[ohaychminus]

I feel like an utter idiot not being able to solve this. I'll show the work I've done so far.

14.41 mL of 0.1569 M aqueous HCl is added to 36.27 mL of 0.2851 M aqueous NaOH at 25°C. Assuming the volumes are additive, what is the molarity of the Na+ ions in the resulting solution?

My first course of action was to find the number of moles of HCl there are, which is 0.01441 x 0.1569 = 0.002261 mols HCl. I then calculated the moles of NaOH, which I found to be 0.03627 x 0.2851 = 0.01034 moles of NaOH.

Where do I go from here? I've gone as far as I know how. 

How, then, would I find the pH of the solution? I know I need to find the concentration of H30+. That's really the problem, though.

[ohaychminus]

I figured it out myself, but my next problem is figuring out [OH-].

[UG]

How does HCl react with NaOH? Will all of the NaOH react? If not, how much of it remains unreacted?

[ohaychminus]

I think I've got it! The left over moles of NaOH should be divided by the total volume of the solution to obtain [OH-], I believe.

From there, I guess I could calculate pOH and pH, and thereby [H+].

[UG]

I think I've got it! The left over moles of NaOH should be divided by the total volume of the solution to obtain [OH-], I believe.

From there, I guess I could calculate pOH and pH, and thereby [H+].

Sounds good! 

[Hunter2]

Why so complicated.

You know the amount of NaOH what is in 36.27 ml, the same amount of sodium is later then  in 50,68 ml solution. It doesnt matter if you have neutralisation.