Titration of 25.00 mL of a monoprotic acid solution with 0.2269 M NaOH is:

Question

Initial NaOH = 3.96 mL
Final NaOH = 38.84 mL (dif is 34.88mL)
Based on data what is the molar concentration of acid?
I did
V x M = 0.03488L NaOH x 0.2269 = 0.079143
Not sure if I should have switched to Liters.
Then M1 x V2 = M2 x V2
M Mor Acid x 0.025 L = 0.2269M NaOH x 0.03488L = 0.07914 / 0.025 = 3.1656

Is this correct or should I have left them in mL

Answer 1

Actually it makes no difference as long as you keep the units straight. However, I think you made a math error of about 10 tims.
0.03488 x 0.2269 = 0.007914 mol NaOH.
mols acid = mols NaOH
M acid = mols acid/L acid = 0.007914/0.025 = aboaut 0.32M but you need to redo, do it more accurately, and answer with the right number of significant figures.

If you to do it in mL, it is done this way.
34.88 x 0.2269M = 7.914 millimoles.
Then M = mmoles/mL = 7.914/25 = about 0.32 M.

Answer 2

Oops. Thanks. I see I copied 0.007943 incorrectly. I'll be more careful in the future.