Finding new pH after NaOH added to buffer solution I have a buffer containing 0.2 M of the acid $\ce{HA}$, and 0.15 M of its conjugate base $\ce{A-}$, with a pH of 3.35. I need to find the pH after 0.0015 mol of $\ce{NaOH}$ is added to 0.5 L of the solution.

I started by converting everything to moles, writing out an equation for $\ce{HA}$ and $\ce{OH-}$ reacting to make $\ce{H2O}$ and $\ce{A-}$, getting 0.0985 mol of $\ce{HA}$ left after all the $\ce{OH-}$ was used up, and 0.015 mol of $\ce{A-}$ produced.

At that point I got a bit stuck. I tried doing things a bunch of different ways after that, such as trying to calculate $\mathrm{p}K\mathrm{_a}$ by using the Henderson-Hasselbalch equation, but I'm not getting anywhere near the answer. I calculated a $\mathrm{p}K\mathrm{_a}$ of 3.47. I added the amount of $\ce{A-}$ made when $\ce{HA}$ reacts with $\ce{OH-}$ to the original moles of $\ce{A-}$ in the buffer (I calculated 0.18 M $\ce{A-}$), and tried putting that, the leftover moles of $\ce{HA}$, and my $\mathrm{p}K\mathrm{_a}$ value into the equation, but that wasn't right either.

What am I doing wrong?
Start by calculating $\mathrm{p}K_\mathrm{a}$ using the Henderson–Hasselbalch equation and the initial data given. $$\mathrm{p}K_\mathrm{a} = 3.35 - \log\frac{0.15}{0.2} = 3.47$$

Calculate the amount of substance of $\ce{HA}$ and $\ce{A- }$ after addition of $\ce{NaOH}$. $$n_{\ce{HA}} = 0.15\times0.5 - 0.0015 = 0.0765$$ $$n_{\ce{A- }} = 0.2\times0.5 + 0.0015 = 0.0985$$

Calculate the final $\mathrm{pH}$. The final volume is not needed because it is the same for both concentrations and so it cancels in the equation. $$\mathrm{pH} = 3.47 + \log\frac{0.0985}{0.0765} = 3.58$$