Question

Calculate the normality of each of the following solutions. a. 0.250\(M \mathrm{HCl}\) b. 0.105\(M \mathrm{H}_{2} \mathrm{SO}_{4}\) c. \(5.3 \times 10^{-2} M \mathrm{H}_{3} \mathrm{PO}_{4}\) d. 0.134 \(\mathrm{M} \mathrm{NaOH}\) e. 0.00521 \(\mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}\) What is the equivalent mass for each of the acids or bases listed above?

Short answer

The normality of each solution is as follows: a. \(HCl\) = 0.250 N b. \(H_2SO_4\) = 0.210 N c. \(H_3PO_4\) = \(15.9 \times 10^{-2}\) N d. \(NaOH\) = 0.134 N e. \(Ca(OH)_2\) = 0.01042 N The equivalent masses of each compound are: a. \(HCl\) = 36.5 g/equiv b. \(H_2SO_4\) = 49 g/equiv c. \(H_3PO_4\) ≈ 32.67 g/equiv d. \(NaOH\) = 40 g/equiv e. \(Ca(OH)_2\) = 37 g/equiv

Step-by-step solution

Identify the number of equivalents (n) for each solution

For each compound, count the moles of ions that can participate in the acid-base reaction, i.e., the number of H+ ions from acids and OH- ions from bases. a. \(HCl\) ⟶ 1 H+ ion; n = 1 b. \(H_2SO_4\) ⟶ 2 H+ ions; n = 2 c. \(H_3PO_4\) ⟶ 3 H+ ions; n = 3 d. \(NaOH\) ⟶ 1 OH- ion; n = 1 e. \(Ca(OH)_2\) ⟶ 2 OH- ions; n = 2

Calculate the normality of each solution

Now, to find the normality of each solution, multiply the molarity by the number of equivalents (n): a. 0.250 M \(HCl\): Normality = 0.250 × 1 = 0.250 N b. 0.105 M \(H_2SO_4\): Normality = 0.105 × 2 = 0.210 N c. \(5.3 \times 10^{-2} M H_3PO_4\): Normality = \(5.3 \times 10^{-2} × 3 = 15.9 \times 10^{-2}\) N d. 0.134 M \(NaOH\): Normality = 0.134 × 1 = 0.134 N e. 0.00521 M \(Ca(OH)_2\): Normality = 0.00521 × 2 = 0.01042 N

Calculate the equivalent mass of each compound

To find the equivalent mass, divide the molar mass of each compound by the number of equivalents (n): a. \(HCl\): Molar mass = 1 + 35.5 = 36.5 g/mol; Equivalent mass = 36.5/1 = 36.5 g/equiv b. \(H_2SO_4\): Molar mass = 2 × 1 + 32 + 4 × 16 = 98 g/mol; Equivalent mass = 98/2 = 49 g/equiv c. \(H_3PO_4\): Molar mass = 3 × 1 + 31 + 4 × 16 = 98 g/mol; Equivalent mass = 98/3 ≈ 32.67 g/equiv d. \(NaOH\): Molar mass = 23 + 16 + 1 = 40 g/mol; Equivalent mass = 40/1 = 40 g/equiv e. \(Ca(OH)_2\): Molar mass = 40 + 2 × (16 + 1) = 74 g/mol; Equivalent mass = 74/2 = 37 g/equiv So, the equivalent masses of \(HCl\), \(H_2SO_4\), \(H_3PO_4\), \(NaOH\), and \(Ca(OH)_2\) are 36.5 g/equiv, 49 g/equiv, 32.67 g/equiv, 40 g/equiv, and 37 g/equiv, respectively.

Key concepts

Molarity

Molarity is a way of expressing the concentration of a chemical solution. Specifically, it measures how many moles of a solute exist in one liter of solution. A mole is a unit in chemistry that represents 6.022 x 10^23 particles, which might be atoms, molecules, or ions.
In chemical reactions, knowing the molarity helps predict how substances will interact, especially in acid-base reactions. It's one of the most common ways to describe solution concentration because it directly relates to the amount of a substance in a known volume.
To calculate molarity, the formula is:

• Molarity (M) = Moles of solute / Liters of solution

Knowing the molarity of a solution, especially for strong acids and bases like HCl and NaOH, is crucial in various laboratory settings as it provides vital information for conducting accurate experiments.

Equivalent Mass

Equivalent mass relates to the stoichiometry of a reaction. It is the mass of a substance that will combine with or displace one mole of hydrogen ions (H+) in a chemical reaction. This concept is especially important in chemical reactions, including those that involve acids and bases.
To determine the equivalent mass of a compound, you should divide its molar mass by the number of equivalents it can produce in a reaction. For instance, with sulfuric acid (H2SO4), which can provide two H+ ions, the molar mass of 98 g/mol is divided by 2, resulting in an equivalent mass of 49 g/equiv.
This calculation helps us understand how much of a compound is needed to interact with a set number of other ions or molecules in a reaction, aiding in the precise formulation of reactions and solutions.

Acid-Base Reactions

Acid-base reactions are fundamental to chemistry and involve the transfer of protons (H+ ions) from acids to bases. These reactions are characterized by their ability to neutralize each other when mixed. When a strong acid like HCl is mixed with a strong base like NaOH, they form water and a salt, exhibiting a classic acid-base reaction.
Understanding acid-base reactions is important because they are prevalent in numerous real-world processes, from digestion in our stomachs to industrial applications. These reactions are important in titrations, where analyzing the concentration of a solution is possible by adding a titrant of a known concentration until reaction completion (neutralization).
These reactions often require precise calculations of normality or molarity to understand how substances interact in solution, highlighting the importance of concepts like molarity and equivalent mass.

HCl

Hydrochloric acid (HCl) is a strong acid commonly found in gastric acid in the human stomach, and widely used in industry. Being a monoprotic acid, it donates one proton per molecule during an acid-base reaction, thus having one equivalency factor.
The molarity of an HCl solution can quickly help define its normality. Since HCl releases just one H+ ion, the normality equal its molarity. This simple relationship makes it straightforward to work with in various chemical processes.
HCl is not only essential in lab experiments but also in the manufacturing industry, for processing metals, in food production, and the formation of certain chemical compounds, making it a vital chemical in many areas.

H2SO4

Sulfuric acid (H2SO4) is another strong acid that plays a vital role in chemistry. It is known as diprotic because it can donate two protons (H+ ions) in a reaction. Thus, when exploring the normality of H2SO4, its equivalency factor is 2. This means normality is twice its molarity.
Calculating its normality involves multiplying the molarity by 2 because it can release two moles of H+ ions per mole of H2SO4. This characteristic makes it highly reactive.
H2SO4 is widely used in the industrial sector, from making fertilizers, refining oil to processing minerals. Its capacity to donate two protons makes it a pivotal acid in both theoretical and practical chemistry applications.