To calculate the normality of nitric acid (HNO₃), you need to determine the number of gram equivalents of the solute present in one liter of the solution. Normality is a measure of concentration that is particularly useful in titration calculations, as it directly relates to the reactive capacity of a substance.
Understanding Normality
Normality (N) is defined as the number of gram equivalents of solute per liter of solution. For an acid like nitric acid, an equivalent represents the amount of the substance that can provide one mole of hydrogen ions (H⁺) in an acid-base reaction.
The general formula for normality is:
$$ \text{Normality (N)} = \frac{\text{Number of equivalents of solute}}{\text{Volume of solution in liters}} $$
Key Components for Calculating Normality of Nitric Acid
To apply the normality formula, you'll need to calculate the "number of equivalents of solute" and determine the "volume of solution."
1. Number of Equivalents of Solute (HNO₃)
The number of equivalents of a solute is calculated using its mass and its equivalent weight:
$$ \text{Number of equivalents of solute} = \frac{\text{Mass of solute}}{\text{Equivalent weight}} $$
2. Equivalent Weight of Nitric Acid (HNO₃)
The equivalent weight of a substance is its molar mass divided by its n-factor (also known as the equivalency factor or valence factor). For acids, the n-factor is the number of acidic hydrogen atoms (protons) that can be donated per molecule in a reaction.
- Molar Mass of HNO₃:
- Hydrogen (H): approx. 1.01 g/mol
- Nitrogen (N): approx. 14.01 g/mol
- Oxygen (O): approx. 16.00 g/mol × 3 = 48.00 g/mol
- Total Molar Mass of HNO₃ ≈ 1.01 + 14.01 + 48.00 = 63.02 g/mol (often rounded to 63 g/mol for calculations).
- n-factor for HNO₃: Nitric acid (HNO₃) is a strong monoprotic acid, meaning it donates one hydrogen ion (H⁺) per molecule. Therefore, its n-factor is 1.
So, the equivalent weight of nitric acid is:
$$ \text{Equivalent weight of HNO}_3 = \frac{\text{Molar mass of HNO}_3}{\text{n-factor}} = \frac{63.02 \text{ g/mol}}{1} = 63.02 \text{ g/equivalent} $$
3. Volume of Solution
The volume of the solution must be in liters. If you have the mass of the solution and its density, you can calculate the volume using the formula:
$$ \text{Volume} = \frac{\text{Mass}}{\text{Density}} $$
Step-by-Step Calculation Process
Here’s a general procedure to calculate the normality of a nitric acid solution:
- Determine the molar mass of HNO₃. (approx. 63.02 g/mol)
- Identify the n-factor for HNO₃. For acid-base reactions, it's 1.
- Calculate the equivalent weight of HNO₃. (Molar mass / n-factor)
- Find the mass of HNO₃ solute in your solution. This might be given directly, or you might need to calculate it from a percentage by weight concentration.
- Calculate the number of equivalents of HNO₃ using the mass of solute and its equivalent weight.
- Determine the volume of the solution in liters. If you have the mass of the solution and its density, convert mass to volume.
- Apply the normality formula by dividing the number of equivalents by the volume of the solution.
Practical Example
Let's illustrate with a common scenario: calculating the normality of a nitric acid solution given its percentage by weight and density.
Scenario: What is the normality of a nitric acid solution that is 21% (w/w) by weight and has a density of 1.12 g/mL?
Solution Steps:
- Assume a basis for calculation: Let's assume we have 1000 mL (1 L) of the HNO₃ solution.
- Calculate the mass of the 1000 mL solution:
- Mass of solution = Density × Volume
- Mass of solution = 1.12 g/mL × 1000 mL = 1120 g
- Calculate the mass of HNO₃ solute in the solution:
- Since the solution is 21% HNO₃ by weight, 21% of the total mass is HNO₃.
- Mass of HNO₃ = 21% of 1120 g = (21 / 100) × 1120 g = 235.2 g
- Determine the equivalent weight of HNO₃:
- Molar mass of HNO₃ ≈ 63 g/mol
- n-factor for HNO₃ = 1
- Equivalent weight of HNO₃ = 63 g/equivalent
- Calculate the number of equivalents of HNO₃:
- Number of equivalents = Mass of HNO₃ / Equivalent weight of HNO₃
- Number of equivalents = 235.2 g / 63 g/equivalent ≈ 3.733 equivalents
- Calculate the normality:
- Normality = Number of equivalents of solute / Volume of solution in liters
- Since we assumed 1000 mL, the volume in liters is 1 L.
- Normality = 3.733 equivalents / 1 L ≈ 3.733 N
Summary of Formulas
For quick reference, here are the key formulas used in calculating normality:
| Concept | Formula |
|---|---|
| Normality (N) | $\text{N} = \frac{\text{Equivalents of solute}}{\text{Volume of solution (L)}}$ |
| Number of Equivalents | $\text{Equivalents} = \frac{\text{Mass of solute}}{\text{Equivalent weight}}$ |
| Equivalent Weight (for Acid) | $\text{Equivalent weight} = \frac{\text{Molar mass}}{\text{n-factor}}$ |
| Density | $\text{Density} = \frac{\text{Mass}}{\text{Volume}}$ |
Calculating the normality of nitric acid involves a clear understanding of these fundamental chemical concepts and their interrelations.
