Answer
$Acidic.$
Work Step by Step
1. Compare the strength of the ions:
$N{H_4}^+:$
$Kb (NH_3) = 1.8 \times 10^{-5}$
$Ka * Kb = 10^{-14}$
$Ka = \frac{10^{-14}}{1.8*10^{-5}}$
$Ka = 5.6 \times 10^{-10}$
$Br^-:$
Since the Ka of $HBr$ is very high (it is a strong acid), the Kb of $Br^-$ is negligible.
$N{H_4}^+$ is the stronger ion, the solution will be acidic.
