If you made up a solution of naoh by adding 0.010 mole of solid naoh to 1.0 liter of distilled water, what would be the concentration of the oh-(aq)?

 The   concentration   of the  OH ^-(aq)   is    0.01 M or  1 x10^-2 M

  calculation
first write the dissociation  equation  for  NaOH

NaOH →  Na^+   +  OH^-

find the molarity of NaOH  used 

molarity = moles/volume in liters=  0.010   moles/ 1L  =  0.01 M

by use of mole  ratio between  NaOH  to OH^- which is 1:1 therefore the   concentration  of   OH^-  is  also   0.01M or  1 x10 ^-2M