Question #5fd98

1 Answer
Jan 10, 2017

#["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.