# Net concentration of sodium hydroxide after addition of phenol red indicator

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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