[bbackes]

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!

[unsu]

There are actually two reactions taking place there at different rates:

CO₂ + H₂O  H₂CO₃    [slow]
H₂CO₃ + OH^-  HCO₃^- + H₂O    [very fast]

Carbon dioxide is present in aqueous solutions mainly as molecular CO₂. When you titrate the solutions of carbon dioxide, NaOH quickly reacts with HCO₃^- and you often get a fading end-point. This is not a true end-point of the titration, because CO₂ is still there, it will slowly react with water to produce more carbonic acid, which acidifies the solution, and the purple colour slowly disappears. Please keep this in mind when you are setting up the experiment.

For more information, review Housecroft and Sharpe Inorganic Chemistry 4ed textbook problem 7.22 on  page 239 and the information on page 462.

The stoichiometry of your calculations look correct. But it is very important to get true titration end-point.
Another thing is I would use lower concentration of NaOH for better accuracy, 0.020 M for example

[Furanone]

There are actually two reactions taking place there at different rates:

CO₂ + H₂O  H₂CO₃    [slow]
H₂CO₃ + OH^-  HCO₃^- + H₂O    [very fast]

Carbon dioxide is present in aqueous solutions mainly as molecular CO₂. When you titrate the solutions of carbon dioxide, NaOH quickly reacts with HCO₃^- and you often get a fading end-point. This is not a true end-point of the titration, because CO₂ is still there, it will slowly react with water to produce more carbonic acid, which acidifies the solution, and the purple colour slowly disappears. Please keep this in mind when you are setting up the experiment.

For more information, review Housecroft and Sharpe Inorganic Chemistry 4ed textbook problem 7.22 on  page 239 and the information on page 462.

The stoichiometry of your calculations look correct. But it is very important to get true titration end-point.
Another thing is I would use lower concentration of NaOH for better accuracy, 0.020 M for example

Wow, great answer here, with references and all