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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 $$\ce{2NaOH + CO2 -> Na2CO3 + H2O}.$$

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.

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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 $$\ce{2 NaOH + CO2 -> Na2CO3 + H2O}$$ 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?

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

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.

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  • $\begingroup$ I have a follow up question: While calculating the number of moles of NaOH required to perform the reaction, We took the conc. of NaOH (in M) and volume of the titrant (which is usually in ml). Shouldn't we convert the $ml$ into $l$? $\endgroup$ – Tyto alba Apr 17 '17 at 12:57
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    $\begingroup$ Simple answer is yes. This is relatively straightforward but is a common stumbling block for many. Molarity, M, is the number of moles of substance in 1 litre. So a 0.5 M solution of NaOH has 0.5 moles in 1 litre of solution. So 1 litre is 1000 mL, so we have 0.5 moles in 1000 mL, or 0.0005 moles in 1 mL. If the volume of titration was 19.5 mL, the number of moles used is 0.0005 moles/mL x 19.5 mL = 0.00975 moles. So make sure units are similar and they should cancel out to give the right unit for the variable you are looking for. $\endgroup$ – Leeser Apr 17 '17 at 23:00
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If your end point is a consistent light pink, around a pH of 8.3 with phenolphthalein then this calculation is wrong. While it does take $\ce{2NaOH}$ to strip both hydrogens off of $\ce{H2CO3}$, at a pH of 8.3 there is little to no carbonate formed. Instead it is nearly all bicarbonate. The reaction from $\ce{NaOH}$ and $\ce{H2CO3}$ to bicarbonate ($\ce{HCO3}$) is 1 to 1. So instead of $\pu{22g/mol}$ you would just use $\pu{44.01g/mol}$ (the molar mass of $\ce{CO2}$)

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