Finding pH of a buffer solution on addition of hydrogen iodide to it
Amount of acid added: 
$$0.002\ \mathrm L \times 0.2\ \mathrm{mol/L} = 0.004\ \mathrm{mol}$$

This will neutralize $0.004\ \mathrm{mol}$ of the benzoate: $$0.50\ \mathrm{mol} - 0.004\ \mathrm{mol} = 0.496\ \mathrm{mol}$$ of benzoate

This will increase benzoic acid by $0.004\ \mathrm{mol}$: $$0.10\ \mathrm{mol} + 0.004\ \mathrm{mol} = 0.104\ \mathrm{mol}$$ of benzoic acid

$$[\ce{H+}] = K_\mathrm a\left(\frac{n(\ce{HA})}{n(\ce{A-})}\right)$$ (yes, it's supposed to be $[\ce{HA}]/[\ce{A-}]$, but volume cancels out leaving amount of substance so long as your numbers leave you inside the titration curve between the starting $\mathrm{pH}$ and the equivalence point $\mathrm{pH}$)

$$[\ce{H+}] = 6.3 \times 10^{-5} \left(\frac{0.104\ \mathrm{mol}}{0.496\ \mathrm{mol}}\right) = 1.32 \times 10^{-5}$$

$$-\log\left(1.32 \times 10^{-5}\right) = \mathrm{pH} = 4.88$$
(actually $4.8794$, so you have a small change in $\mathrm{pH}$, as you would expect with a buffer with so much of the conjugate base in solution, but not enough to make a significant difference.)

I'm not at all clear why people are married to the Henderson–Hasselbalch equation for buffer problems. The equilibrium expression covers just about any situation I can think of, there's no formula to remember if you know how to do an equilibrium expression.