# What are the reactions occurring during the titration of a carbon dioxide contaminated water sample with diluted sodium hydroxide solution?

During estimation of dissolved $\ce{CO2}$ in water, we took a water sample, added few drops of phenolphthalein and titrated it against dil $\ce{NaOH}$.

## Questions

What are the reactions that happen:

1. initially in the water sample?
2. after addition of phenolphthalein (if any)?

## My speculations

1. Inside the sample water

I think dissolved $\ce{CO2}$ reacts with $\ce{H2O}$ to form $\ce{H2CO3}$. But then does all $\ce{CO2}$ get converted into $\ce{H2CO3}$?

In an acidic medium the equilibrium shifts to the left and the weak acid phenolphthalein ($\ce{HIn}$) mostly remains undissociated and hence colourless.

$$\ce{HIn <- H+ (excess) + In-}$$

1. After addition of $\ce{NaOH}$

According to my book after $\ce{NaOH}$ is added to the sample the following occurs: \begin{align} \ce{2NaOH + CO2 &= Na2CO3 + H2O}\\ \ce{Na2CO3 + H2O + CO2 &= 2NaHCO3}\\ \end{align} From my understanding \begin{align} \ce{NaHCO3 &= Na+ + HCO3-}\\ \ce{HCO3- + H2O &= H2CO3 + OH-}\\ \ce{OH- + H+ + In- &= H2O + In-}\\ \end{align} As the $\ce{H+}$ concentration decreases equilibrium shifts to the right $$\ce{HIn -> H+ + In-}$$

This $\ce{In-}$ ion imparts the pink colour which at pH 8.3 becomes susceptible to our eyes.

At the beginning some of the dissolved $\ce{CO2}$ got converted into $\ce{H2CO3}$. They remain beyond the scope of estimation, isn't it?

Are my speculations correct?

What you are effectively titrating is carbonic acid, or more precise the hydrogen carbonate ion. Carbonic acid is not very stable in water, an most of the carbon dioxide stays the way it is, i.e. dissolved and interaction via van der Waals forces. The main equilibria are as follows (and derivations thereof): \begin{align} \ce{CO2 + 2H2O &<=> HCO3- + H3O+}\tag1\label{major}\\ \ce{HCO3- + H2O &<=> H2CO3 + OH-}.\tag2\label{minor}\\ \end{align} In a contaminated water sample $\eqref{major}$ is by far the more predominant reaction.
When you now add sodium hydroxide solution you shift $\eqref{major}$ to the right side, as you are removing hydrogen carbonate ions via $$\ce{HCO3- + OH- <=> CO3^2- + H2O}\tag3\label{conversion}$$
As you correctly identified, phenolphtalein is a weak acid by itself and the colour change happens via $$\ce{H2In + 2OH- <=> In^2- + 2H2O}.\tag4\label{indicator}$$
Once you have added enough sodium hydroxide solution to completely shift the equilibrium $\eqref{conversion}$ to the right, you will start deprotonating the indicator molecule via $\eqref{indicator}$.