If you titrate some hydrogen bromide solution (0.1 M) to some sodium hydroxide in order to determine the exact concentration of hydrogen bromide (0.1 M), and you use anthocyanins as a pH indicator. This anthocyanin will be red if you have a low pH value and purple if it is slightly acidic until it is slightly basic. It will be blue if the solution is basic.

Do you have to titrate until the color of the solution is blue or purple?

Equilibrium from red to purple: $$\ce{H2X+ <=> HX}$$

$K_a = 1.047*10^{-3}$

Equilibrium from purple to blue: $$\ce{HX <=>X^{−}}$$
$K_a = 3.16 * 10^{-8}$


2 Answers 2


If you titrate some hydrogen bromide solution to some sodium hydroxide [...]

It's probably a bad idea to start with the anthocyane in strongly alkaline solution. Like other pyrylium compounds, flavylium cations do not tolerate (strongly) akaline conditions very well.

The nucleophilic opening of the pyrylium ring to a chalcone is irreversible under the conditions of the titration.


First of all, I don't understand why you would titrate the acid of unknown concentration to your base of known concentration, I've learned it the other way around.

Now, following my proposal, we have our solution of $\ce{HBr}$ waiting to be titrated with $\ce{NaOH}$. Since sodium hydroxide is a fairly strong base, we can assume total neutralization: $$ \ce{HBr + NaOH -> NaBr + H2O}$$

Operating under this assumption, equilibrium is reached at $\text{pH}=7$, so we have to know when that happens. How do we find out where the indicator changes colour?

Well, we calculate the $\mathrm{pK_a}$ values of the indicator: $$ \mathrm{pK_a} = -\log \mathrm{K_a}$$

The second equilibrium (from purple to blue) switches at $\text{pH} = 7.5$, which is about where we need it to. So you need to titrate until you get the blue colour.

Note: It is not always easy to see these colour transitions, I myself have overtitrated quite a few solutions because the colour shift was too subtle (and my eyes too untrained).

  • $\begingroup$ How did you calculate pH at which the second equilibrium swithes did you use Henderson–Hasselbalch equation? $\endgroup$
    – G M
    Jan 25, 2014 at 14:32
  • $\begingroup$ The pKa is 7.5, so at pH=7.5 the switch happens. $\endgroup$
    – tschoppi
    Jan 25, 2014 at 14:33
  • $\begingroup$ I'm sorry I didn't catch it, I thought pKa indicates how much an acid is dissociate not at which pH a specie dissociate... $\endgroup$
    – G M
    Jan 25, 2014 at 14:43
  • $\begingroup$ It's the pKa of the indicator. The "a" merely refers to its property as an acid. $\endgroup$
    – tschoppi
    Jan 25, 2014 at 14:45
  • $\begingroup$ mmm I know that is the pKa of the indicator for this I'm not sure how you can correlate it with a system composed from this with additional $H^{+}$ ions from another acid... I will meditate about it! Thanks however! $\endgroup$
    – G M
    Jan 25, 2014 at 15:03

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