bbackes
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Help with CO2 dissolved in H2O
Hello everyone, I'm wondering if someone could glance at this and let me know if everything is correct. I'm setting up an experiment to calculate how
many grams of CO2 are dissolved in a 25mL sample of carbonated water. Here is the equation:
CO2(aq) + NaOH(aq) -> NaHCO3(aq)
I will be using phenolphthalein as an indicator(pink) and 0.2M NaOH. Assuming I used 1.2mL of 0.2M NaOH would this math be correct:
0.2M NaOH X 0.0012L = 0.00024 moles NaOH
Since CO2 to NaOH is a 1:1 molar ratio this would mean 0.00024 moles of CO2 were present so:
0.00024 moles CO2 X 44.01g/mole = 0.012 grams of CO2 dissolved in the 25mL sample?
If someone can verify my work I would appreciate it. Thanks!
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Milan
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| Quote: |
I will be using phenolphthalein as an indicator(pink) and 0.2M NaOH. Assuming I used 1.2mL of 0.2M NaOH would this math be correct:
0.2M NaOH X 0.0012L = 0.00024 moles NaOH
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I'm not sure why would you multiply moles with liters, but if you put 0.2 M of NaOH you're still going to have 0.2 M. Assuming that those 0.2 M were
enough to react with all CO2 (you could always have excess CO2) then you have:
0.2 moles CO2 X 44.01 g/mole = 8.802 grams of CO2
Hope it helps! 
[Edited on 31-3-2015 by Milan]
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bbackes
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Quote: Originally posted by Milan  |
| Quote: |
I will be using phenolphthalein as an indicator(pink) and 0.2M NaOH. Assuming I used 1.2mL of 0.2M NaOH would this math be correct:
0.2M NaOH X 0.0012L = 0.00024 moles NaOH
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I'm not sure why would you multiply moles with liters, but if you put 0.2 M of NaOH you're still going to have 0.2 M. Assuming that those 0.2 M were
enough to react with all CO2 (you could always have excess CO2) then you have:
0.2 moles CO2 X 44.01 g/mole = 8.802 grams of CO2
Hope it helps!
[Edited on 31-3-2015 by Milan] |
M is molarity; I'm using a 0.2M solution of NaOH. moles=Molarity X volume(in Liters)
sorry I should have specified.
[Edited on 31-3-2015 by bbackes]
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Milan
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Oh, I see. I know what molarity is but I was never sure what was meant by that "M" (I thought it was moles all along). Well I've never encountered
that way of putting it in school (most probably difference in education system), I'm more used to putting it as moles/meter squared or moles/liter.
Thanks for explaining it (that "M" really made quite a lot of problems to me).
Just a question, what unit does it represent? Is it moles/liter?
Also if M is molarity then I think it is 0.011 g of CO2 that is dissolved.
[Edited on 31-3-2015 by Milan]
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bbackes
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Yes, Molarity= moles of solute/liters of solution
Cool, I calculated 0.012g CO2 so it should be correct.
[Edited on 31-3-2015 by bbackes]
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blogfast25
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I suspect this is a homework question disguised as an experiment but anyway.
Your math is correct.
The 'experiment' would benefit in accuracy by using 0.02 M NaOH, or even 0.015 M, because the titrant volume would be higher.
[Edited on 31-3-2015 by blogfast25]
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