Acid-Base Neutralization Reactions and Solution Concentrations

4.3 Acids, Bases, and Neutralization Reactions

Introduction to Acid-Base Reactions

Acid-base reactions are a fundamental class of chemical reactions in which an acid reacts with a base to produce water and an ionic salt. These reactions are commonly referred to as neutralization reactions and are classified as double displacement reactions.

• Acid: A substance that increases the concentration of H+ ions in solution.

• Base: A substance that increases the concentration of OH- ions in solution.

• Neutralization: The process in which H+ ions from the acid react with OH- ions from the base to form water.

General Equation for Neutralization

When an acid and base react:

• General form: $\text{Acid} + \text{Base} \rightarrow \text{Water} + \text{Salt}$

Examples of Acid-Base Reactions

• Strong Acid and Strong Base: $\mathrm{HCl}_{(aq)} + \mathrm{NaOH}_{(aq)} \rightarrow \mathrm{H_2O}_{(l)} + \mathrm{NaCl}_{(aq)}$ Strong acid: HCl Strong base: NaOH Products: Water and ionic salt (NaCl)

• Weak Acid and Weak Base: $\mathrm{C_2H_5OOH}_{(aq)} + \mathrm{NH_3}_{(aq)} \rightarrow \mathrm{H_2O}_{(l)} + \mathrm{NH_4C_2H_5OO}_{(aq)}$ Weak acid: C2H5OOH (acetic acid) Weak base: NH3 (ammonia) Products: Water and ionic salt (ammonium acetate)

Neutralization Reaction Mechanism

During a neutralization reaction, the concentration of H+ ions in water decreases as they react with OH- ions to form H2O. This process results in a solution that is less acidic or basic, depending on the initial concentrations.

• Acids: Characterized by a higher concentration of H+ ions.

• Bases: Characterized by a higher concentration of OH- ions.

• Pure Water: Neutral, with equal concentrations of H+ and OH- ions.

• pH Scale: Used to rank the acidity or basicity of a solution.

4.5 Concentrations of Solutions

Introduction to Solution Concentration

The concentration of a solution describes how much solute is dissolved in a given amount of solvent. It is a key concept in chemistry for preparing and analyzing solutions.

• Solute: The substance being dissolved (e.g., NaCl).

• Solvent: The substance doing the dissolving (e.g., water).

• Molarity (M): The most common unit for expressing concentration, defined as moles of solute per liter of solution.

Molarity: Definition and Formula

Molarity is calculated using the following formula:

• $\text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in Liters}}$

• Units: mol/L = M

Example: A 6.0 M solution of HCl contains 6.0 mol of HCl per 1.0 L of solution.

Calculating Molarity: Worked Example

Example 4.9: Calculate the molarity of a solution made by dissolving 23.4 g of sodium sulfate (Na2SO4) in enough water to form 125 mL of solution.

• Molecular Weight (MW) of Na2SO4: $2 \times 22.990 \text{ g/mol} + 32.066 \text{ g/mol} + 4 \times 15.999 \text{ g/mol} = 142.042 \text{ g/mol}$

• Convert mass to moles: $\text{mol Na}_2\text{SO}_4 = \frac{23.4 \text{ g}}{142.042 \text{ g/mol}} = 0.16474 \text{ mol}$

• Convert volume to liters: $125 \text{ mL} \times \left( \frac{1 \text{ L}}{1000 \text{ mL}} \right) = 0.125 \text{ L}$

• Calculate molarity: $M = \frac{0.16474 \text{ mol}}{0.125 \text{ L}} = 1.32 \text{ M}$

Calculating Mass of Solute for a Desired Molarity

Example 4.10: How many grams of NaOH are needed to make 1.5 L of a 3.5 M solution?

• Step 1: Calculate moles needed: $\text{mol} = \text{Molarity} \times \text{Volume}$ $\text{mol} = 3.5 \text{ mol/L} \times 1.5 \text{ L} = 5.25 \text{ mol}$

• Step 2: Convert moles to grams: $\text{grams} = 5.25 \text{ mol} \times \frac{39.997 \text{ g}}{1 \text{ mol}} = 210 \text{ g}$

• Conclusion: You need 210 g of NaOH to make 1.5 L of a 3.5 M solution.

Summary Table: Solution Preparation Calculations

Calculation Step	Formula	Example Value
Convert mass to moles	$\text{mol} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}$	$\frac{23.4}{142.042} = 0.16474$
Convert mL to L	$\text{L} = \text{mL} \times \frac{1}{1000}$	$125 \times \frac{1}{1000} = 0.125$
Calculate Molarity	$M = \frac{\text{mol}}{\text{L}}$	$\frac{0.16474}{0.125} = 1.32$
Calculate grams needed	$\text{grams} = \text{mol} \times \text{molar mass}$	$5.25 \times 39.997 = 210$

Additional info: The examples provided illustrate the stepwise approach to solution preparation, including unit conversions and the use of molar mass for calculations.