Answer
This mixture results in a buffer.
Work Step by Step
$$HF(aq) + KOH(aq) \rightarrow H_2O(l) + KF(aq)$$
$HF$ and $F^-$ is a conjugate acid-base pair. Therefore, if a certain quantity of $HF$ remains in solution, it will act as a buffer. In order for this to happen, $HF$ must be the excess reactant.
Calculate the number of moles of each compound:
$HF: 165.0 \space mL \times \frac{0.10 \space M}{1 \space L} \times \frac{1 \space L}{1000 \space mL} = 0.0165 \space moles$
$KOH: 135.0 \space mL \times \frac{0.050 \space M}{1 \space L} \times \frac{1 \space L}{1000 \space mL} = 0.00675 \space moles$
$0.0165 \gt 0.00675$. $HF$ is the excess reactant. Therefore, this mixture will result in a buffer.
