What is the molarity of a NCL solution if 37.50 mL of the acid is needed to neutralize 25.00 mL of 0.0725 M NaOH?

I'm not quite sure how to go about doing this one.

And also this one thing.

The following technical errors were committed in standardizing the NaOH. What will be the effect of each of each on the calculated value of the molarity of NaOH?
~75 mL of water was used to dissolve KHP instead of recommended 50 mL

I have never heard of two of the chemicals you have mentioned: NCl and KHP. Please verify that you have written the formulas correctly.

Did you mean HCl instead of NCl? And KHP is potassium hydrogen phthalate. The answer to the second part of the question is "essentially no error is caused by the added 25 mL water."

My apologies for the confusion. Yes, I assumed that you meant HCl instead of NCl since NCl is not a commonly used chemical formula. And you are correct, KHP stands for potassium hydrogen phthalate. Thank you for clarifying.

Now, let's address your first question about determining the molarity of an HCl solution.

To find the molarity (M) of a solution, you can use the formula:

M1 * V1 = M2 * V2

Where:
- M1 is the molarity of the first solution (0.0725 M NaOH)
- V1 is the volume of the first solution used (25.00 mL or 0.025 L)
- M2 is the molarity of the second solution (unknown, HCl)
- V2 is the volume of the second solution used (37.50 mL or 0.0375 L)

Substituting the known values into the equation:

(0.0725 M NaOH) * (0.025 L) = M2 * (0.0375 L)

Now, rearranging the equation to solve for M2:

M2 = (0.0725 M NaOH * 0.025 L) / (0.0375 L)

M2 ≈ 0.04833 M

Therefore, the molarity of the HCl solution is approximately 0.04833 M.

Moving on to your second question regarding the errors in standardizing the NaOH solution, specifically the use of 75 mL of water instead of the recommended 50 mL.

In the standardization process, the purpose of dissolving KHP is to determine the exact molarity of the NaOH solution. The volume of water used in dissolving KHP doesn't impact the molarity calculation directly, as long as the correct amount of KHP was used. The additional 25 mL of water may dilute the resulting solution slightly, but since the exact mass of KHP used is known, it can still be used to accurately determine the molarity of NaOH.

Thus, the effect of using 75 mL of water instead of 50 mL is minimal and should not significantly impact the calculated value of the molarity of NaOH.