Normality (N) is another way to quantify solution concentration. It is similar to molarity but uses the gram-equivalent weight of a solute in its expression of solute amount in a liter (L) of solution, rather than the gram molecular weight (GMW) expressed in molarity. A 1N solution contains 1 gram-equivalent weight of solute per liter of solution.
Expressing gram-equivalent weight includes the consideration of the solute's valence. The valence is a reflection of the combining power of an element often as measured by the number of hydrogen atoms it can displace or combine with. A 1.0 gram-equivalent weight is the amount of a substance that will combine with or displace 1 atom of hydrogen.
To determine gram-equivalent weight of a substance:
Divide the GMW (formula weight) of a solute by the valence (number of hydrogen ions that can be displaced).
Example:
The normality of a 1.0 liter NaCl solution that contains 1.0 gram-equivalent weight will be the GMW of NaCl divided by the valence of NaCl:
(atomic weight of Na = 22.99; atomic weight of Cl = 35.45)
GMW of NaCl = 22.99 + 35.45 = 58.44 g
N = GMW/valence (the valence for NaCl is 1.0)
58.44 g/1.0 = 58.44 g = 1.0 gram-equivalent weight of NaCl = 1N solution of NaCl
In this situation, because NaCl has a valence of one, the molarity and normality of the solution are the same.
Some compounds, however, will not have the same normality as molarity, as in the case of H2SO4:
Example:
The normality of a 1.0 liter solution of H2SO4 containing 1.0 gram-equivalent weight will be the molecular weight of H2SO4 divided by the valence of H2SO4:
(atomic weight of H = 1; atomic weight of S = 32.06; atomic weight of O = 16)
GMW of H2SO4 = 1(2) + 32.06 + 16(4) = 98 g
N = GMW/valence (the valence for H2SO4 is 2.0, as there are 2.0 H ions that could be displaced)
98 g /2 = 49 g = 1.0 gram-equivalent weight of H2SO4 = 1N solution of H2SO4
The molarity of this 1N solution of H2SO4 would be 0.5 (M = g /GMW per liter or 49g/98g = 0.5)
To simply calculate the amount or weight of a substance needed for a desired normal solution, the following formula may be used:
Weight in grams = desired normality x volume needed in liters x GMW/valence
(W = N x V x GMW/valence)
Example:
500 mL of a 0.1N solution of NaOH is needed for a procedure. Calculate the amount of solute (NaOH) needed to prepare the solution. (atomic weights: Na = 22.99; O = 16; H = 1) Valence = 1
X g= 0.1N x 500 mL (0.5 L) x GMW 39.99 / 1.0
X = 0.1 x 0.5 x 39.99/1.0
X = 1.99
1.99 g of NaOH must be diluted to 500 mL to prepare a 0.1N solution.
