Answer
$d. 0.1M KOH$ and $0.2M CH_3NH_3Cl$
Work Step by Step
Since there are a weak acid and a strong base in the solution, this reaction will occur:
$KOH(aq) + CH_3N{H_3}^+(aq) -- \gt K^+(aq) + CH_3NH_2(aq) + H_2O(l)$
$KOH$ is the limiting reactant, because it has the lower concentration $(0.1M)$.
Therefore, these are the final concentrations in the solution:
$[KOH] = 0.1M - 0.1M = 0$
$[CH_3N{H_3}^+] = 0.2M - 0.1M = 0.1M$
$[CH_3NH_2] = 0.1M$ (Produced)
As we can see, we have $CH_3N{H_3}^+$ and $CH_3NH_2$ in equal concentrations, so, this is a buffer.
