Normality (N)

It is a concentration unit used in chemistry, particularly in acid-base and redox reactions. Unlike molarity (M), which measures the concentration of moles of solute per liter of solution, normality takes into account the number of equivalents of solute per liter of solution.

The equivalent is a measure of the reactive capacity of a substance. It depends on the type of reaction being considered. For acids and bases, one equivalent is defined as the amount that can donate or accept one mole of hydrogen ions (H+).

Mathematically, normality (N) is calculated using the following formula:

N = n / V

Where:

  • N = Normality of the solution (in equivalents per liter, eq/L or N)
  • n = Number of equivalents of solute
  • V = Volume of the solvent in liters

It's important to note that normality is only used for substances that have the potential to donate or accept multiple equivalents, such as acids and bases.

Example Numericals:

Problem 1:

Calculate the normality of a solution that contains 0.2 moles of sulfuric acid (H2SO4) dissolved in 0.5 liters of water.

Solution: Given:

  • Number of moles of H2SO4 (n) = 0.2 moles
  • Volume of solution (V) = 0.5 L

Since H2SO4 can donate two equivalents of H+, the number of equivalents is 2 * n: Number of equivalents = 2 * 0.2 eq = 0.4 eq

Using the normality formula: N = n / V = 0.4 eq / 0.5 L = 0.8 N

The normality of the solution is 0.8 N.

Problem 2:

What volume of a 1.5 N hydrochloric acid (HCl) solution contains 0.3 equivalents of HCl?

Solution: Given:

  • Normality of HCl solution (N) = 1.5 N
  • Number of equivalents of HCl (n) = 0.3 eq

Rearranging the normality formula to solve for volume: V = n / N = 0.3 eq / 1.5 N = 0.2 L

The volume of the HCl solution required is 0.2 liters.

Problem 3:

Dilute 100 mL of a 2.0 N sulfuric acid (H2SO4) solution to a final volume of 0.5 L. What is the normality of the diluted solution?

Solution: Given:

  • Initial normality of H2SO4 solution (N1) = 2.0 N
  • Initial volume of H2SO4 solution (V1) = 100 mL = 0.1 L
  • Final volume of diluted solution (V2) = 0.5 L

Since H2SO4 can donate two equivalents of H+, the number of equivalents is 2 * n: Number of equivalents = 2 * 0.1 eq = 0.2 eq

The relationship between initial and final equivalents of solute in a dilution is given by: N1V1 = N2V2

Solving for N2: N2 = N1V1 / V2 = 2.0 N * 0.1 L / 0.5 L = 0.4 N

The normality of the diluted H2SO4 solution is 0.4 N.

Problem 4:

You want to prepare 250 mL of a 1.2 N hydrochloric acid (HCl) solution. How many equivalents of HCl are needed to make this solution?

Solution: Given:

  • Desired normality of HCl solution (N) = 1.2 N
  • Volume of solution (V) = 250 mL = 0.25 L

Rearranging the normality formula to solve for equivalents: n = NV = 1.2 N * 0.25 L = 0.3 eq

You need 0.3 equivalents of HCl to prepare the solution.

Practice Problems:

  1. Calculate the normality of a solution containing 0.25 moles of hydrochloric acid (HCl) dissolved in 0.3 liters of water.

  2. What volume of a 0.9 N sulfuric acid (H2SO4) solution contains 0.4 equivalents of H2SO4?

  3. Dilute 150 mL of a 1.8 N hydrochloric acid (HCl) solution to a final volume of 0.6 L. What is the normality of the diluted solution?

  4. Calculate the normality of a solution prepared by dissolving 0.35 moles of acetic acid (CH3COOH) in enough water to make a final volume of 0.4 liters.

  5. What volume of a 2.5 N potassium hydroxide (KOH) solution contains 0.6 equivalents of KOH?

  6. Dilute 120 mL of a 3.0 N nitric acid (HNO3) solution to a final volume of 0.9 L. Calculate the normality of the diluted solution.

  7. You have 0.5 moles of aluminum sulfate (Al2(SO4)3) and want to prepare a solution with a normality of 1.2 N. What volume of solution should you make?

  8. Calculate the normality of a solution prepared by dissolving 0.2 moles of potassium hydroxide (KOH) in enough water to make a final volume of 0.5 liters.

  9. What volume of a 1.5 N hydrochloric acid (HCl) solution contains 0.7 equivalents of HCl?

  10. Dilute 180 mL of a 4.5 N sodium hydroxide (NaOH) solution to a final volume of 1.2 L. Calculate the normality of the diluted solution.