Answer:
Approximately [tex]8.41[/tex] (assuming that the solution is at [tex]\rm 25^\circ C[/tex], under which [tex]K_{\rm w} = 10^{-14}[/tex].)
Explanation:
Let [tex]{\rm [H^{+}]}[/tex] and [tex]{\rm [OH^{-}]}[/tex] denote the concentration of [tex]\rm H^{+}[/tex] and [tex]\rm OH^{-}[/tex] respectively.
Let [tex]K_{\rm w}[/tex] denote the self-ionization constant of water. The exact value of [tex]K_{\rm w}\![/tex] depends on the temperature of the solution. [tex]K_{\rm w} =10^{-14}[/tex] at [tex]\rm 25^\circ C[/tex].
The product of [tex]{\rm [H^{+}]}[/tex] and [tex]{\rm [OH^{-}]}[/tex] in a solution (with [tex]\rm M[/tex], or moles per liter, as the unit) is supposed to be equal to the [tex]K_{\rm w}[/tex] value of that solution at the corresponding temperature. In other words:
[tex]{\rm [H^{+}]} \cdot {\rm [OH^{-}]} = K_{\rm w}[/tex].
Rearrange to obtain an expression for [tex]{[\rm OH^{-}]}[/tex]:
[tex]\begin{aligned}{\rm [OH^{-}]} &= \frac{K_{\rm w}}{[\rm H^{+}]}\end{aligned}[/tex].
Assume that the solution in this question is at [tex]\rm 25^\circ C[/tex] (for which [tex]K_{\rm w} =10^{-14}[/tex].) For this solution:
[tex]\begin{aligned}{\rm [OH^{-}]} &= \frac{K_{\rm w}}{[\rm H^{+}]} \\ &= \frac{10^{-14}}{2.6 \times 10^{-6}}\approx 3.85\times 10^{-9}\; \rm M\end{aligned}[/tex].
Hence, the [tex]\rm pOH[/tex] of this solution would be:
[tex]\begin{aligned}\rm pOH &= -\log_{10}{\rm [OH^{-}]} \\&\approx -\log_{10} (3.85 \times 10^{-9}) \approx 8.41 \end{aligned}[/tex].
