[Anon]

A 0.397 g sample of KHC₈H₄O₄ (molar mass=204.44 g/mol)  is dissolved w/ 50mL of deionized water in a 125mL flask.  It is titrated to the phenolphthalein endpoint w/ 16.22 mL of a NaOH solution.  What is the molar concentration of the NaOH?

I know that M=n/V_L.  What I don't know is how to correctly set up the problem. 
Through various calculations and formulas I came with an answer of .01197 M NaOH.  Is there someone that could double check to see if my answer is correct?

Thanks in advance for any help that is given!



(Chemistry is my Kryptonite)

[DevaDevil]

almost correct; an order of magnitude too low.

The correct answer is 0.120 M to be precise (3 significant figures: 0.397 g was your starting weight)

0.397 g / 204.44 g/mol = 1.94 ^. 10^-3 mol
1.94 ^. 10^-3 mol / 16.22 ^. 10^-3 l = 0.120 mol / l

Of course assumed that the KHC₈H₄O₄ fully dissociates into K⁺ and  HC₈H₄O₄^-