Answer
This mixture results in a buffer.
Work Step by Step
$$HCl(aq) + CH_3NH_2(aq) \rightarrow CH_3{NH_3}^+(aq) + Cl^-(aq)$$
$CH_3NH_2$ and $CH_3N{H_3}^{+}$ is a conjugate acid-base pair. Therefore, if a certain quantity of $CH_3NH_2$ remains in solution, it will act as a buffer. In order for this to happen, $CH_3NH_2$ must be the excess reactant.
Calculate the number of moles of each compound:
$$HCl: 95.0 \space mL \times \frac{0.10 \space M}{1 \space L} \times \frac{1 \space L}{1000 \space mL} = 0.0095 \space mole$$ $$CH_3NH_2: 105.0 \space mL \times \frac{0.15 \space M}{1 \space L} \times \frac{1 \space L}{1000 \space mL} = 0.01575 \space mole$$
$0.01575 \gt 0.0095$. $CH_3NH_2$ is the excess reactant. Therefore, this mixture will result in a buffer.
