Acid Base Notes

Influence of Acid Structure on Acidity

• Metal oxides can be basic or acidic in water, depending on the metal cation's properties.
  • Basic oxides: M-O-M + H-OH → KOH + OH⁻, resulting in an increase in hydroxide ions (OH⁻) and a basic solution.
  • Acidic oxides: High charge on the metal removes electron density from the oxygen, stabilizing the structure.
    • Example: Fe^{3+} in Fe(H2O)6^{3+} , where the 3+ charge on iron makes it acidic in water.
    • Fe(H2O)6^{3+} loses a proton (H⁺) to form Fe(H2O)5(OH)^{2+}.

pH and pOH Calculations

• pH is calculated as the negative logarithm of the hydrogen ion concentration: pH = -log[H^+].
• pOH is related to pH by the equation: pH + pOH = 14, where 14 is treated as an exact number.

Example 1: Calculating pH of a Strong Acid (HCl) Solution

• Given: 0.25 M HCl solution.
• HCl is a strong acid and dissociates completely: HCl → H^+ + Cl^-.
• [H^+] = 0.25 M.
• pH = -log(0.25) = 0.60.

Example 2: Calculating pH of a Strong Base (NaOH) Solution

• Given: 0.025 M NaOH solution.
• NaOH is a strong base and dissociates completely: NaOH → Na^+ + OH^-.
• [OH^-] = 0.025 M.
• pOH = -log[OH^-] = -log(0.025) = 1.60.
• pH = 14 - pOH = 14 - 1.60 = 12.40.

Weak Acid Problem: Acetic Acid

• Acetic acid (CH₃COOH) is a weak acid with an acid dissociation constant (Ka) of 1.8 × 10^{-5}.
• The dissociation equilibrium is: CH3COOH ⇌ CH3COO^- + H^+.

Setting up the ICE Table:

• Initial: [CH3COOH] = 0.175 M, [CH3COO^-] = 0, [H^+] = 0.
• Change: [CH3COOH] = -x, [CH3COO^-] = +x, [H^+] = +x.
• Equilibrium: [CH3COOH] = 0.175 - x, [CH3COO^-] = x, [H^+] = x.

Using the Ka Expression:

• Ka = [CH3COO^-][H^+] / [CH3COOH].
• 1.8 × 10^{-5} = x^2 / (0.175 - x).

Solving for x:

• Assuming x is small compared to 0.175, the equation simplifies to 1.8 × 10^{-5} = x^2 / 0.175.
• x = [H^+] = √(0.175 × 1.8 × 10^{-5}) = 1.77 × 10^{-3} M.

Calculating pH:

• pH = -log(1.77 × 10^{-3}) ≈ 2.752.

Weak Acid Problem (Approximation Method)

• Acetic acid has a K_a of 1.8 x 10^{-5}.

• Equilibrium: CH3COOH ⇌ CH3COO^- + H^+

• Initial concentrations: [CH3COOH]i = 0.175 M, [CH3COO^-]i = 0, [H^+]_i = 0

• Approximation: Assume change in [CH_3COOH] is small relative to its initial concentration.

• Ka = [CH3COO^-] [H^+] / [CH_3COOH] = x^2 / 0.175 = 1.8 × 10^{-5}

• x = √(0.175 * 1.8 × 10^{-5}) = 1.77 × 10^{-3} M

• Percent dissociation: (1.77 × 10^{-3} / 0.175) * 100 = 1.01 %

Validity Check:

• The approximation is valid if the percent dissociation is less than 5%.

pH Calculation:

• pH = -log(1.77 × 10^{-3}) = 2.752

Approximation vs Non-approximations

• For a 0.10 M pyridinium solution with K_a = 5.6 × 10^{-6}:

Full Solution:

• Equilibrium: PyH^+ ⇌ Py + H^+

• K_a = [H^+] [Py] / [PyH^+] = x^2 / (0.10 - x)

• Solving the quadratic equation: x^2 + 5.6 × 10^{-4}x - 5.6 × 10^{-7} = 0

Approximation Method:

• [H^+] = √(Ka * [PyH^+]i) = √(5.6 × 10^{-6} * 0.10) = 7.5 × 10^{-4} M

pH Calculation:

