Understanding the reactions of bases with acids and ammonium salts is fundamental in the study of chemistry, particularly in the Cambridge IGCSE syllabus under the unit "Acids, Bases, and Salts." These reactions are pivotal in various industrial processes, environmental applications, and biological systems. Mastery of these concepts not only aids in academic success but also in practical real-world applications.

Bases are substances that can accept protons (H⁺ ions) or donate electron pairs. They typically have a bitter taste, slippery texture, and can change the color of indicators like litmus paper to blue. Common examples include sodium hydroxide (NaOH) and ammonia (NH₃). Acids, on the other hand, are substances that can donate protons or accept electron pairs. They usually have a sour taste, can conduct electricity, and turn indicators like litmus paper red. Examples include hydrochloric acid (HCl) and sulfuric acid (H₂SO₄). The reaction between an acid and a base is known as a neutralization reaction, producing water and a salt: $$ \text{Acid} + \text{Base} \rightarrow \text{Salt} + \text{Water} $$ For example: $$ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} $$

When bases react with acids, they undergo neutralization to form water and a corresponding salt. This reaction is exothermic, releasing heat energy. The general equation is: $$ \text{Base} + \text{Acid} \rightarrow \text{Salt} + \text{Water} $$ **Example Reaction:** $$ \text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} $$ In this reaction, sodium hydroxide (a strong base) reacts with hydrochloric acid (a strong acid) to produce sodium chloride (common salt) and water. **Indicator Changes:** During the reaction, indicators like phenolphthalein change color. Phenolphthalein turns from pink in basic solutions to colorless in acidic solutions, signaling the completion of neutralization. **Applications:** - **Titration:** A method used to determine the concentration of an acid or base by reacting it with a base or acid of known concentration. - **Antacid Tablets:** Contain basic substances that neutralize excess stomach acid to relieve heartburn. **Equilibrium Considerations:** In aqueous solutions, the extent of ionization of acids and bases affects the position of equilibrium. Strong acids and bases fully ionize, leading to complete neutralization, whereas weak acids and bases partially ionize, resulting in incomplete neutralization.

Ammonium salts, such as ammonium chloride (NH₄Cl), react with bases to produce ammonia gas and water. This reaction is a form of deprotonation where the base removes a proton (H⁺) from the ammonium ion (NH₄⁺). **General Reaction:** $$ \text{Ammonium Salt} + \text{Base} \rightarrow \text{Ammonia} + \text{Water} + \text{Salt} $$ **Specific Example:** $$ \text{NH}_4\text{Cl} + \text{NaOH} \rightarrow \text{NH}_3 \uparrow + \text{H}_2\text{O} + \text{NaCl} $$ In this reaction, ammonium chloride reacts with sodium hydroxide to produce ammonia gas, water, and sodium chloride. **Characteristic Features:** - **Ammonia Gas Production:** The release of NH₃ can be detected by its pungent odor and its ability to turn damp red litmus paper blue. - **Endothermic Reaction:** The reaction absorbs heat from the surroundings, making it feel cold to the touch. **Applications:** - **Lehmann’s Fern Manufacture:** Uses ammonium salts and bases to grow ferns. - **Calibration of pH Meters:** Ammonia gas is used to generate basic conditions. **Safety Considerations:** Handling ammonia gas requires caution due to its corrosive nature and irritant effects on the respiratory system.

**Lewis and Brønsted-Lowry Definitions:** While the Brønsted-Lowry definition focuses on proton transfer, the Lewis definition broadens the concept to include electron pair donors and acceptors. In the context of acid-base reactions: - **Lewis Acid:** Electron pair acceptor. - **Lewis Base:** Electron pair donor. This dual perspective allows for a more comprehensive understanding of the mechanisms behind acid-base interactions, especially in complex reactions involving coordination compounds. **Mathematical Derivations:** The strength of an acid or base can be quantified using the equilibrium constant expressions: - **For Acids (Ka):** $$ K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} $$ - **For Bases (Kb):** $$ K_b = \frac{[\text{BH}^+][\text{OH}^-]}{[\text{B}]} $$ The relationship between Ka and Kb for conjugate acid-base pairs is given by: $$ K_a \times K_b = K_w = 1.0 \times 10^{-14} $$ where \( K_w \) is the ion-product constant of water. **pH Calculations:** The pH of solutions resulting from acid-base reactions can be determined using: $$ \text{pH} = -\log[\text{H}^+] $$ For neutralization reactions, if equimolar amounts of strong acids and bases are mixed, the resulting solution is neutral (pH 7). Deviations occur with excess acid or base.

