What is the #"normality"# of a #4.0*g# mass of sodium hydroxide dissolved in a #1.0*L# volume of water?

1 Answer
Apr 8, 2017

Answer:

#"Normality"=1.0*N#

Explanation:

#"Normality"# is defined similarly to #"molarity"#.

#"Normality"=("Moles of "H_3O^+"or "HO^-)/"Volume of solution"#

Here, #"normality"=((4.0*g)/(40.0*g*mol^-1))/(0.100*L)=1.0*N#.

This will react quantitatively with one equiv of acid. One the other hand, a solution of diprotic #H_2SO_4(aq)# that is #0.50*mol*L^-1# with respect to #H_2SO_4#, but necessarily #1.0*mol*L^-1# with respect to #H_3O^+# (why?), would also be regarded as #1*N#.