bare with it =/)

I'm going to take a stab at this and say it ends up being HO(C6H4)COO- + HCl since the OH ions hanging out there is more likely to react with the HCl than the Na ion. The Na from the CH3COONa will hydrolyze the water in the solution though creating more H and OH ions in the solution. So when you titrate you are finding the amount of HCl is needed to neutralize both the HO(C6H4)COO- and CH3COONa and since the hydrolysis that created these was a 1:1 ratio it's the same concentration of the original acetylsalicylic acid...

Hope this made sense. This titration stuff is currently killing all 5 of us for the last week :(

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Hi! :)

I really need your help!! I have a test due tomorrow :coffee: and I have a terrible problem trying to understand entropy and spontaneity. I am able to cram the rules and all but I would really like to understand all this as I like chemistry, I would be very grateful!

Thank you,

Matthew

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Ummm...any specific part you don't understand? Hess's law? Conversions using enthalpy values? dH-dST=dG?

Can't teach you a whole topic through a forum, that's your teacher's job :P

Maybe without entropy - just the sponaneity topic - gibbs free energy change and the universe depending on the surroundings and the system,then the gibbs free energy of formation - I'm just having trouble with understanding the whole topic, I will apreciate any help, really.

Thanks,

Matthew

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Hey guys, still talking about this http://www.ibsurvival.com/topic/10834-chemistry-hlsl-help/page__view__findpost__p__104259 lab on aspirin

I have read this guide here:

http://chemistry-react.org/go/Tutorial/Tutorial_21681.html

and it answers my questions quite clearly

anyhow I am now confused

After the back titration I did a standardisation.

I put 25 mL of 1.0 M of NaOH in a volumetric flask and added distilled water till I get 250 mL of a dilute NaOH

Then I titrated 25 mL of the diluted NaOH (from the volumetric flask) with 0.1 M of HCl

What is the significance of this standardisation? Just to check the concentration or what?

Because in the main experiment I diluted the hydrolysed aspirin thing into 250 mL too. But I only had NaO(C6H4)COONa and the extra NaOH (b/c I added excess NaOH in the beginning) and probably the CH3COONa too from reaction 1 in that ^ guide but I don't really see the connection.. So what for do you think the standardisation is?

Thank you~

EDIT: I think I get it now. but I still want a reply in case if I understand wrongly

Edited by Desy ♫
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Normally its to dilute the solution down so that the titration is more effective. Sometimes is just used to increase the size of the sample so more tests can be done. I don't think the standardisation has any major significance apart from the dilution factor.

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  • 2 weeks later...

In my design lab I am investigating the rate of a reaction by measuring the volume of gas collected in a given period of time. Below is my diagram of setup.

bFxaq.jpg

My teacher requires me to write an explanation for each design lab. Don't ask why and don't tell me that I don't need to do it.

So anyway... I need to explain why the delivery tube is not submerged in the solution. I roughly know why but I can't form a good sentence to explain that. Can anybody help? Thanks before!!

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If the delivery tube was submerged into the solution, there is not pathway between the gaseous substance in the flask and the instrument which aims to measure the change of volume as both the delivery tube and the dense liquid create a barrier. The gas released would build up as pressure rather than push the syringe out to create volume. Thus if the delivery tube is submerged into the solution, the device is rendered useless.

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In an experiment, I reacted crushed aspirin tablets with some stuffs in order to find the mass of acetyl salicyclic acid in each aspirin tablet.

I crushed 3 aspirin tablets. From the powder, I only took 1.7 g of it. So 3 aspirin tablets actually weigh more than 1.7 g since there was still some powder remaining.

With a lot of effort, I found the mass of the acetyl salicyclic acid in total, from the 1.7 g aspirin powder.

The problem is that I cannot find the mass of acetyl salicyclic acid in each aspirin tablet! Not even the mass of each aspirin tablet!! Because I didn't crush exactly 3 tablets.

I actually asked my teacher via email (didn't have the chance to meet her for some reasons) what I could do. I even asked her if I can just find the percentage of acetyl salicyclic acid in aspirin. But then her reply was just:

mass of each aspirin tablet = 0.6 g +/- 0.01 g

mass of acetyl salicyclic acid per tablet = 0.5 g

I do not get it.. I think she still wants me to find the mass of acetyl salicyclic acid in each aspirin tablet... but how?

If the mass of each aspirin tablet=0.6 g and I used 1.7 g powder, so I had 2.83 aspirin tablets.

And so I divide my mass of the acetyl salicyclic acid by 2.83 to get the mass of acetyl salicyclic acid in each aspirin tablet? Does that make sense?

Summary:

My question is does that make sense?

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