Sciencemadness Discussion Board

Help with CO2 dissolved in H2O

bbackes - 31-3-2015 at 10:37

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!

Milan - 31-3-2015 at 12:36
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


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]

bbackes - 31-3-2015 at 12:57

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


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]

Milan - 31-3-2015 at 13:20

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]

bbackes - 31-3-2015 at 13:33

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]

blogfast25 - 31-3-2015 at 13:35

Quote: Originally posted by bbackes  
If someone can verify my work I would appreciate it. Thanks!


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]