Hard Normality Practice Questions

Concept Explanation

Normality (N) is a measure of concentration equal to the number of gram equivalent weights of solute per liter of solution, specifically used to describe the reactive capacity of a chemical species. Unlike molarity, which measures the number of moles of solute per liter, normality accounts for the stoichiometry of a reaction, such as the number of protons (H+) or electrons (e-) exchanged. The fundamental relationship between these two units is Normality = Molarity × n-factor, where the n-factor represents the valence factor or the number of equivalents per mole. For a thorough review of concentration basics, you might find it helpful to compare molarity vs molality before tackling these advanced calculations.

To calculate normality, you must identify the specific reaction occurring. In acid-base chemistry, the n-factor is the number of replaceable hydrogen or hydroxide ions. In redox chemistry, it is the total change in oxidation state per molecule. This makes normality highly context-dependent; a single solution of potassium permanganate might have different normalities depending on whether it is used in an acidic, basic, or neutral medium. Mastering this concept is essential for complex acid-base titration problems frequently encountered in analytical chemistry. You can further explore this at Wikipedia's guide on equivalent concentration.

Solved Examples

1. Calculate the normality of a solution containing 15.8 grams of KMnO4 in 250 mL of solution, assuming the reaction occurs in an acidic medium where MnO4- is reduced to Mn2+.

   1. Determine the molar mass of KMnO4: K(39.1) + Mn(54.9) + 4O(16.0) = 158.0 g/mol.

   2. Calculate the number of moles: 15.8 g / 158.0 g/mol = 0.1 moles.

   3. Identify the n-factor: In acidic medium, Mn changes from +7 to +2, meaning it gains 5 electrons. n-factor = 5.

   4. Calculate equivalents: 0.1 moles × 5 equivalents/mole = 0.5 equivalents.

   5. Calculate normality: 0.5 equivalents / 0.250 L = 2.0 N.

2. A 0.5 M solution of H3PO4 is used in a reaction where it forms Na2HPO4. What is its normality?

   1. Identify the number of protons replaced: In the product Na2HPO4, two hydrogens have been replaced by sodium.

   2. Determine the n-factor: Since two protons were donated, the n-factor is 2.

   3. Apply the formula: Normality = Molarity × n-factor.

   4. Calculate: 0.5 M × 2 = 1.0 N.

3. How many grams of Na2CO3 (molar mass = 106 g/mol) are required to prepare 500 mL of a 0.1 N solution for a reaction where CO3^2- reacts with 2 H+ ions?

   1. Identify the n-factor: Since the carbonate ion reacts with 2 protons, the n-factor is 2.

   2. Calculate equivalents needed: Normality × Volume = 0.1 N × 0.5 L = 0.05 equivalents.

   3. Calculate moles needed: Equivalents / n-factor = 0.05 / 2 = 0.025 moles.

   4. Calculate mass: 0.025 moles × 106 g/mol = 2.65 grams.

Practice Questions

1. Calculate the normality of a solution prepared by dissolving 4.9 grams of H2SO4 in enough water to make 100 mL of solution. Assume complete dissociation.

2. A sample of 0.25 N NaOH is used to titrate 50 mL of an unknown H2C2O4 (oxalic acid) solution. If 40 mL of NaOH is required to reach the equivalence point, what is the normality of the acid?

3. Find the normality of a 0.15 M solution of Ba(OH)2 when it reacts completely with HCl. For more practice on basic concentration, see our molarity formula guide.

4. In a redox titration, K2Cr2O7 is reduced to Cr3+ in an acidic medium. Calculate the normality of a solution containing 14.7 grams of K2Cr2O7 in 1.0 L of solution. (Molar mass of K2Cr2O7 = 294 g/mol).

5. A solution is labeled 3.0 N HCl. If you dilute 100 mL of this solution to a final volume of 1.5 L, what is the new normality?

6. Calculate the mass of Ca(OH)2 needed to prepare 250 mL of a 0.5 N solution. (Molar mass of Ca(OH)2 = 74.1 g/mol).

7. If 25 mL of 0.1 N HCl is mixed with 75 mL of 0.2 N HCl, what is the normality of the resulting mixture?

8. A 0.2 M solution of Fe2(SO4)3 is used in a reaction where Fe3+ is reduced to Fe2+. What is the normality of this solution with respect to the redox reaction?

