Question

Question #92c8a

Answer 1

`["OH"^-]=1.8\times10^{-10}" M"`.

Explanation:

Use the autoionization equilibrium constant for water:

`2"H"_2"O"->"H"_3"O"^++"OH"^-`

`"K"_w="[H"_3"O"^+]["OH"^-]=1.0\times10^{-14}" M"^2` at `25°"C"`