# What is a Normal Solution?

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What is a Normal Solution?

**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 H

_{2}SO_{4}:**Example:**

The normality of a 1.0 liter solution of H

_{2}SO_{4 }containing 1.0 gram-equivalent weight will be the molecular weight of H_{2}SO_{4 }divided by the valence of H_{2}SO_{4}:(atomic weight of H = 1; atomic weight of S = 32.06; atomic weight of O = 16)

GMW of H

_{2}SO_{4 }= 1(2) + 32.06 + 16(4) = 98 gN = GMW/valence (the valence for H

_{2}SO_{4 }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 H**_{2}SO_{4 }= 1N solution of H_{2}SO_{4}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.**