Estimation of CO2 by titrimetric method using phenolphthalein indicator

Question

We were estimating dissolved $\ce{CO2}$ in water by American Public Health Association (APHA) method. It was a titrimetric method using phenolphthalein indicator.

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

$$\ce{CO2 (g) ⇌ CO2 (aq)}$$ $$\ce{CO2 aq + H2O ⇌ H2CO3 ⇌ H+ + HCO3^2-} \label{rxn:1}\tag{1}$$

(Only $10\%$ of dissolved $\ce{CO2}$ get converted into $\ce{H2CO3}$.)

On addition of $\ce{NaOH}$:

\begin{align} \ce{2NaOH + CO2 &-> Na2CO3 + H2O} \label{rxn:2}\tag{2} \\ \ce{Na2CO3 + H2O + CO2 &-> 2NaHCO3} \label{rxn:3}\tag{3}\\ \ce{NaHCO3 &-> Na+ + HCO3-} \label{rxn:4}\tag{4} \\ \ce{\underset{from \eqref{rxn:1} and \eqref{rxn:4}}{HCO3-} + H2O &-> H2CO3 + OH-} \\ \ce{OH- + H+ &-> H2O} \\ \ce{HIn &⇌ H+ + In-} \end{align}

(The equilibrium shift to the right and when $[\ce{HIn}]:[\ce{In^-}]=1:10$ pink colour appears.)

But when we calculate the concentration of $\ce{CO2}$ we consider only reaction $\eqref{rxn:2}$, which means we are calculating only a portion of the molecular $\ce{CO2}$ while the $\ce{CO2}$ that was present in the solution and reacted in step 3 and the $\ce{CO2}$ (the $10\%$) that was directly converted into $\ce{H2CO3}$ (Rxn. $\eqref{rxn:1}$) are not taken into account.

So it means we are calculating only a part of the free $\ce{CO2}$ ($45\%$) and not all.

From this can we conclude that APHA's method is not very accurate at estimating the free $\ce{CO2}$ in water?

P. S. I read the procedure and also the calculations in a practical book.

Answer 1

But when we calculate the concentration of $\ce{CO2}$ we consider only reaction 2 which means we are calculating only a portion of the molecular $\ce{CO2}$ while the $\ce{CO2}$ that was present in the solution and reacted in Step 3 and the CO2 (the $\pu{10\%}$) that was directly converted into $\ce{H2CO3}$ (eq 1) are not taken into account.

Almost $\ce{H2CO3}$ in the sample will be converted to $\ce{HCO3-}$ as you proceed with the phenolpthalein endpoint. Search for "Speciation of Carbonate as function of pH" (Sorry I do want to immediately link the graph but it is only a http and stackexchange won't accept it).

At ~8.3 or so, all carbonate species is at the form of bicarbonate. All carbonic acid in the solution will be transformed as bicarbonate. The amount of carbonate ion is almost nonexistent. The $\ce{H+}$ released will be neutralized by NaOH, so aside from reacting directly with $\ce{CO2}$, it will also react with the $\ce{H+}$ released by $\ce{H2CO3}$, so (1) is accounted on the amount of NaOH that will be used.

Also, the sum of all equations (2) and (3) is just

$$\ce{2 NaOH + 2CO2 -> 2NaHCO3}$$

So in essence, for 1 mole of NaOH you react with one mole of $\ce{CO2}$ to produce one mole sodium bicarbonate.

In totality, for every one mole $\ce{H2CO3}$, you react with one mole NaOH. For every direct reaction of NaOH with $\ce{CO2}$, the ratio is $1:1$.

So I guess that's all reactions accounted for.