Question

Calculate the normality of each of the following solutions. a. $0.250M\mathrm{HCl}$ b. $0.105M{\mathrm{H}}_2{\mathrm{SO}}_4$ c. $5.3\times {10}^{-2}M{\mathrm{H}}_3{\mathrm{PO}}_4$ d. $0.134M\mathrm{NaOH}$ e. $0.00521M\mathrm{Ca}(\mathrm{OH}{)}_2$ What is the equivalent mass for each of the acids or bases listed above?

Short answer

In summary, the normalities and equivalent masses are: a. HCl: 0.250 N, 36.5 b. H₂SO₄: 0.210 N, 49 c. H₃PO₄: $1.59\times {10}^{-1}$ N, 32.67 d. NaOH: 0.134 N, 40 e. Ca(OH)₂: 0.01042 N, 37

Step-by-step solution

Find the n factor for each solution

For each given acid and base, we need to identify how many H+ ions (protons) or OH- ions (hydroxides) are present to find the n-factor. a. $\mathrm{HCl}$ has one H+, so n = 1. b. ${\mathrm{H}}_2{\mathrm{SO}}_4$ has two H+ ions, so n = 2. c. ${\mathrm{H}}_3{\mathrm{PO}}_4$ has three H+ ions, so n = 3. d. $\mathrm{NaOH}$ has one OH-, so n = 1. e. $\mathrm{Ca}(\mathrm{OH}{)}_2$ has two OH- ions, so n = 2.

Calculate the normality of each solution.

The normality (N) can be calculated using the formula N = Molarity * n, where n is the number of equivalents given above and Molarity is given in the exercise. a. The normality for HCl: N = 0.250 * 1 = 0.250 N b. The normality for H₂SO₄: N = 0.105 * 2 = 0.210 N c. The normality for H₃PO₄: N = $5.3\times {10}^{-2}\ast 3=1.59\times {10}^{-1}$ N d. The normality for NaOH: N = 0.134 * 1 = 0.134 N e. The normality for Ca(OH)₂: N = 0.00521 * 2 = 0.01042 N

Calculation of equivalent mass for each acid and base

The equivalent mass is given by the ratio of molar mass (M) to the number of equivalents n. a. Equivalent mass of HCl: $M(\mathrm{HCl})/n=36.5/1=36.5$ b. Equivalent mass of H2SO4: $M({\mathrm{H}}_2{\mathrm{SO}}_4)/n=98/2=49$ c. Equivalent mass of H3PO4: $M({\mathrm{H}}_3{\mathrm{PO}}_4)/n=98/3=32.67$ d. Equivalent mass of NaOH: $M(\mathrm{NaOH})/n=40/1=40$; e. Equivalent mass of Ca(OH)₂: $M(\mathrm{Ca}(\mathrm{OH}{)}_2)/n=74/2=37$ In summary, the normalities and equivalent masses are: a. HCl: 0.250 N, 36.5 b. H2SO4: 0.210 N, 49 c. H3PO4: $1.59\times {10}^{-1}$ N, 32.67 d. NaOH:0.134 N , 40 e. Ca(OH)₂: 0.01042 N, 37

Key concepts

Normality of Solutions

Normality is a measure of concentration that’s particularly important in titrations and stoichiometric chemistry. It emphasizes the chemical reactivity or the number of reactive units in a solution. For acids and bases, this means considering the number of hydrogen ions (H+) or hydroxide ions (OH-) the substance can provide or accept.

To calculate normality, you need to understand the concept of equivalent factors or the n-factor. It tells us the amount of an acid or base that can furnish or react with one mole of hydrogen or hydroxide ions, respectively. Once you’ve determined the n-factor, normality (N) is simply the molarity (M) of the solution multiplied by this n-factor. For example, hydrochloric acid (HCl) gives one H+ ion, so its n-factor is 1, while sulfuric acid (H₂SO₄) can release two H+ ions, so its n-factor is 2.

It’s critical not to confuse normality with molarity. While molarity is the moles of solute per liter of solution, normality accounts for the moles of reactive units. This distinction is especially important when dealing with polyvalent acids and bases.

Equivalent Mass Calculation

Equivalent mass represents the mass that will combine with or replace one mole of hydrogen atoms in an acid-base reaction. It's directly linked to the n-factor. You calculate equivalent mass by dividing the molar mass (M) of a substance by its n-factor. This concept is sturdy as it ties together the physical mass and its chemical capability to react.

For instance, the equivalent mass of HCl is its molar mass divided by one (since it has an n-factor of 1), while for H₂SO₄ with n-factor of 2, you divide its molar mass by two. The equivalent mass becomes very useful in stoichiometry, allowing chemists to quickly understand the relation between reactants in a balanced chemical equation.

Remember, the equivalent mass varies depending on the type of reaction in which the substance is involved, and this underscores the versatility and utility of the concept in different chemical contexts.

Molarity to Normality Conversion

The conversion from molarity to normality is a common step in various chemical calculations, particularly in titration. To perform the conversion, you need to multiply the molarity by the n-factor of the solute, which depends on the number of active ions or atoms involved in the chemical reaction.

This is straightforward for substances like HCl and NaOH, where the n-factor is 1 because they provide one ion each per molecule in solution. Consequently, for these substances, the normality and molarity are equal. For polyprotic acids like H₂SO₄ and bases such as Ca(OH)₂, where multiple ions per formula unit are reactive, the normality will be different from the molarity. It's essential to comprehend this concept to avoid miscalculations in experiments that rely on precise reactions, such as titrations.

Stoichiometry in Chemistry

Stoichiometry is the heart of chemical reactions. It deals with the quantitative relationships between the reactants and products involved in a chemical reaction. It's often visualized in the form of a balanced equation, indicating the proportion of molecules needed to react completely.

Understanding stoichiometry is critical to calculating normality and equivalent mass because it clarifies the ratio in which substances react. This includes the concept of limiting reactants, theoretical yield, and percent yield. In stoichiometric calculations, normality can be incredibly insightful when comparing reactants that participate in a reaction in different volume ratios.

Stoichiometry isn't just a theoretical construct; it's practically applied in laboratory work, industrial chemical production, and even in understanding biological systems. It demands precision and a clear grasp of the underlying principles to be effectively used in real-world applications.