How to determine the equivalent weight of CO2 We were estimating dissolved $\ce{CO2}$ in water by American Public Health Association method. It was a titrimetric method using phenolpthalein indicator.

Titrant used was $\ce{NaOH}$ and analyte was sample water. 

After the end point was reached and the volume of titrant estimated the ultimate calculation was done to find out the amount of $\ce{CO2}$ present in the sample per litre. The formula mentioned in the book was

$$\ce{CO2} \mathrm{(mg/l)} =\frac{\text{[Volume of NaOH} \times \text{Conc. of NaOH (in Normality)}\times 22 \times 1000]}{\text{Volume of sample water}}$$

$22$ here is the equivalent weight of $\ce{CO2}$. My question is how's it so?

NOTE: 

Our teacher said that since 2 molecules of $\ce{NaOH}$ react with 1 molecule of $\ce{CO2}$ to form $\ce{Na2CO3}$ so $n=2$ here. 

Is it right? I doubt it because I have always known $n =$ the number of $\ce{H+}$ or $\ce{OH-}$ released by an acid /base or the number of electrons released or accepted in a redox reaction.

The reaction here is
$$\ce{2NaOH + CO2 -&gt; Na2CO3 + H2O}.$$

But as $\ce{CO2}$ is not an acid or base and even doesn't undergo redox so how do we determine its equivalent weight?

P.S. I'm a Zoology student and have learned chemistry for XII + 2 years.
This is the danger of using formulae from books without fully understanding where the formulae came from.



So let's take it step by step. 

We have a titration of an unknown concentration solution with a known (standardised) solution.

The reaction $$\ce{2 NaOH + CO2 -&gt; Na2CO3 + H2O}$$ gives us the stoichiometric relationship between $\ce{NaOH}$ and $\ce{CO2}$, i.e. 2 moles of $\ce{NaOH}$ react with 1 mole of $\ce{CO2}$.

How much $\ce{CO2}$ Was used? Half the amount of $\ce{NaOH}$!

How much $\ce{NaOH}$ was used? 

$$\begin{align}
c(\ce{NaOH})  &amp;= \frac{\text{amount of substance}}{volume}\\
              &amp;= \frac{n_{\ce{NaOH}}}{V}\\
n_{\ce{NaOH}} &amp;= c(\ce{NaOH})\times V\\
n_{\ce{CO2}}  &amp;= \frac{1}{2} n_{\ce{NaOH}} \\
              &amp;= \frac{1}{2} c(\ce{NaOH})\times V\\
\end{align}$$

What is the concentration of $\ce{CO2}$ in sample?

$$c(\ce{CO2}) = \frac{n_{\ce{CO2}}}{\text{Volume of sample}} = \frac{1}{2}\frac{c(\ce{NaOH})\times V}{\text{Volume of sample}}$$

Concentration is also expressed as $\mathrm{mg/L}$. 1 mole of $\ce{CO2}$ weighs $44~\mathrm{g}$ or $44\times 1000$ mg so sticking this into the above

$$c(\ce{CO2})\mathrm{(mg/L)} = \frac{1}{2}\frac{c(\ce{NaOH})\times V}{\text{Volume of sample}} \times 44 \times 1000 = \frac{c(\ce{NaOH})\times V\times 22\times 1000}{\text{Volume of sample}}$$

That's where the $22$ came from.
This may not address exactly where the equivalent weight of $\ce{CO2}$ comes from but at least it shows you the derivation of how the $22$ got into the formula.