Getting the wrong pH value, according to data and online calculator I'm doing a titration problem with a week acid and strong base, I used experiment data and an online calculator to check my answer, but my pH I calculated is a bit off. Can anyone see where I messed up? I can't figure out where...

AcOH
0.0984M
0.02500L
0.00246mol
Ka 1.8x10^-5

NaOH
0.096M
0.02580L
0.00025mol

After rxn:

0mol AcOH left

0.0025mol - 0.00246mol = 4.0x10^-5 mol OH excess

0.00246mol AcO-

4.0x10^-5mol OH / 0.0508 L total soln = 7.87x10^-4 M OH-

-log([OH-]) = pOH = 3.104

pH = 10.90

The pH should be 10.52 according to my data and the online calculator.
As you can see from the point that $n_\ce{OH} &gt; n_\ce{AcOH}$, there will be an excess of hydroxide ions. What you want to calculate is the pH for this amount of hydroxide ions within the total volume at this point.

Thus the calculation looks like:
\begin{align}
[\ce{OH-}] &amp;= \frac{n_\ce{NaOH}-n_\ce{AcOH}}{V_\text{a} + V_\text{b}}\\
\Leftrightarrow \mathrm{pOH} &amp;= -\lg ([\ce{OH-}])\\
\Leftrightarrow \mathrm{pH} &amp;= 14+\lg ([\ce{OH-}])
\end{align}

Your error is based on a rounding error ... please don't round intermediate result as all rounding errors will sum up. You just need to round the final results to a proper amount of decimal places.

The correct calculation:

$$\mathrm{pH} = 14 + \lg \left(\frac{0.0024768 - 0.00246}{0.025 + 0.0258}\right) = 10.5194 \approx 10.52$$

vs. your calculation:

$$\mathrm{pH} = 14 + \lg \left(\frac{0.0025 - 0.00246}{0.025 + 0.0258}\right) = 10.8962 \approx 10.90$$