You would do better to post your question instead of parts of it.
In general, however, in a titration of NaOH and HCl, I assume you have the molarity of NaOH and are trying to find the molarity of HCl.
mols NaOH = molarity NaOH x Liters NaOH.
mols HCl = mols NaOH (from the equation).
mols HCl = M x L. (You will know mols and you will know liters; calculate molarity).
In regard to the last part, no, water added to the titration vessel during titration does not change the results for exactly the reason you cite; i.e., the reaction is mol for mol and water dilutes BOTH acid and base at the same time. However, any water added to the HCl BEFORE the titration WILL change the molarity of HCl.
When you complete a acid base titration of NaOH and HCl and how to you find the molarity/concentration of HCl. I used 9 mL of HCl with 50 mL of water. But when solving for concentration, do I use 9 mL or 59 mL (water + HCl)?
Also, does the water in the HCl affect the outcome of the titration? I don't think so since the amount of moles of HCl remains the same but I'm not too sure.
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