# Estimation of CO2 by titrimetric method using phenolphthalein indicator

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.

• The equations are all a bit convoluted, but remember that CO2 + H2O and hydrogen carbonate + hydronium are also in equillibrium. – Karl Dec 17 '16 at 10:04
• In equation 1, you have to check the pH of your water sample and check the pKa of carbonic acid, bicarbonate and carbonate equilibrium. – Another.Chemist Dec 20 '16 at 7:27

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.