Answer:
Approximately [tex]3.6 \times 10^{-10}\; {\rm M}[/tex], assuming that the solution is dilute and is at room temperature (such that [tex]K_\text{w} = 10^{-14}[/tex].)
Explanation:
Look up the ion product constant of water:
[tex]K_{\text{w}} \approx 10^{-14}[/tex].
In a dilute solution where water is the solvent, the product of [tex][{\rm {H_{3}O}^{+}}][/tex] and [tex][{\rm {OH}^{-}}][/tex] (concentration of [tex]\rm {H_{3}O}^{+}}[/tex] ions and [tex]{\rm {OH}^{-}}[/tex] ions) is constantly equal to [tex]K_{\text{w}}[/tex]. (The unit of both [tex][{\rm {H_{3}O}^{+}}]\![/tex] and [tex][{\rm {OH}^{-}}]\![/tex] need to be [tex]{\rm M}[/tex].) In other words:
[tex][{\rm {H_{3}O}^{+}}]\, [{\rm {OH}^{-}}] = K_{\text{w}}[/tex].
Rearrange this equation to find [tex][{\rm {OH}^{-}}][/tex] in terms of [tex]K_{\text{w}}[/tex] and [tex][{\rm {H_{3}O}^{+}}][/tex]:
[tex]\begin{aligned}[] [ {\rm OH^{-}} ] &= \frac{K_{\text{w}}}{[{\rm {H_{3}O}^{+}}]} \\ &= \frac{10^{-14}}{2.8 \times 10^{-5}} \\ &\approx 3.6 \times 10^{-10}\end{aligned}[/tex].