**Problem 1: Titration Curve Analysis** A 25.0 mL sample of hydrochloric acid is titrated with 0.100 M sodium hydroxide. Calculate the pH after the addition of 15.0 mL of NaOH. **Solution:** 1. **Calculate moles of HCl:** $$ n_{\text{HCl}} = 0.100 \, \text{M} \times 0.025 \, \text{L} = 0.0025 \, \text{mol} $$ 2. **Calculate moles of NaOH added:** $$ n_{\text{NaOH}} = 0.100 \, \text{M} \times 0.015 \, \text{L} = 0.0015 \, \text{mol} $$ 3. **Determine excess reactant:** $$ n_{\text{HCl remaining}} = 0.0025 - 0.0015 = 0.0010 \, \text{mol} $$ 4. **Total volume after addition:** $$ V_{\text{total}} = 25.0 \, \text{mL} + 15.0 \, \text{mL} = 40.0 \, \text{mL} = 0.0400 \, \text{L} $$ 5. **Concentration of excess HCl:** $$ [\text{H}^+] = \frac{0.0010 \, \text{mol}}{0.0400 \, \text{L}} = 0.025 \, \text{M} $$ 6. **Calculate pH:** $$ \text{pH} = -\log(0.025) \approx 1.60 $$ **Problem 2: Buffer Solution Preparation** Prepare 500 mL of a buffer solution with a pH of 4.75 using acetic acid (Ka = $1.8 \times 10^{-5}$) and its sodium salt. **Solution:** 1. **Henderson-Hasselbalch equation:** $$ \text{pH} = \text{p}K_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) $$ 2. **Calculate pKa:** $$ \text{p}K_a = -\log(1.8 \times 10^{-5}) \approx 4.74 $$ 3. **Set up equation:** $$ 4.75 = 4.74 + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) $$ $$ 0.01 = \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) $$ $$ \frac{[\text{A}^-]}{[\text{HA}]} = 10^{0.01} \approx 1.023 $$ 4. **Assume total concentration (C):** Let \( [\text{HA}] + [\text{A}^-] = C \). $$ \frac{[\text{A}^-]}{[\text{HA}]} = 1.023 \Rightarrow [\text{A}^-] = 1.023 [\text{HA}] $$ $$ [\text{HA}] + 1.023[\text{HA}] = C \Rightarrow 2.023[\text{HA}] = C \Rightarrow [\text{HA}] = \frac{C}{2.023} $$ $$ [\text{A}^-] = 1.023 \times \frac{C}{2.023} = \frac{1.023C}{2.023} $$ 5. **Choose C (e.g., 0.10 M):** $$ [\text{HA}] = \frac{0.10}{2.023} \approx 0.0495 \, \text{M} $$ $$ [\text{A}^-] = \frac{1.023 \times 0.10}{2.023} \approx 0.0505 \, \text{M} $$ 6. **Prepare solutions:** - **0.0495 M Acetic Acid:** $$ \text{Molar mass of CH}_3\text{COOH} \approx 60.05 \, \text{g/mol} $$ $$ \text{Mass} = 0.0495 \, \text{M} \times 0.500 \, \text{L} \times 60.05 \, \text{g/mol} \approx 1.49 \, \text{g} $$ - **0.0505 M Sodium Acetate:** $$ \text{Molar mass of NaCH}_3\text{COO} \approx 82.03 \, \text{g/mol} $$ $$ \text{Mass} = 0.0505 \, \text{M} \times 0.500 \, \text{L} \times 82.03 \, \text{g/mol} \approx 2.07 \, \text{g} $$ 7. **Mix both in 500 mL solution.** **Problem 3: Decomposition of Ammonium Chloride** Calculate the temperature change when 50.0 g of ammonium chloride reacts with excess sodium hydroxide. **Solution:** 1. **Balanced Reaction:** $$ \text{NH}_4\text{Cl} + \text{NaOH} \rightarrow \text{NH}_3 \uparrow + \text{H}_2\text{O} + \text{NaCl} $$ 2. **Molar Masses:** - NH₄Cl: 53.49 g/mol 3. **Calculate moles of NH₄Cl:** $$ n = \frac{50.0 \, \text{g}}{53.49 \, \text{g/mol}} \approx 0.935 \, \text{mol} $$ 4. **Enthalpy Change (ΔH):** Suppose ΔH = +15.1 kJ/mol (endothermic) 5. **Total Heat Absorbed:** $$ q = n \times \Delta H = 0.935 \times 15.1 \approx 14.13 \, \text{kJ} $$ 6. **Temperature Change (ΔT):** Assume specific heat capacity (c) = 4.18 J/g°C, mass (m) = 100 g (solution) $$ q = m \times c \times \Delta T $$ $$ 14,130 \, \text{J} = 100 \, \text{g} \times 4.18 \, \text{J/g°C} \times \Delta T $$ $$ \Delta T = \frac{14,130}{418} \approx 33.8 \, \text{°C} $$ Since the reaction is endothermic, the temperature decreases by approximately 33.8°C.

**Environmental Science:** Acid-base reactions are crucial in understanding acid rain formation. When atmospheric pollutants like sulfur dioxide (SO₂) react with water, they form sulfuric acid (H₂SO₄), which can neutralize in the soil, affecting plant growth and water bodies. **Biology:** Buffers, which are solutions that resist pH changes, are essential in maintaining the pH of blood and cellular environments. The bicarbonate buffer system in blood is a prime example, involving the equilibrium between carbonic acid (H₂CO₃) and bicarbonate ions (HCO₃⁻). **Industrial Chemistry:** Neutralization reactions are employed in wastewater treatment to neutralize acidic or basic effluents before discharge. Additionally, the Haber process for ammonia synthesis involves managing acid-base equilibria. **Materials Science:** The formation of soaps (salts of fatty acids) involves the neutralization of fatty acids with bases like sodium hydroxide. Understanding these reactions is vital for developing sustainable and biodegradable materials. **Pharmaceuticals:** Many drugs are formulated as salts to enhance their stability and solubility. The interaction between active pharmaceutical ingredients (APIs) and excipients often involves acid-base chemistry.

• Neutralization reactions between bases and acids produce water and salts.
• Bases react with ammonium salts to release ammonia gas, water, and salts.
• Understanding acid-base equilibria is essential for applications in environmental science, biology, and industry.
• Proper handling and calculation of these reactions are crucial for safe and effective chemical processes.