Lecture Notes on Acids, Bases, and Equilibria

Polyprotic Acid Problem

• Oxalic acid is a diprotic acid with two acid dissociation constants: $K<em>{a1} = 6.5 \times 10^{-2}$ and $K</em>{a2} = 6.1 \times 10^{-5}$. The problem aims to determine the pH and concentrations of all species in a 1.0 M oxalic acid solution.

Step 1

• The dissociation of oxalic acid ((OXH2)) in water is considered: $OXH</em>2 \rightleftharpoons OXH^- + H^+$

• Initial concentration of (OXH_2) is 1.0 M.

• Change in concentration is represented by -x for (OXH_2) and +x for (OXH^-) and (H^+).

• Equilibrium concentrations are: ([OXH_2] = 1.0 - x), ([OXH^-] = x), and ([H^+] = x).

• The expression for (K_{a1}) is:

  $K<em>{a1} = \frac{[OXH^-][H^+]}{[OXH</em>2]} = \frac{x^2}{1.0 - x} = 6.5 \times 10^{-2}$

• Solving the quadratic equation:

  $x^2 + 6.5 \times 10^{-2}x - 6.5 \times 10^{-2} = 0$

• The positive root is x = 0.22 M.

• Equilibrium concentrations:

  • [OXH_2] = 1.0 - 0.22 = 0.8 M
  • [OXH^-] = 0.22 M
  • [H^+] = 0.22 M

Step 2

• The second dissociation step is:

  $OXH^- \rightleftharpoons OX^{2-} + H^+$

• Initial concentrations are ([OXH^-] = 0.22 M) and ([H^+] = 0.22 M).

• Change in concentration is -x for (OXH^-) and +x for (OX^{2-}) and (H^+).

• Equilibrium concentrations are: ([OXH^-] = 0.22 - x), ([OX^{2-}] = x), and ([H^+] = 0.22 + x).

• The expression for (K_{a2}) is:

  $K_{a2} = \frac{[OX^{2-}][H^+]}{[OXH^-]} = \frac{x(0.22 + x)}{0.22 - x}$

• Approximation: Assuming x is much smaller than 0.22, the equation simplifies to:

  $x \approx K_{a2} = 6.1 \times 10^{-5}$

• Equilibrium concentrations:

  • [OXH_2] = 0.80 M
  • [OX^{2-}] = 6.1 \times 10^{-5} M
  • [H^+] = 0.22 M

• pH Calculation:

  $pH = -\log(0.22) = 0.66$

Common Ion Effect

• The common ion effect describes the influence on equilibrium when one or more species in a reaction is shared with another equilibrium process.
• Important in various aspects of chemical equilibria including:
  • Acid-Base chemistry
  • Solubility

Example

• Consider the equilibrium of hydrofluoric acid (HF) in water:

  $HF(aq) \rightleftharpoons H^+(aq) + F^-(aq)$

• Sodium fluoride (NaF) dissociates in water as follows:

  $NaF(aq) \rightarrow Na^+(aq) + F^-(aq)$

• Fluoride ((F^-)) is the common ion.

• Its presence in the HF equilibrium will influence the NaF reaction and vice versa.

• Also:

  $F^-(aq) + H_2O(l) \rightleftharpoons HF + OH^-(aq)$

Quantitative Common Ion Effect

• Problem: Determine the pH of a 0.500 M solution of acetic acid ((K_a = 1.8 \times 10^{-5})). Then, recalculate the pH after adding sodium acetate to increase the initial acetate concentration by 0.100 M and 0.500 M.

Part 1: Acetic Acid Only

• Equilibrium: (CH3COOH \rightleftharpoons CH3COO^- + H^+)

• Initial concentrations: ([CH3COOH] = 0.500 M), ([CH3COO^-] = 0), ([H^+] = 0)

• Change: (-x) for (CH3COOH), (+x) for (CH3COO^-) and (H^+)

• Equilibrium concentrations: ([CH3COOH] = 0.500 - x), ([CH3COO^-] = x), ([H^+] = x)

• (K_a) expression:

  $K<em>a = \frac{[CH</em>3COO^-][H^+]}{[CH_3COOH]} = \frac{x^2}{0.500 - x}$

• Approximation: Assume (x << 0.500 M), so (0.500 - x \approx 0.500)

  $1.8 \times 10^{-5} = \frac{x^2}{0.500}$

  $x = [H^+] = \sqrt{1.8 \times 10^{-5} \times 0.500} = 0.0030 M$

• pH Calculation:

  $pH = -\log[H^+] = -\log(0.0030) = 2.52$

• Check Approximation:

  $\frac{0.0030}{0.500} \times 100 = 0.6\%$

  • The approximation is valid.

Part 2: Acetic Acid with 0.100 M Acetate

• Initial concentrations: ([CH3COOH] = 0.500 M), ([CH3COO^-] = 0.100 M), ([H^+] = 0)

• Change: (-x) for (CH3COOH), (+x) for (CH3COO^-) and (H^+)

• Equilibrium concentrations: ([CH3COOH] = 0.500 - x), ([CH3COO^-] = 0.100 + x), ([H^+] = x)

• (K_a) expression:

  $K<em>a = \frac{[CH</em>3COO^-][H^+]}{[CH_3COOH]} = \frac{(0.100 + x)x}{0.500 - x}$

• Approximation: Assume (x << 0.100 M), so (0.100 + x \approx 0.100) and (0.500 - x \approx 0.500)

  $1.8 \times 10^{-5} = \frac{0.100 \times x}{0.500}$

  $x = [H^+] = \frac{1.8 \times 10^{-5} \times 0.500}{0.100} = 9.0 \times 10^{-5} M$

