A buffer holds a weak acid AND its conjugate base in the same flask, so acid added is mopped up by the base and base added is mopped up by the acid. Take the Ka expression, solve it for $[\mathrm{H^+}]$ and run p over the lot, and the ICE table's algebra reduces to one line: $\mathrm{pH} = \mathrm{p}K_a + \log\dfrac{[\mathrm{A^-}]}{[\mathrm{HA}]}$ — read aloud: pH equals pK-a plus log of A-minus over H-A. Naming every symbol: $\mathrm{p}K_a$ is the weak acid's own constant on the p-scale (a pure number), $[\mathrm{A^-}]$ is the conjugate base concentration in mol/L, $[\mathrm{HA}]$ is the undissociated acid concentration in mol/L, and whichever of the three the question leaves blank is the one you solve for.

Two facts do most of the work. When the pair is split evenly the log is zero and the pH SITS ON the pKa — which is how a chemist picks a buffer: choose an acid whose pKa is near the pH you want. And because it is a RATIO, diluting a buffer with water barely moves its pH at all; both concentrations fall together and the ratio survives. The base-side twin runs in the mirror: $\mathrm{pOH} = \mathrm{p}K_b + \log\dfrac{[\mathrm{BH^+}]}{[\mathrm{B}]}$, with $[\mathrm{B}]$ the free weak base and $[\mathrm{BH^+}]$ its conjugate acid, both in mol/L — and it answers in pOH, so one hop across the 14 still remains. Getting the ratio upside down is the classic; ask yourself which way MORE base should push the pH before you write anything.