Question: Justify the use of phenolphthalein $(\mathrm{p}K_\ce{in}=9.4)$ as a indicator for the titration of benzoic acid $(K_\mathrm{a}=6.3\times 10^{-5})$ with $\ce{NaOH}$.

My attempt:

I can calculate the $K_\mathrm{b}$ of $\ce{NaOH}$ from the above data.

And I also know the formula

$$\mathrm{pH} = \mathrm{p}K_\mathrm{b} + \log \frac{\ce{In}}{\ce{HIn}}$$

I know that this is the titration of a strong base and a weak acid. So the equivalent point will be at a $\mathrm{pH}>7$ and hence phenolphthalein is a good choice.

But how can I conclude mathematically that this is a correct choice?

  • 1
    $\begingroup$ I am slightly confused. Your Henderson-Hasselbalch equation should include the most relevant species which are benzoic acid and benzoate. Also, what you did was calculate and thus it is a mathematical proof, isn’t it? $\endgroup$
    – Jan
    Jun 5, 2016 at 20:15
  • 1
    $\begingroup$ Geez simply compare pKb of acid with pKind. If it's similar then OK. $\endgroup$
    – Mithoron
    Jun 6, 2016 at 1:14

3 Answers 3


You have to take two reactions into account.

First the protolysis of benzoic acid

$$\ce{HA + H2O <=> H3O+ + A-}$$ $$k_A = \frac{[\mathrm{H_3O^+}] [\mathrm{A^-}]}{[\mathrm{HA}]} \tag{1}$$

and second the (simplified) protolysis of phenolphthalein

$$\ce{HInd + H2O <=> H3O+ + Ind-}$$ $$k_{Ind} = \frac{[\mathrm{H_3O^+}] [\mathrm{Ind^-}]}{[\mathrm{HInd}]} \tag{2}$$

The color change is around the point where the concentrations $[\mathrm{HInd}]$ (colorless) and $[\mathrm{Ind^-}]$ (colored) are equal, i.e. $$k_{Ind} = [\mathrm{H_3O^+}] \tag{3}$$

If you substitute (3) into (1) and take the negative logarithm to base 10 you get

$$\mathrm{pK_A} - \mathrm{pK_{Ind}} = -\log\frac{[\mathrm{A^-}]}{[\mathrm{HA}]} = 4.4 - 9.4 = -5.2$$

what shows that the benzoic acid has been completely titrated at this point.

  • 1
    $\begingroup$ One needs to compare pKb with pKind here $\endgroup$
    – Mithoron
    Jun 6, 2016 at 1:12
  • $\begingroup$ This is what the difference $\mathrm{pK_A} - \mathrm{pK_{Ind}}$ does. $\endgroup$
    – aventurin
    Jun 6, 2016 at 15:18

The answers given by SchrodingersCat and aventurin are both bogus.

The gist is that for an titration with sodium hydroxide we'd like;

  1. A blank to use essentially zero volume of the titrant.
  2. The acid to be "completely" neutralized

We can assume a fairly standard situation for the analysis. We can assume 0.1000 molar NaOH as the titrant and want to use about 30 ml of a 50 ml burette. That would mean weighing out about 3.0 millimoles of benzoic acid (0.37 grams) which would be dissolved in 25 ml of distilled water.

Significant Figures

All calculations done in chemistry involve errors. We can assume that at best a 50 ml burette must deliver at 0.01 ml to be detectable. We'd normally use about 30 ml of titrant. Thus any error or bias less than 0.33 parts per thousand is insignificant.

To give ourselves a bit of leeway, we can assume a desired precision of 0.1% overall, or about 1 part per thousand.

You have to be careful here since 10 factors with -0.1 parts per thousand bias each would introduce a -1.0 ppt error overall which would be significant.

Preparation of Phenolphthalein Indicator (Acid/Base Indicator)

Use a fairly standard "recipe" for this.

