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

1 Answer
anor277
Apr 8, 2017

"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.

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