Question

Calculate the pH of the resulting solution if 28.0 mL of 0.280 M HCl(aq) is added to 18.0 mL of 0.380 M NaOH(aq).
I keep getting a pH = 12.337 but its wrong. I found the individual number of moles for each concentration and found the limiting reagent by subtraction, and divided it by the total volume. I found the pH using -log, but I keep getting it wrong.

Answer 1

When you have a problem like this you would do better to show your work, not just HOW you worked it. I can find the error in you calculations in a third of the time it takes me to work the problem from scratch
HCl + NaOH ==> NaCl + H2O
millimoles HCl = mL x M = 28.0 x 0.280 = 7.84
millimoles NaOH = 18.0 x 0.380 = 6.84
So you have 7.84 - 6.84 = 1.00 millimoles HCl in excess.
M HCl = millimoles/mL = 1.00/46.0 = 0.0217 M
pH = 1.66