  1. Weigh out 0.5g of phenolphthalein.
  2. Prepare 50% ethyl alcohol solution contained of 50mL ethanol and 50mL water.
  3. Now, dissolve the phenolphthalein in the 50% ethyl alcohol solution.
  4. Store in dropper bottle for use.

Let's say that 2 drops of the indicator would be used for a titration. Then that would be:

moles indicator = $\dfrac{2}{20*100}\dfrac{0.50\ \mathrm{grams}}{318.328\ \mathrm{g/mol}} = 1.6\times10^{-6}$

Blank Value

Now for the blank we'd just titrate 25 ml of distilled water with two drops of the indicator. The notion here is that if 0.01 ml of titrant would change the indicator then we can ignore the necessity of using a blank. (Think of the blank as "noise" in a signal-to-noise measurement. No noise means we're only measuring signal.)

Now 0.01 ml of 0.1 molar NaOH is $1\times10^{-6}$ moles of base ($\ce{OH^-}$). In 55 ml of solution that yields a pH of 9.3. The pKa of phenolphthalein is 9.4 so almost half of the phenolphthalein would converted to the colored form so the color change should be very detectable at that point.

Here we have ignored how much base that would react with the indicator itself. This is mitigated because we only have $1.6\times10^{-6}$ moles of indicator in solution. If we assume half of the base reacts with the indicator we still have a pH of about 9.0 for the solution.

Assuming that half the indicator must change color to be detectable puts us on the ragged edge of needing a blank correction. Since we used $3\times10^{-3}$ moles of acid, but $1.6\times10^{-6}$ of indicator that gives a ratio of about 0.3 parts per thousand. In reality the color change should be detectable at a pH of about 8.5, so much less than half of the indicator will react to make the colored form.

Thus all in all no blank should be needed.

"Total" Reaction of the Acid

For any acid , $\ce{A}$, we can assume the ionization reaction to be:

$\ce{HA + H2O <=> H3O+ + A-}$


$K_a = \dfrac{[\ce{H_3O^+}] [\ce{A^-}]}{[\ce{HA}]}$

Now at the pKa we know that $\ce{[A^-] = [HA]}$. We can't ever remove all of the $\ce{HA}$ form since we can always calculate the relative amounts of $\ce{HA}$ and $\ce{A^-}$ based on the pH. So let's calculate pH at which:

$\ce{[HA] = 1\times10^{-3}[A^-]}$

That turns out to be at:

$\ce{[H^+] = 1\times10^{-3}K_a}$

or at $\text{pH(99.9% conversion)} = \mathrm{p}K_a + 3.00$

Given that benzoic acid has a pKa value of 4.20 that means that at a pH of 7.20 99.9% will be in the unprotonated form.

We can assume that the amount of acid necessary to take the solution from a pH of 7.00 to a pH of 7.20, or even about 8.6 is negligible from our blank calculations.


As far as I have understood, I have to compare the $\mathrm{p}K_\mathrm{b}$ of sodium benzoate with the $\mathrm{p}K_\mathrm{ind}$ of phenolphthalein.

Thus what I get is:

$K_\mathrm{b}$ of sodium benzoate $=\frac{K_\mathrm{w}}{K_\mathrm{a}}=1.59\times 10^{-10}$

$\mathrm{p}K_\mathrm{b}$ of sodium benzoate $=9.79$

$\mathrm{p}K_\mathrm{ind}$ of phenolphthalein $=9.4$

So, $\mathrm{p}K_\mathrm{ind}$ lies in between the values $\mathrm{p}K_\mathrm{b} \pm 1$.

Hence phenolphthalein is a good indicator for the titration.

Thanks all for helping.

  • $\begingroup$ This reasoning is totally wrong. You want a difference between the pKa's of the acid and the indicator. Using the pKb (14 -pKa) of the acid is just wrong. Consider if the pKa of the acid was 2. Then this reasoning says that the pH of the final solution should be +/- 1 pH unit of pH 12. That would introduce a significant blank value which would then need to be corrected for. $\endgroup$
    – MaxW
    Jan 10, 2018 at 19:05

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