You have a beaker containing $\pu{542 mL}$ of a solution of $\pu{0.48 M}$ acetate buffer ($\mathrm{p}K_\mathrm{a} = 4.76$) at $\ce{pH 4.2}$. To this solution, you add $\pu{342 mL}$ of pure water. What is the $\ce{pH}$ of the resulting solution?
The answer is $4.6$. However, I do not know how to arrive at that answer. I know the H-H equation must be used:
$$\ce{pH} = \mathrm{p}K_\mathrm{a} + \log\frac{[\ce{A-}]}{[\ce{HA}]}$$