9. Determine the normality of 98% (w/w) H2SO4 with a density of 1.84 g/mL.

10. How many milliliters of 6.0 N H2SO4 are required to prepare 500 mL of 0.25 N H2SO4?

Answers & Explanations

1. 1.0 N: Molar mass of H2SO4 = 98 g/mol. Moles = 4.9 / 98 = 0.05. H2SO4 has 2 replaceable protons, so n-factor = 2. Equivalents = 0.05 × 2 = 0.1. Normality = 0.1 / 0.1 L = 1.0 N.

2. 0.2 N: Use the titration formula N1V1 = N2V2. (0.25 N)(40 mL) = (N_acid)(50 mL). N_acid = 10 / 50 = 0.2 N.

3. 0.3 N: Barium hydroxide Ba(OH)2 has an n-factor of 2 because it provides 2 OH- ions. Normality = 0.15 M × 2 = 0.3 N.

4. 0.3 N: Moles = 14.7 / 294 = 0.05. In acidic medium, Cr2O7^2- goes from +6 to +3. Since there are 2 Cr atoms, the total change is 6 electrons per molecule. n-factor = 6. Normality = (0.05 moles × 6) / 1.0 L = 0.3 N.

5. 0.2 N: Use the dilution formula N1V1 = N2V2. (3.0 N)(0.1 L) = (N2)(1.5 L). N2 = 0.3 / 1.5 = 0.2 N.

6. 4.63 g: Normality = 0.5 N, Volume = 0.25 L. Equivalents = 0.5 × 0.25 = 0.125. For Ca(OH)2, n-factor = 2. Moles = 0.125 / 2 = 0.0625. Mass = 0.0625 × 74.1 = 4.631 g.

7. 0.175 N: Total equivalents = (0.1 N × 0.025 L) + (0.2 N × 0.075 L) = 0.0025 + 0.015 = 0.0175. Total volume = 0.1 L. Normality = 0.0175 / 0.1 = 0.175 N.

8. 0.4 N: Each Fe2(SO4)3 molecule contains 2 Fe3+ ions. Each Fe3+ gains 1 electron to become Fe2+. Total electrons per molecule = 2. n-factor = 2. Normality = 0.2 M × 2 = 0.4 N.

9. 36.8 N: 98% w/w means 98g solute in 100g solution. Volume of 100g solution = Mass/Density = 100 / 1.84 = 54.35 mL = 0.05435 L. Moles of H2SO4 = 98 / 98 = 1. n-factor = 2. Equivalents = 2. Normality = 2 / 0.05435 = 36.8 N.

10. 20.83 mL: Using N1V1 = N2V2. (6.0 N)(V1) = (0.25 N)(500 mL). V1 = 125 / 6 = 20.83 mL.

Quick Quiz

Frequently Asked Questions

What is the difference between normality and molarity?

Molarity measures the concentration of molecules or ions per liter, while normality measures the reactive capacity or equivalents per liter. Normality is always equal to or greater than molarity because the n-factor is always a positive integer.

Can the normality of a solution change depending on the reaction?

Yes, the normality of a solution is context-dependent because the n-factor depends on the specific chemical reaction. For example, H3PO4 can have a normality of 1N, 2N, or 3N depending on how many of its three protons are replaced in a reaction.

Why is normality used more often in titrations than molarity?

Normality simplifies titration calculations because at the equivalence point, the equivalents of one reactant always equal the equivalents of the other (N1V1 = N2V2). This avoids the need for complex stoichiometric ratios required when using molarity.

How do you calculate the equivalent weight for a redox agent?

The equivalent weight in a redox reaction is calculated by dividing the molar mass of the substance by the total change in oxidation number per molecule. This change represents the number of electrons transferred during the reaction. You can find more detailed data on molar masses at PubChem.

Is normality still widely used in modern chemistry?

While the International Union of Pure and Applied Chemistry (IUPAC) recommends using molarity to avoid ambiguity, normality remains common in analytical chemistry, clinical labs, and water quality testing. It provides a direct way to express the concentration of reactive species like acids and bases.

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