Question

Question #5fd98

Answer 1

`["OH"^(-)] = 2.24 * 10^(-5)"M"`

Explanation:

Your starting point here will be the definition of a solution's `"pOH"`, which as you know is defined as the negative log base `10` of the concentration of hydroxide anions, `"OH"^(-)`

`color(blue)(ul(color(black)("pOH" = - log(["OH"^(-)]))))`

Now, in order to solve for the concentration of hydroxide anions, you must rearrange this equation as

`log(["OH"^(-)]) = - "pOH"`

and rewrite it using exponents of base `10`

`10^log(["OH"^(-)]) = 10^(-"pOH")`

This is equivalent to

`color(blue)(ul(color(black)(["OH"^(-)] = 10^(-"pOH"))))`

Now all you have to do is to plug in the value given to you for the `"pOH"` of the solution

`["OH"^(-)] = 10^(-4.65) = color(darkgreen)(ul(color(black)(2.24 * 10^(-5)"M")))`

I'll leave the answer rounded to three sig figs.