An unknown amount of water is mixed with

Question

350 mL of a 6 M solution of NaOH solution.
A 75 mL sample of the resulting solution is
titrated to neutrality with 51.2 mL of 6 M
HCl. Calculate the concentration of the diluted NaOH solution.
Answer in units of M

What volume of water was added to the
350 mL of NaOH solution? Assume volumes
are additive. Answer in units of mL

Answer 1

This is the corrected copy. An equal sign I typed as a - sign.

millimoles HCl = 6M x 51.2 = 3072.
mmoles NaOH solution (75 mL) = 3072.
M NaOH soln = mmoles/mL = 3072/75 = 4.096M That's the molarity of the diluted solution.

The diluted solution started out as 6M and it was diluted and is now 4.096M. Let x = mL water added to the 6M NaOH.
6M x [350/(350+x) = 4.096
I calculated x to be approximately 163 mL so the final volume would be about 350 + 163 or about 513 mL. You need to clean up the numbers and watch the number of significant figures but this is the way you do it.