You're on the right track and you ALMOST have it.
Would you just do this:
0.100M X 0.025 L = ?? mols
and then ?? mols / 18.65 mL?
remember definition of M = mols/Liter. Therefore, you must change 18.65 mL to liters. Otherwise, you have it down ok but see the note in italics that follows.
Technically you can't work the problem nor solve for the molarity of the solution without an equation. You may have one that was given to you and you just didn't copy it into this post. If you know what the acid is (HCl for example and I think that was the previous post) then you can write it
NaOH + HCl ==> NaCl + H2O.
The idea here, and what I'm cautioning you against, is that we assumed above that the mols of the acid and the mols of the base were the same and that works as long as the mole ratio in the equation is 1:1 as it is for HCl and NaOH. BUT, if you have an equation in which the ratio is NOT 1:1, for example,
H2SO4 + 2NaOH ==> Na2SO4 + 2H2O THEN
you must convert using the equation from mols of one reactant to mols of the other. In this case you were given standard acid of 0.100 M. Let's supposer that was H2SO4 and you have the equation I wrote above. Then 0.100 x 0.025 L = 0.0025 mols H2SO4.
Then 0.0025 mols H2SO4 x (2 mol NaOH/1 molH2SO4) = 0.0050 mols NaOH.
Then M NaOH = mol NaOH/L = 0.0050 mols/0.01865 L = ??molarity NaOH.
that did make sense, and my little mistake was the conversion from mL to L. i referenced into some of my material and i believe the standard acid is probably H2SO4 so i will try it out and see my result.
sorry to keep bugging you, but if it makes you feel better i am understanding it more.
what if you have the same numbers, but you are trying to find the concentration of an UNKNOWN acid. all you have is:
Volume of Unknown Acid
Volume of Stock Solution
Concentration of Stock solution..
how would you determine that? same process?