[ohaychminus]

Hey, all. I posted here yesterday asking a similar question. This time, though, I'm having problems calculating the pH of a solution. Here's the question:

19.21 mL of 0.1679 M aqueous HCl is added to 20.68 mL of 0.2752 M aqueous NaOH at 25°C. Assuming the volumes are additive, what is pH of the resulting solution?

I started off by calculating the number of moles of HCl and NaOH (0.003225 and 0.005091, respectively).

Shouldn't I be able to calculate [OH-] like so: (0.005091 - 0.003225)/0.03989 = 0.04677

Then, since I have [OH-], I could do this: -log(0.04677) to obtain pOH, and then 14 - pOH to get pH.

The answer I get from this isn't being accepted.

[Hunter2]

Calculate the moles of HCl and NaOH
then do the difference of it. What is left OH^- or H⁺. The moles what you obtained calculate to the new volume.

[ohaychminus]

Isn't that what I did, though?

[Borek]

Check your math.

[ohaychminus]

Ah, I see I made a mistake in calculating the moles of NaOH.

Is my methodology for calculating pH this way correct, though?

[Borek]

Yes, this is a simple limiting reagent question.

http://www.chembuddy.com/?left=pH-calculation-questions&right=pH-of-mixture-q1