Decarbonation of soft drink (2 Viewers)

[bEAbEA]

Hi can anyone help me out with this dot point, a breif sketch of the prac is all i need. Thanks

Identify data plan and perform a first hand investigation to decarbonate soft drink and gather data to measure the mass changes involved and calculate the volume of gas released at 25oC and 100kPa

 

[Dreamerish*~]

Do you mean a sketch - literally? Because that would just be drawing some scales and a big plastic drink bottle.

http://www.geocities.com/jamesyscience_chemistry/hsc/tae/exp.html

That should help.

 

[bmc]

Hi can anyone help me out with this dot point, a breif sketch of the prac is all i need. Thanks

Identify data plan and perform a first hand investigation to decarbonate soft drink and gather data to measure the mass changes involved and calculate the volume of gas released at 25oC and 100kPa

when sketching it draw ur soft drink can on a tripod and gauze mat and with a bunsen. Heating increases the likelihood of all the gas being removed. just include in ur method to weight the can b4 and after heating.

 

[COCO_moiselle]

Hi Kate

I did this degassing prac for my chemistry assessment for which i got 19/20

I have the document (hypothesis, aim, method, conclusion, discussion) saved on my computer. I can send it to you by email if you like

 

[jamesy_1988]

when sketching it draw ur soft drink can on a tripod and gauze mat and with a bunsen. Heating increases the likelihood of all the gas being removed. just include in ur method to weight the can b4 and after heating.

You have to be really careful when you decarbonate soft drinks using this method as there is a possibility that water may be lost to the atmosphere when it is heated, even just slightly above room temperature.

The best method of decaqrbonating soft drink i have found is to either shake it until you have a consistent mass or to add a known mass of sodium chloride to the soft drink. Either method works fine and reduces the possibility of experimental error

Jamesy

http://www.geocities.com/jamesyscience

 

[tango400]

what does the sodium chloride actually do?

 

[wah]

There are 3 methods to achieve the aim:

-heating the soft drink gently such that the evaporation of the water (and other stuff) is at a minimum. This means no boiling of the soft drink.

-Shake the soft drink

-Use NaCl to salt it out

though the first two methods are more common.

 

[Tess653]

i've got the same prac.

would freezing work? i read in my text book that CO2 is less soluble in ice.

 

[Dreamerish*~]

i've got the same prac.

would freezing work? i read in my text book that CO2 is less soluble in ice.

You should never freeze a bottle of soft drink because the solubility of CO₂ in ice is very low. So almost all the CO₂ is going to come out of solution as the water freezes out, and pack into the small space under the cap. This would cause a significant build-up of pressure, and possibly an explosion.

However, I suppose if you were to place the drink in the freezer without the lid (or with the lid loosely on), the CO₂ could escape from the bottle. But that would certainly take more than one period at school, and you usually have just one lesson to do it.

 

[Riviet]

I once froze a can of coke in the freezer at home.

It exploded of course. And I think I heard it.

 

[femuu]

when adding salt to the soda water, i know CO2 comes out because of a disruption in the equilibrium , but wot exactly does the salt do??????
does it act as a catalyst???

 

[Dreamerish*~]

when adding salt to the soda water, i know CO2 comes out because of a disruption in the equilibrium , but wot exactly does the salt do??????
does it act as a catalyst???

I don't know exactly why, but salt causes dehydration, and since water is a reactant, dehydration of the equilibrium will shift it to the left, producing more water and carbon dioxide.

After thinking about it, I'm guessing that the salt in solution is stopping the water from reacting with the carbon dioxide.

As you know, when a system is at equilibrium, the rate of the forward reaction is equal to the rate of the backward reaction. That is to say, some carbon dioxide is reacting with water to form carbonic acid. Salts, being ionic substances ionise in solution, so by adding NaCl, you are adding sodium and chlorine ions to the solution. From year 11 you learned that ionic substances dissolve well in water because water as a polar molecule has oxygen being slightly negative and hydrogen being slightly positive. So when NaCl dissolves in water, the positive Na⁺ is attracted to the negative oxygen atom while the negative Cl^- is attracted to the positive hydrogen atom.

So I'm guessing that when water molecules are "wrapped" in anions and cations, they can't react with carbon dioxide to form carbonic acid. Making a reactant unable to participate in the reaction is as good as removing it. Hence the equilibrium will shift to the left, producing more water and carbon dioxide.

My best guess.

