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I am planning to use an acid-base reaction with an indicator for some flow studies. I will be using aqueous solutions of sodium hydroxide (NaOH) and hydrochloric acid (HCl) with phenol red as the indicator. My plan is to use a 20% stoichiometric excess of acid (phenol red will convert from red to yellow after reaction).

I want to confirm that if I add phenol red to aqueous NaOH (prior to the reaction with HCl), then in order to get the actual concentration of NaOH available for subsequent reaction with HCl, I need to subtract the amount that was needed to convert the phenol red to it's basic form. As I understand it, as the pH increases, the phenol red's hydroxy group loses its proton (Wikipedia). It takes hydroxide ion in order for this proton to be lost, so that decreases the net concentration of NaOH available to react with HCl.

So the calculations would be as follows:

Initially want 0.01M NaOH as the reactant. Total volume of aqueous NaOH reactant = 10 liters. Therefore, the number of moles of NaOH = 0.1 moles of NaOH

Desire to add 4 grams of phenol red to NaOH. MW of phenol red = 354.38 grams per mole. Amount of moles of phenol red = 0.011287 moles.

Subtracting the moles of phenol red from moles of NaOH yields a net of 0.088713 moles of NaOH. So, if I want a net of 0.1 moles to be available to react with HCl, I need to increase the original amount of NaOH from 0.1 moles to 0.111287 moles. Alternatively, I could accept that the amount is 0.088713. Since the plan is to use a 20% stoichiometric excess of HCl, I would base the calculated amount of HCl needed on 0.088713 and not 0.1. The important thing is that I want 20% stoichiometric excess of acid.

I just want confirmation that I am thinking about this correctly. That is the need to correctly calculate the net amount of NaOH available for reaction so that I can correctly determine the amount of HCl to prepare.

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1 Answer 1

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Indicators used for acid-base titrations are dissolved in the acidic or basic solution in such a high dilution that the amount of acid or base necessary to produce a change of color is negligible. In your message you dissolve 4 g of phenol red into a NaOH solution. 4 g is such a huge amount that the phenol red is not working as an indicator any more. You will never use such a huge amount of indicator in a titration.

A substance may be called an indicator when its concentration in a solution is a couple of nanomoles per liter. The amount of base or of acid necessary to produce a color change is of the same order of magnitude (some nanomoles). This is negligible with respect to the amount of base and of acid to be titrated.

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  • $\begingroup$ I'm thinking micromoles per liter, not nanomoles. Neutral water autoionization is 100 nanomoles per liter, and you want your indicator capacity to exceed this. Micromoles is still negligible, though. $\endgroup$ Commented 2 days ago
  • $\begingroup$ I'm not sure I understand. As long as there is more of the reactants by an order of magnitude or so, then all of the indicator will change color. I want a stoichiometric excess of acid of 20% and I just want to make sure I am calculating it correctly because some of the NaOH will be needed to create the basic ionic form of the phenol red. Just because I am using a significantly higher concentration than usual, doesn't mean that the phenol red won't change color. I know for a fact that it will. I need a relatively high conc. because I'm working with thin films and it is hard to see color. $\endgroup$
    – rdemo
    Commented 2 days ago
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    $\begingroup$ Also, this is not a simple titration in a beaker. I started my post by saying "some flow studies". I'm using the neutralization reaction (i.e. mixing a flowing stream of dilute NaOH with indicator with a flowing stream of aqueous HCl) to get a rough idea on the speed of mixing within the system. The indicator will tell me when most of that mixing has occurred. $\endgroup$
    – rdemo
    Commented 2 days ago
  • $\begingroup$ One last thing, the actual concentration of phenol red in the final diluted base stream will be 200 ppm (by weight). The concentration of base will be about 0.0044 M and acid about 0.0053 M. $\endgroup$
    – rdemo
    Commented 2 days ago
  • $\begingroup$ Typical amount of used indicator during ordinary titrations is 1-2 drops of 1000 ppm w/w solution, what would be roughly 30-60 micrograms, with M about 300 g/mol it would be 100-200 nmol. if in 50 mL, then 2-4 microM. 200 ppm w/w seems too much, like adding 5 mL of stock ind. solution to 20 mL of sample, instead of few drops. Roughly 100 times more. Could be justified if evaluated across very short light path. $\endgroup$
    – Poutnik
    Commented 2 days ago

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