Question

For each of the following solutions, calculate the normality. a. \(25.2 \mathrm{~mL}\) of \(0.105 \mathrm{M} \mathrm{HCl}\) diluted with water to a total volume of \(75.3 \mathrm{~mL}\) b. \(0.253 M \mathrm{H}_{3} \mathrm{PO}_{4}\) c. \(0.00103 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}\)

Short answer

a. Normality of diluted HCl = \(\frac{(0.105 M)(25.2 \mathrm{~mL})}{75.3 \mathrm{~mL}}\) b. Normality of H\(_{3}\)PO\(_{4}\) = (0.253 M) × 3 c. Normality of Ca(OH)\(_{2}\) = (0.00103 M) × 2

Step-by-step solution

a. Calculate new concentration for HCl solution after dilution

First, we need to find the new concentration after diluting the HCl solution. The original concentration was 0.105 M, and the volume was increased with water from 25.2 mL to 75.3 mL. We will use the equation: \[C_{1}V_{1} = C_{2}V_{2}\] Where \(C_{1}\) is the initial concentration, \(V_{1}\) is the initial volume, \(C_{2}\) is the final concentration, and \(V_{2}\) is the final volume. In our case: \[(0.105 M)(25.2 \mathrm{~mL}) = C_{2}(75.3 \mathrm{~mL})\] Now solve for \(C_{2}\): \[C_{2} = \frac{(0.105 M)(25.2 \mathrm{~mL})}{75.3 \mathrm{~mL}}\]

b. Calculate the equivalence factor for HCl

To find the normality of the diluted HCl solution, we need to know how many H+ ions are being donated by the HCl molecule. Since HCl donates only one H+ ion, the equivalence factor is 1.

c. Calculate the normality of diluted HCl solution

Now we can calculate the normality by multiplying the new concentration by the equivalence factor: Normality = New Concentration × Equivalence Factor Normality = \(C_{2}\) × 1 Normality = \(\frac{(0.105 M)(25.2 \mathrm{~mL})}{75.3 \mathrm{~mL}}\)

a. Calculate the equivalence factor for H3PO4

To calculate the normality of the given 0.253 M H3PO4 solution, we must first find the equivalence factor, i.e., how many H+ ions can be donated by the H3PO4 molecule. H3PO4 donates three H+ ions, so the equivalence factor is 3.

b. Calculate the normality of H3PO4 solution

Now we can calculate the normality for the H3PO4 solution by multiplying the concentration by the equivalence factor: Normality = Concentration × Equivalence Factor Normality = (0.253 M) × 3

a. Calculate the equivalence factor for Ca(OH)2

To calculate the normality of the given 0.00103 M Ca(OH)2 solution, first find the equivalence factor, i.e., how many OH- ions can be donated by the Ca(OH)2 molecule. Ca(OH)2 donates two OH- ions, so the equivalence factor is 2.

b. Calculate the normality of Ca(OH)2 solution

Now calculate the normality for the Ca(OH)2 solution by multiplying the concentration by the equivalence factor: Normality = Concentration × Equivalence Factor Normality = (0.00103 M) × 2

Key concepts

Dilution Equation

The dilution equation, \(C_1V_1 = C_2V_2\), is essential in understanding how to adjust concentrations in solutions by adding a solvent, usually water. This process doesn't change the amount of solute but changes the concentration. Let's break down what each term means:

• \(C_1\): Initial concentration of the solution before dilution.
• \(V_1\): Initial volume of the solution before dilution.
• \(C_2\): Final concentration after dilution.
• \(V_2\): Final volume after dilution, when more solvent is added.

To use this formula, multiply the initial concentration by the initial volume to find their product. Then divide this product by the final volume to find the new concentration, \(C_2\). This is crucial when preparing solutions of a desired concentration, as seen in chemistry labs and calculations.

Equivalence Factor

The equivalence factor simplifies calculating normality by connecting molarity with the number of reactive particles a solute can donate or accept. In acid-base chemistry, it’s vital to understand this since it determines how solutions interact. The equivalence factor is defined as the number of moles of reactive units in one mole of a substance. Here's how it applies to different molecules:

• HCl (Hydrochloric acid): Donates one H⁺ ion, so the equivalence factor is 1.
• H₃PO₄ (Phosphoric acid): Donates three H⁺ ions, giving it an equivalence factor of 3.
• Ca(OH)₂ (Calcium hydroxide): Provides two OH^- ions, resulting in an equivalence factor of 2.

Understanding and finding the equivalence factor allow us to accurately convert molarity to normality, enhancing our grasp of chemical reactions and solution characteristics.

Acid-Base Chemistry

Acid-base chemistry is fundamental to understanding reaction dynamics, from titrations to pH calculations. Acids are substances that release H⁺ ions, while bases give OH^- ions in a solution. Key Concepts of Acid-Base Chemistry:

• pH Scale: Measures acidity or basicity of a solution. Lower pH indicates an acid, while higher pH shows a base.
• Neutralization: Occurs when acids and bases mix, usually producing water and a salt.
• Strong vs Weak Acids/Bases: Strong acids/bases dissociate completely in water, while weak ones do not.

Knowing acid-base properties helps us predict behaviors in chemical reactions and real-world scenarios like digestion or environmental science. These principles underpin many scientific processes and applications.