 

[(enigma)]

when adding salt to the soda water, i know CO2 comes out because of a disruption in the equilibrium , but wot exactly does the salt do??????
does it act as a catalyst???

apparently the salt reduces the liquids ability (not sure if thats the right word for it) to allow gas to dissolve into it, but im not really sure why. does anyone else know?

 

[Riviet]

We don't need to know exactly why the salt causes dehydration, it's not in the syllabus and even someone as knowledgeable as Dreamerish doesn't fully know the answer. I'm not trying to say that we shouldn't find out about these outside questions, but you could ask your chem teacher.

 

[+Po1ntDeXt3r+]

I don't know exactly why, but salt causes dehydration, and since water is a reactant, dehydration of the equilibrium will shift it to the left, producing more water and carbon dioxide.

After thinking about it, I'm guessing that the salt in solution is stopping the water from reacting with the carbon dioxide.

As you know, when a system is at equilibrium, the rate of the forward reaction is equal to the rate of the backward reaction. That is to say, some carbon dioxide is reacting with water to form carbonic acid. Salts, being ionic substances ionise in solution, so by adding NaCl, you are adding sodium and chlorine ions to the solution. From year 11 you learned that ionic substances dissolve well in water because water as a polar molecule has oxygen being slightly negative and hydrogen being slightly positive. So when NaCl dissolves in water, the positive Na⁺ is attracted to the negative oxygen atom while the negative Cl^- is attracted to the positive hydrogen atom.

So I'm guessing that when water molecules are "wrapped" in anions and cations, they can't react with carbon dioxide to form carbonic acid. Making a reactant unable to participate in the reaction is as good as removing it. Hence the equilibrium will shift to the left, producing more water and carbon dioxide.

My best guess.

ure kinda right.. but its not a dehydration per say.. because thats a specific physiological effect in medicine rather than the chemical change ure describing..

time to school the chem tutor

salting causes.. the diverse ion effect. its 1st yr uni chem.. so its right up ure ally
http://www.psigate.ac.uk/newsite/reference/plambeck/chem2/p01193.htm

here has a description.. both a oversimplified and accurate version

 

[Dreamerish*~]

ure kinda right.. but its not a dehydration per say.. because thats a specific physiological effect in medicine rather than the chemical change ure describing..

time to school the chem tutor

salting causes.. the diverse ion effect. its 1st yr uni chem.. so its right up ure ally
http://www.psigate.ac.uk/newsite/reference/plambeck/chem2/p01193.htm

here has a description.. both a oversimplified and accurate version

I can't say I understood all of it , but thanks for providing an answer.

Maybe this is why high-school teachers don't bother explaining the reason behind adding salt to aerated drinks.

 

[XcarvengerX]

I did this experiment at school laboratory by heating the soft drink. But, the liquid just spurt out uncontrollably...:mad1:

 

[eli_1988_1]

Hey
ok so some of u might recall helping me during the holidays with some questions i had about the decarbonating soft drink prac....
one question was about Le Chatelier's Principle and to give a specific example in nature. After hearing many of your opinions about the example in nature, i decided to write about a lake and its acidity. When i got my assessment back, the teacher has taken a mark off me for this, telling me i am wrong because lakes are not acidified (they have a "negligible amount"). I'm just wondering who is correct in this case???

Also, I used the heating method for the actual prac, where i made a water bath, with the bunsen, tripod, etc. and heated a bottle of soft drink. to compare the mass loss due to evaporation, i also heated a bottle of water at the same time, under exactly the same conditions. The water was my control as i subtracted this mass loss from evaporation to the mass loss of the soft drink to get the mass loss of carbon dioxide. My teacher has taken marks from me saying that this is wrong because it has to be the same drink. I have spoken to my tutor about this (a HSC chemistry teacher) who says i am correct, and also found a past hsc question with the same method as i used, showed my teacher the question and yet i am still wrong.
is she correct or do i have grounds to appeal??
An extra 2marks (1 for each of these things) will move my rank from 6th to 1st so it could potentially mean alot if I can get this sorted out.

Thanks in advance for your help!!
Any advice is appreciated!
Eli

 

[Dreamerish*~]

Hey
ok so some of u might recall helping me during the holidays with some questions i had about the decarbonating soft drink prac....
one question was about Le Chatelier's Principle and to give a specific example in nature. After hearing many of your opinions about the example in nature, i decided to write about a lake and its acidity. When i got my assessment back, the teacher has taken a mark off me for this, telling me i am wrong because lakes are not acidified (they have a "negligible amount"). I'm just wondering who is correct in this case???

