Question #5fd98
1 Answer
Explanation:
Your starting point here will be the definition of a solution's
#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^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
#["OH"^(-)] = 10^(-4.65) = color(darkgreen)(ul(color(black)(2.24 * 10^(-5)"M")))#
I'll leave the answer rounded to three sig figs.