• pH Calculation:

  $pH = -\log[H^+] = -\log(9.0 \times 10^{-5}) = 4.05$

Part 3: Acetic Acid with 0.500 M Acetate

• Initial concentrations: ([CH3COOH] = 0.500 M), ([CH3COO^-] = 0.500 M), ([H^+] = 0)

• Change: (-x) for (CH3COOH), (+x) for (CH3COO^-) and (H^+)

• Equilibrium concentrations: ([CH3COOH] = 0.500 - x), ([CH3COO^-] = 0.500 + x), ([H^+] = x)

• (K_a) expression:

  $K<em>a = \frac{[CH</em>3COO^-][H^+]}{[CH_3COOH]} = \frac{(0.500 + x)x}{0.500 - x}$

• Approximation: Assume (x << 0.500 M), so (0.500 + x \approx 0.500) and (0.500 - x \approx 0.500)

  $1.8 \times 10^{-5} = \frac{0.500 \times x}{0.500}$

  $x = [H^+] = 1.8 \times 10^{-5} M$

• pH Calculation:

  $pH = -\log[H^+] = -\log(1.8 \times 10^{-5}) = 4.74$

• Note: (pKa = -\log(Ka) = -\log(1.8 \times 10^{-5}) = 4.74)

• The solution now acts as a buffer.

Aqueous Equilibria

• Topics covered include:
  • Buffers
  • Titrations
  • Solubility ((K_{sp}))

Textbook Questions

• Recommended textbook questions for:
  • Buffers: 21, 23, 27, 29, 33, 35, 37, 39, 41
  • Acid-Base Titrations: 57, 63, 65, 71, 73
  • Polyprotic Acid Titrations: 87, 91
  • Solubility Equilibria: 95, 97, 101, 103, 105, 107, 109, 117, 121

pH Buffers

• pH Buffers: Solutions that resist changes in pH upon addition of acids or bases.

• Examples:

  • Blood (carbonic acid/bicarbonate buffer): pH = 7.4
  • Sea Water (carbonic acid/bicarbonate buffer): pH = 8.1

• Living organisms that produce (CaCO_3) skeletons are relevant in this context.

  $HCO<em>3^- \rightleftharpoons CO</em>3^{2-} + H^+ \rightleftharpoons CO_2 + O^{2-}$

• A buffer solution is resistant to pH change. It resists changes to added (H^+) or (OH^-).

• pH buffers take advantage of the common ion effect and are made of acid/base conjugate pairs (weak acid or base with a common ion).

How pH Buffers Work

• Buffer solution:

  $HA + H<em>2O \rightleftharpoons A^- + H</em>3O^+$

• Added (H^+) reacts with (A^-) to make HA (weak acid).

  $A^- + H_2O \rightleftharpoons HA + OH^-$

• Added (OH^-) reacts with HA to make (A^-).

• Henderson-Hasselbalch Equation:

  $pH = pK_a + \log(\frac{[A^-]}{[HA]})$

• If ([HA] = [A^-]), then (pH = pK_a).

• Considerations:

  • Will the buffer be exposed to (H^+) or (OH^-)? Prepare accordingly for both cases.
  • Aim for an acid with (pK_a = pH).
  • Use a large excess of HA and (A^-) (e.g., ~100-fold excess).

The Henderson-Hasselbalch Equation

• Equilibrium:

  $HA(aq) \rightleftharpoons A^-(aq) + H^+(aq)$

• If ([HA]i) and ([A^-]i) are comparable and we can assume that changes (\Delta[HA]) or (\Delta[A^-]) are small:

  $K<em>a = \frac{[A^-][H^+]}{[HA]} = \frac{[A^-]</em>i[H^+]}{[HA]_i}$

  $[H^+]<em>e = K</em>a \frac{[HA]<em>i}{[A^-]</em>i}$

• pH Calculation:

  $pH = -\log[H^+] = -\log K<em>a - \log \frac{[HA]</em>i}{[A^-]_i}$

  $pH = pK<em>a + \log \frac{[A^-]</em>i}{[HA]_i}$

Buffer Solutions Example

• Objective: Prepare a buffer solution with a target pH of 3.20 using either (HNO2/NaNO2) ((Ka = 4.0 \times 10^{-4})) or (AcOH/AcONa) ((Ka = 1.8 \times 10^{-5})). Generate 0.100 M buffer solutions (in acid concentration) and calculate the pH following the addition of 0.0010 M HCl.

Using AcOH / AcONa

• Henderson-Hasselbalch equation:

  $pH = pK_a + \log \frac{[A^-]}{[HA]}$

  $3.20 = -\log(1.8 \times 10^{-5}) + \log(x)$

  $3.20 = 4.74 + \log(x)$

  $\log(x) = -1.54$

  $x = \frac{[AcO^-]}{[AcOH]}$

  $x[AcOH] = [AcO^-]$

• Given ([AcOH] = 0.100 M), then:

  $[AcO^-] = 0.0028 M$

• Equilibrium: (AcOH \rightleftharpoons AcO^- + H^+)

• Initial concentrations: ([AcOH] = 0.100 M), ([AcO^-] = 0.0028 M)

• Addition of 0.0010 M HCl:

  • (AcOH) increases by 0.0010 M (to 0.101 M)
  • (AcO^-) decreases by 0.0010 M (to 0.0018 M)

• pH after HCl addition:

  $pH = 4.74 + \log \frac{0.0018}{0.101}$

  $pH = 2.99$

• There is a significant change in pH (one's place).