Biochem - 9-6-2005 at 07:02
Ex: CO2 + H20 --> H2CO3
H2CO3 + NaOH --> NaCO3 + H20
Would it be because the formation of H2CO3 being an acid would screw up the titration with NaOH or that As CO2 gas is removed from the closed bottle,
the pressure inside the bottle decreases. The atmospheric pressure on the outside remains the same and will collapse the bottle.
Referenced from: chemlearn.chem.indiana.edu/demos/naohco2.htm
[Edited on 9-6-2005 by Biochem]
sparkgap - 9-6-2005 at 07:07
It's more of the carbon dioxide-carbonate equilibrium; although weakly acidic, it can be sufficient to skew the titration results.
That's why NaOH solutions are prepared from boiled water; and standardization and titrations are usually done on the same day.
sparky (^_^)
Biochem - 9-6-2005 at 07:32
You did not use boiled distilled water to make your KHP solution. What effect might this have on your standardization – would it raise or lower the
concentration you determined from your NaOH solution?
so that is to say if water, NaOH wasn't boiled on the same day or if NaOH is left out and then used to titrate an unknown acid the calculated
concentration of NaOH would be less than origianally?
My work: Moles of KHP x (1NaOh/1KHP) x (1/vol. of NaOH(L)) = (M) of NaOH
According to the equation if NaOH were affected by H2CO3 it would be less basic therefore more NaOH would be added and since vol of NaOH added is in
the denominator it would lower the end value...is my thinking correct?
[Edited on 9-6-2005 by Biochem]
sparkgap - 9-6-2005 at 07:48
KHP solution is slightly acidic; Le Chatelier's principle applies, so carbon dioxide would tend not to go into solution. 
"...so that is to say if water, NaOH wasn't boiled on the same day or if NaOH is left out and then used to titrate an unknown acid the
calculated concentration of NaOH would be less than originally?"
Yes. That is why we standardize before titrating.
sparky (~_~)