• pH = -log(7.5 × 10^{-4}) = 3.12

• Validity Check: (7.5 × 10^{-4} / 0.10) * 100 = 0.75 % < 5% (Approximation is valid)

Approximation vs Non-approximations (Citric Acid)

• For a 0.10 M citric acid solution with K_a = 8.4 × 10^{-4}:

Full Solution:

• K_a = [c^-] [H^+] / [CH] = x^2 / (0.10 - x)

• Solving the quadratic equation: x^2 + 8.4 × 10^{-4}x - 8.4 × 10^{-5} = 0

Approximation Method:

• [H^+] = √(Ka * [CH]i) = √(8.4 × 10^{-4} * 0.10) = 9.2 × 10^{-3} M

• pH = -log(9.2 × 10^{-3}) = 2.03

• Solving quadratic: x = [H^+] = 8.8 x 10^{-3} M, pH= 2.06

Approximation vs Non-approximations (Sulfurous Acid)

• For a 0.10 M sulfurous acid solution with K_a = 1.6 × 10^{-2}:

Full Solution:

• Ka = [H^+] [HSO3^-] / [H2SO3] = x^2 / (0.10 - x)

• Solving the quadratic equation: x^2 + 1.6 × 10^{-2}x - 1.6 × 10^{-3} = 0

Approximation Method:

• [H^+] = √(Ka * [H2SO3]i) = √(1.6 × 10^{-2} * 0.10) = 4.0 × 10^{-2} M

• pH = -log(4.0 × 10^{-2}) = 1.40

• Solving quadratic: x = [H^+] = 3.3 x 10^{-2} M, pH= 1.48

When to Use Approximations

• As K_a becomes larger, the approximation fails because x (change in initial concentration) becomes significant relative to the initial concentration.

• 5% Rule: If x/[HA]_i < 0.05, then the approximation is valid.

• Approximation valid for weak acids and bases.

• [HA]i should be 1000x bigger than Ka or [B]i 1000x bigger than Kb

The pH Measurement and Activity

• Glass electrode setup for pH measurement.

• pH = -log[H^+], but in reality, pH = -log(a{H^+}) where a{H^+} is activity.

• Activity is strongly influenced by ionic concentration.

• pH is defined by the electrochemical potential of H⁺.

Weak Base Problem: Ammonia

• Ammonia (NH3) has a K_b of 1.8 × 10^{-5}.

• Equilibrium: NH3(aq) + H2O(l) ⇌ NH_4^+(aq) + OH^-(aq)

• Kb = [NH4^+] [OH^-] / [NH_3] = x^2 / (0.30 - x) ≈ x^2 / 0.30

• x = [OH^-] = √(1.8 × 10^{-5} * 0.30) = 2.3 × 10^{-3} M

Validity Check:

• (2. 3 × 10^{-3} / 0.30) * 100 = 0.77 % < 5% (Approximation is valid)

pOH Calculation:

• pOH = -log(2.3 × 10^{-3}) = 2.63

pH Calculation:

• pH = 14 - 2.63 = 11.37

Conjugate Base Problem: Sodium Chlorite

• Chlorous acid (HClO₂) has a K_a of 1.1 × 10^{-2}.

• Given: 0.30 M NaClO₂ solution.

• Hydrolysis of chlorite ion: ClO2^-(aq) + H2O(l) ⇌ HClO_2(aq) + OH^-(aq)

• Kb = Kw / K_a = (1 × 10^{-14}) / (1.1 × 10^{-2}) = 9.09 × 10^{-13}

• Kb = [HClO2] [OH^-] / [ClO_2^-] = x^2 / (0.30 - x) ≈ x^2 / 0.30

• x = [OH^-] = √(9.09 × 10^{-13} * 0.30) = 5.2 × 10^{-7} M

Validity Check:

• (5.2 × 10^{-7} / 0.30) * 100 ≈ 1.73 × 10^{-4} % < 5% (Approximation is valid)

pOH Calculation:

• pOH = -log(5.2 × 10^{-7}) = 6.28

pH Calculation:

• pH = 14 - 6.28 = 7.72