Why phenolphthalein cannot show the endpoint for the complete titration of Na2CO3 with only phenolphthalein?

What's the reason that we consider only half of the moles of sodium carbonate when titrating with phenolphthalein indicator?

  • $\begingroup$ LOL - There is a difference between a solid theoretical answer and sheer speculation. If you won't accept acid-base theory I don't know how to help you. $\endgroup$
    – MaxW
    Commented Jan 4, 2016 at 6:48
  • $\begingroup$ I removed such stuff from the question. $\endgroup$
    – Mithoron
    Commented Jan 4, 2016 at 18:38

1 Answer 1


Suppose you want to find out the amount of sodium carbonate in a solution of sodium hydroxide. Let's use phenolphthalein as the indicator and HCl for titrating the mixture.

The first reaction that takes place is $$\ce{NaOH + HCl -> NaCl + H2O}$$ The pH at this point isn't alkaline enough for phenolphthalein to become colourless, since there's still base ($\ce{Na2CO3}$) present in the solution.

Then $\ce{Na2CO3}$ reacts. When all the $\ce{CO3^2-}$ becomes $\ce{HCO3^-}$, the pH is given by:

$$\mathrm{pH}=\frac{\mathrm{p}K_{\mathrm{a}1} + \mathrm{p}K_{\mathrm{a}2}}{2}$$

which is approximately $8.3$. (The above expression is for amphiprotic salts.) Phenolphthalein changes colour at approximately pH $8.2$.

So when the colour of phenolphthalein disappears, it means that all the carbonate is converted to bicarbonate. So if $x$ is the volume of HCl required to neutralize NaOH and $y$ is the volume of HCl required for complete neutralization, of carbonate, then phenolphthalein changes colour when volume of HCl used is $\displaystyle x+\frac{y}{2}$.

Remember that for complete neutralization, two moles of HCl is required for every mole of carbonate. For half neutralization only one mole of HCl is required. (Which is half the amount required for complete reaction).


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