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.

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.

To find the molarity/concentration of HCl in the acid-base titration, you need to know the volume and concentration of NaOH used, as well as the balanced chemical equation for the reaction between NaOH and HCl. Let's break down the steps:

Step 1: Write the balanced chemical equation for the reaction. In this case, the reaction is:

NaOH + HCl → NaCl + H2O

Step 2: Determine the volume and concentration of NaOH used in the titration. Let's assume you used 20 mL of NaOH with a concentration of 0.1 M (moles per liter).

Step 3: Convert the volume of NaOH used to liters:

20 mL = 0.02 L

Step 4: Use the balanced equation to determine the stoichiometry between NaOH and HCl. From the equation, we see that one mole of NaOH reacts with one mole of HCl.

Step 5: Calculate the number of moles of NaOH used:

moles of NaOH = volume of NaOH (L) × concentration of NaOH (M)
= 0.02 L × 0.1 M
= 0.002 moles of NaOH

Step 6: Based on the stoichiometry, we can see that 1 mole of NaOH reacts with 1 mole of HCl. Therefore, the number of moles of HCl used is also 0.002 moles.

Now, let's address your specific questions:

1. To find the concentration of HCl, you should only consider the volume of HCl used, which in your case is 9 mL. The water is not relevant for calculating the concentration of HCl because it does not participate in the reaction and does not affect the concentration of the acid.

2. You are correct in your understanding. The water in the HCl does not affect the outcome of the titration since it does not react with NaOH. The number of moles of HCl used will remain the same regardless of the amount of water present.

Remember, to accurately determine the concentration of HCl, you need to know the volume and concentration of NaOH used, as well as perform the necessary calculations based on the balanced chemical equation and stoichiometry of the reaction.