Lesson 8.4: Acid–Base Equilibrium: Calculations Involving Acidic Solutions

Calculations Involving Acidic Solutions

Overview of Acid–Base Equilibria

Acid–base equilibria involve the quantitative analysis of the species present in acidic solutions and their interactions. Understanding the major entities in solution and their reactions is essential for solving acid–base problems, especially when calculating pH, ion concentrations, and equilibrium constants.

Calculations Involving Solutions of Strong Acids

Strong Acid Ionization and pH Calculation

Strong acids, such as hydrochloric acid (HCl) and nitric acid (HNO3), ionize completely in water. This means the concentration of hydrogen ions, [H+], is equal to the initial concentration of the acid. The autoionization of water is negligible in these cases.

• Key Point 1: For a strong acid, [H+] = initial acid concentration.

• Key Point 2: The pH is calculated using the formula: $pH = -\log [H^+]$

• Key Point 3: The concentration of hydroxide ions, [OH–], can be found using the water dissociation constant: $K_w = [H^+][OH^-] = 1.0 \times 10^{-14}$ at 25°C

Example: For a 1.0 mol/L HCl solution, [H+] = 1.0 mol/L, so pH = 0.

Sample Calculation: Nitric Acid Solution

• Given: 0.25 mol/L HNO3 (strong acid)

• [H+] = 0.25 mol/L

• pH = $-\log(0.25) = 0.60$

• [OH–] = $\frac{1.0 \times 10^{-14}}{0.25} = 4.0 \times 10^{-14}$ mol/L

• pOH = $-\log(4.0 \times 10^{-14}) = 13.40$

Calculations Involving Solutions of Weak Acids

Weak Acid Ionization and Percentage Ionization

Weak acids do not ionize completely in water. The percentage ionization describes the fraction of acid molecules that ionize:

• Key Point 1: Percentage ionization is calculated as: $\text{percentage ionization} = \frac{[\text{ionized acid}]}{[\text{initial acid}]} \times 100\%$

• Key Point 2: The acid dissociation constant, $K_a$, quantifies the extent of ionization: $K_a = \frac{[H^+][A^-]}{[HA]}$

Example: For a 0.10 mol/L ethanoic acid solution with 1.3% ionization: $[H^+] = 0.013 \times 0.10 = 1.3 \times 10^{-3}$ mol/L

Sample Calculation: Percentage Ionization from pH

• Given: pH = 2.38 for 0.10 mol/L methanoic acid

• $[H^+] = 10^{-2.38} = 4.17 \times 10^{-3}$ mol/L

• Percentage ionization = $\frac{4.17 \times 10^{-3}}{0.10} \times 100\% = 4.2\%$

Sample Calculation: Determining $K_a$ from Percentage Ionization

• Given: 0.1000 mol/L ethanoic acid, 1.3% ionized

• Change in [HC2H3O2] = $0.1000 \times 0.013 = 0.0013$ mol/L

• At equilibrium: [H+] = [C2H3O2–] = 0.0013 mol/L, [HC2H3O2] = 0.0987 mol/L

• $K_a = \frac{(0.0013)^2}{0.0987} = 1.7 \times 10^{-5}$

Calculating the pH of Weak Acid Solutions

Using the ICE Table and $K_a$

To calculate the pH of a weak acid solution, set up an ICE (Initial, Change, Equilibrium) table and use the $K_a$ expression. For most weak acids, the autoionization of water is negligible compared to the acid's ionization.

• Key Point 1: Set up the equilibrium expression using the ICE table.

• Key Point 2: Use the "hundred rule" to simplify calculations: if $\frac{[HA]_{initial}}{K_a} > 100$, then $[HA]_{eq} \approx [HA]_{initial}$.

• Key Point 3: Solve for $x = [H^+]$ and calculate pH.

Sample Calculation: pH of Hydrofluoric Acid Solution

• Given: [HF] = 1.00 mol/L, $K_a = 6.6 \times 10^{-4}$

• ICE table: Let $x$ be the amount ionized. $K_a = \frac{x^2}{1.00 - x} \approx \frac{x^2}{1.00}$

• $x^2 = 6.6 \times 10^{-4}$, $x = 2.6 \times 10^{-2}$ mol/L

• pH = $-\log(2.6 \times 10^{-2}) = 1.59$

Sample Calculation: pH of Hypochlorous Acid Solution

• Given: [HClO] = 0.100 mol/L, $K_a = 3.5 \times 10^{-8}$

• ICE table: $K_a = \frac{x^2}{0.100 - x} \approx \frac{x^2}{0.100}$

• $x^2 = 3.5 \times 10^{-9}$, $x = 5.9 \times 10^{-5}$ mol/L

• pH = $-\log(5.9 \times 10^{-5}) = 4.23$

Polyprotic Acids

Definition and Ionization Steps

Polyprotic acids have more than one ionizable hydrogen atom. Each ionization step has its own acid dissociation constant ($K_{a1}$, $K_{a2}$, etc.), and typically, $K_{a1} > K_{a2} > K_{a3}$.

• Key Point 1: Monoprotic acids have one ionizable hydrogen; diprotic have two; triprotic have three.

• Key Point 2: The first ionization usually contributes most to the [H+] and thus to the pH.

• Key Point 3: For most calculations, only the first ionization is considered unless high accuracy is needed.

Sample Calculation: pH of a Polyprotic Acid (Ascorbic Acid)

• Given: [H2C6H6O6] = 0.10 mol/L, $K_{a1} = 7.9 \times 10^{-5}$, $K_{a2} = 1.6 \times 10^{-12}$

• ICE table for first ionization: $K_{a1} = \frac{x^2}{0.10 - x} \approx \frac{x^2}{0.10}$

• $x^2 = 7.9 \times 10^{-6}$, $x = 2.8 \times 10^{-3}$ mol/L

• pH = $-\log(2.8 \times 10^{-3}) = 2.55$

• Equilibrium concentrations: [H2C6H6O6] = 0.097 mol/L [HC6H6O6–] = 2.8 \times 10–3 mol/L [C6H6O62–] = $K_{a2}$ = 1.6 \times 10–12 mol/L

Summary Table: Acid Ionization Constants for Polyprotic Acids

Acid	Formula	Ka1	Ka2	Ka3
Oxalic acid	H2C2O4	5.4 × 10–2	5.4 × 10–5	–
Ascorbic acid	H2C6H6O6	7.9 × 10–5	1.6 × 10–12	–
Sulfuric acid	H2SO4	very large	1.0 × 10–2	–
Phosphoric acid	H3PO4	7.1 × 10–3	6.3 × 10–8	4.2 × 10–13

Summary

• Percentage ionization describes the degree of ionization of a weak acid.

• Polyprotic acids ionize one hydrogen at a time, each with its own $K_a$ value.

• For most polyprotic acids, only the first ionization significantly affects pH.