Answer:
Option A) 2.6 × 10⁻¹¹ M, acidic.
Explanation:
The ion-product constant of water [tex]K_{\rm w}[/tex] gives a relationship between [tex]\rm [H_3O^{+}][/tex] and [tex]\rm [OH^{-}][/tex]:
[tex]K_{\rm w} = \rm [H_3O^{+}]\cdot [OH^{-}][/tex].
The exact value of [tex]K_{\rm w}[/tex] depends on the temperature. Under [tex]\rm 25^{\circ}C[/tex], [tex]K_{\rm w} \approx 1^{-14}[/tex].
The question states that [tex]\rm [H_3O^{+}] = 3.9\times 10^{-4}\; M[/tex]. As a result,
[tex]\begin{aligned} {\rm [OH^{-}]} &= \frac{K_{\rm w}}{[{\rm H_3O^{+}}]}\\ &\approx\frac{1^{-14}}{3.9\times 10^{-4}} && \rm {\leftarrow Only~under~25^{\circ}C.}\atop{}\\&\approx \rm 2.6\times 10^{-11}\; M\end{aligned}[/tex].
[tex]\rm [H_3O^{+}] > [OH^{-}][/tex] for this solution. As a result, this solution is acidic.