Answer:
The [OH−] is 5.88x10^-7M.
Explanation:
1st) It is necessary to calculate the pH of the solution using the pH formula and replacing the concentration of H3O+:
[tex]\begin{gathered} pH=-log\left[H_3O+\right] \\ pH=-log(1.7×10−8) \\ pH=7.77 \end{gathered}[/tex]Now we know that the pH is 7.77.
2nd) Now we can calculate the pOH of the solution using the relation between pH and pOH:
[tex]\begin{gathered} pH+pOH=14 \\ 7.77+pOH=14 \\ pOH=14-7.77 \\ pOH=6.23 \end{gathered}[/tex]Now we know that the pOH of the solution is 6.23.
3rd) Finally, we can calculate the [OH−] using the pOH formula and replacing the value of pOH:
[tex]\begin{gathered} pOH=-log\left[OH−\right] \\ 6.23=-log\left[OH−\right] \\ 10^{(-6.23)}=\left[OH−\right] \\ 5.88*10^{-7}=\left[OH−\right] \end{gathered}[/tex]So, the [OH−] is 5.88x10^-7M.
