What is the #"normality"# of a solution prepared by dissolving a #5*g# mass of sodium hydroxide in a #1*L# volume of water?

1 Answer
Aug 19, 2017

Answer:

#"Normality"=0.125*mol*L^-1#.

Explanation:

Here #"molarity"# #-=# #"normality"#. Why? Because hydroxide anion does not speciate to any degree in aqueous solution.

And #"molarity"# #-=# #"moles of solute"/"volume of solution"#

#=((5*g)/(40.0*g*mol^-1))/(1*L)=(0.125*mol)/(1*L)=0.125*mol*L^-1# with respect to #NaOH(aq)#.

What is #pH# of this solution? We know that #pH+pOH=14# under standard conditions.