Are you talking about a buffering solution in lakes? I wouldn't say lakes are "acidified" - some of them have a buffering system which consists of comparable amounts of a weak acid and its conjugate base - in this case, H₂CO₃ and HCO₃^-. The equation is:

H₂CO₃ + H₂O

H₃O⁺ + HCO₃^-

When acid is added, the concentration of H₃O⁺ is increased. The equilibrium adjusts to this change by shifting to the left to absorb some of it. When base is added, the OH^- reacts with H₃O⁺ to form water, removing the H₃O⁺ from the products side of the reaction. Furthermore, the water produced increases the concentration of H₂O, which is a reactant. The equilibrium then shifts to the right to absorb the extra water and to produce more H₃O⁺ to replace the loss.

There is a small amount of the buffer solution in lakes and rivers. I wouldn't say it's negligible, but I think your teacher was talking about something else. Anyhow, a simpler equilibrium reaction in nature would be the carbon dioxide equilibrium. Some carbon dioxide from the atmosphere is dissolved in rain water, resulting in the rain water being slightly acidic, even when there is no atmospheric pollution:

CO₂ + H₂O

H₂CO₃

Also, I used the heating method for the actual prac, where i made a water bath, with the bunsen, tripod, etc. and heated a bottle of soft drink. to compare the mass loss due to evaporation, i also heated a bottle of water at the same time, under exactly the same conditions. The water was my control as i subtracted this mass loss from evaporation to the mass loss of the soft drink to get the mass loss of carbon dioxide. My teacher has taken marks from me saying that this is wrong because it has to be the same drink. I have spoken to my tutor about this (a HSC chemistry teacher) who says i am correct, and also found a past hsc question with the same method as i used, showed my teacher the question and yet i am still wrong.
is she correct or do i have grounds to appeal??
An extra 2marks (1 for each of these things) will move my rank from 6th to 1st so it could potentially mean alot if I can get this sorted out.

Thanks in advance for your help!!
Any advice is appreciated!
Eli

I think your argument is justified. Water has the same boiling point whether it is a pure solution or a mixture - in this case a soft drink.

 

[Mountain.Dew]

Hey
ok so some of u might recall helping me during the holidays with some questions i had about the decarbonating soft drink prac....
one question was about Le Chatelier's Principle and to give a specific example in nature. After hearing many of your opinions about the example in nature, i decided to write about a lake and its acidity. When i got my assessment back, the teacher has taken a mark off me for this, telling me i am wrong because lakes are not acidified (they have a "negligible amount"). I'm just wondering who is correct in this case???

Also, I used the heating method for the actual prac, where i made a water bath, with the bunsen, tripod, etc. and heated a bottle of soft drink. to compare the mass loss due to evaporation, i also heated a bottle of water at the same time, under exactly the same conditions. The water was my control as i subtracted this mass loss from evaporation to the mass loss of the soft drink to get the mass loss of carbon dioxide. My teacher has taken marks from me saying that this is wrong because it has to be the same drink. I have spoken to my tutor about this (a HSC chemistry teacher) who says i am correct, and also found a past hsc question with the same method as i used, showed my teacher the question and yet i am still wrong.
is she correct or do i have grounds to appeal??
An extra 2marks (1 for each of these things) will move my rank from 6th to 1st so it could potentially mean alot if I can get this sorted out.

Thanks in advance for your help!!
Any advice is appreciated!
Eli

okay...heres my 2 cents

about le chatelier's principle...when people were talking about its relevance in nature, i think it has implied buffer systems, such as the amount of carbon dixoide in blood, or the amount of a weak acid in wine. yes, i agree that a lake will have some acidity, but i agree with the teacher that it has such a 'neglible amount' that the HSC doesnt factor in such small measures. the HSC always trys to simplify things to make ideas and concepts easier for students to understand. and that does mean sacrificing a degree of needed detail and accuracy. and, there is also the case that some rivers are not always acidic...some are basic, and some are neutral as well.

about ur decarbonating prac...you do have good grounds for appeal. its ridiculous that ur teacher is rejecting a perfectly valid method...i guess it just wasnt in the scope of what ur teacher taught you, and so, has rejected it.

but there is one word of advice. sometimes, IN THE HSC -and only in the HSC- for science subjects, it is not wise to always think outside the square.