Were were estimating dissolved $\ce{CO2}$ in water by American Public Health Association method. It was a titrimetric method using phenolphthalein indicator.
Titrant used was $\ce{NaOH}$ and analyte was sample water.
Our instructor said that on adding phenolpthalein if the water sample turns pink prior to the experiment (addition of $\ce{NaOH}$) then it can be concluded that no free $\ce{CO2}$ is present . But how's it so?
Phenolphtalein's color change does appears at around pH=8.2.
If you just bubble $\ce{CO2}$ in water, the most of the $\ce{CO2}$ is dissolved into the water and only a small fraction converts to $\ce{H2CO3}$. But if the solution is very basic then all the $\ce{CO2}$ will react via a series of reactions and there is no free $\ce{CO2}$ left in water.
$\ce{CO2(aq) + H2O <=> H2CO3}$
$\ce{H2CO3 + 2OH^- <=> CO3^2- + 2H2O}$
Well more truthfully there is an insignificantly small amount of $\ce{CO2}$ in basic solutions since we discussing chemical equilibria. So $\ce{[CO3^{2-}] >> [CO2]}$
You're doing a titration of the carbonic acid $\ce{H_2CO_3}$ which is the product of the dissolution of $\ce{CO_2}$ in water by a base here the sodium hydroxide. According to wikipedia for carbonic acid, $\ce{pKa_1=3.6}$, $\ce{pKa_2=6.3}$ and $\ce{pKa_3=10.32}$.
When the carbon dioxyde is in water, the solution is acid then your phenolphtalein is colorless. During your titration while not all the carbonic acid as react then the solution will still be colorless because phenolphtalein's color change appears at around $\ce{pH=8.2}$ (Ref). Then when you finish you're titration the color of the solution will change.
You need to make some prediction to find the theorical pH at the equivalence. In this case at the equivalence $\ce{[H^+]=\sqrt{Ka_1Ka_2}}=1.26 \times 10^{-10}$ then $$\ce{pH_{eq}=-\log_{10}(1.26 \times 10^{-10})=9.9}$$
So at this point the phenolphtalein must be pink and then no more acid (so carbone dioxyde) will be in your solution. You'll have only carbonate ion which will not react with sodium hydroxyde.
