We were estimating the amount of dissolved $\ce{CO2}$ in water using the American Public Health Association method. It was a Titrimetric Method using phenolphthalein indicator.

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

The following questions came to my mind:

  1. Why does the phenolphthalein $\ce{[HIn]}$ solution change colour at pH 8.2-10 when addition of a little titrant $\ce{NaOH}$ could result into production of some $\ce{In−}$ ions, the ions that give the colour to the solution. Why does it show colour change in only when there's $10^{-5.8}$ to $10^{-4}$ free $\ce{OH-}$ in the water?

  2. Is the source of this resulting basic pH, the titrant $\ce{NaOH}$? I think so because all the $\ce{OH-}$ formed from the reaction of titrant $\ce{NaOH}$ and analyte $\ce{CO2}$ reacts with the $\ce{H+}$ ions of phenolpthalien. So its only possible that the shift from pH 7 to pH 8.2 is due to the $\ce{OH-}$.


I am not absolutely sure I understood the question completely, but I will try to shed some light on the colour change of an indicator.

As a rule of thumb the human eye can make out a change in colour when there is a 10:1 ratio of the components. Since phenolphtalein, $\ce{HIn}$, is without colour, and its deprotonated form $\ce{In-}$ is pink, it means that a solution with

  1. $\displaystyle \frac{n(\ce{HIn})}{n(\ce{In-})} = \frac{10}{1}$ will appear clear, while
  2. $\displaystyle \frac{n(\ce{HIn})}{n(\ce{In-})} = \frac{1}{10}$ will appear pink.

Since phenolphtalein is a weak acid, you will see, that the colour change will theoretically happen around equilibrium constant, i.e. $\mathrm{pH}=\mathrm{p}K_\mathrm{a}$. (Which is also a simplification, because phenolphtalein has multiple acidic protons.) At that point you will probably not notice significant changes. Though phenolphtalein is a bit of an exception here, because it has only one colour. The human eye can probably see a smaller surplus of the colourful component. (It is much more difficult for an indicator like methyl red.)

If you substitute the above fractions into the Henderson-Hasselbalch equation, you can approximate the range in which the colour change will happen: \begin{align} \mathrm{pH} &= \mathrm{p}K_\mathrm{a} - \lg\left(\frac{n(\ce{HIn})}{n(\ce{In-})}\right)\\ &=% \begin{cases} \mathrm{p}K_\mathrm{a} - \lg\left(\frac{10}{1}\right); & \text{clear}\\ \mathrm{p}K_\mathrm{a} - \lg\left(\frac{1}{10}\right); & \text{pink} \end{cases}\\ \mathrm{pH} &= \mathrm{p}K_\mathrm{a} \pm 1 \end{align}

When you are estimating the amount of carbon dioxide dissolved in water, then the following equilibrium will be present: $$\ce{CO2 + H2O <=> HCO3- + H+ <=> H2CO3}$$

Adding hydroxide ions you are shifting this equilibrium to the right, essentially deprotonating the acid. Once you have fully (in first approximation) converted it to carbonate, you start deprotonating the indicator until you notice the colour change.

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