Chapter 17.1-3

Chapter 16 Review and Practice

Calculations for Strong and Weak Acids

• Practice questions include:

  1. Calculations for strong acids

  2. Calculations for weak acids

  3. Calculations for polyprotic acids and salts

• Consideration of Ka (acid dissociation constant) and Kb (base dissociation constant).

Chapter 17 Overview: Buffers and Titration

Section 17.1: The Common-Ion Effect

• The common-ion effect refers to the shift in equilibrium that occurs when an ion common to the equilibrium is added, affecting the solubility of salts and dissociation of weak acids/bases.

• Example: Dissolving NaClO in solution results in the equilibrium:
  $\text{NaClO(aq)} \rightarrow \text{Na}^+ (aq) + \text{ClO}^- (aq)$

Calculating Concentration Example

1. Determine concentration of ClO⁻:

   • $\text{mol ClO}^- = \frac{(0.074 \, g) \times (1 \, mol/74 \, g)}{0.100 \, L} = 0.010 \, M$

2. Balanced equation for weak electrolyte dissociation:

   • $\text{HClO(aq) + H}_2\text{O(l)} \rightleftharpoons \text{H}_3\text{O}^+(aq) + \text{ClO}^-(aq)$

3. Setup ICE table for concentrations:

   • | Species | Initial | Change | Equilibrium |
     | --- | --- | --- | --- |
     | HClO | 0.35 | -X | 0.35 - X |
     | H₃O⁺ | 0 | X | X |
     | ClO⁻ | 0.010 | +X | 0.010 + X |

Section 17.2: Buffers

Definition of Buffered Solutions

• A buffered solution consists of a weak acid/conjugate base pair, or a weak base/conjugate acid pair.

• When acid or base is added, the solution resists significant changes in pH.

• Common examples:

  • Acetic acid (CH₃COOH) and acetate (CH₃COO⁻)

  • Hydrofluoric acid (HF) and fluoride (F⁻)

  • Ammonium hydroxide (NH₃) and ammonium (NH₄⁺)

• Natural examples: Blood and oceans exhibit buffer properties.

Making a Buffer

1. Mix a weak acid (e.g., HF) with its conjugate base (e.g., NaF).

2. Mix a weak base (e.g., NH₃) with its conjugate acid (e.g., NH₄Cl).

3. Combine a weak acid (e.g., HF) with a strong base (e.g., NaOH).

4. Combine a weak base (e.g., NH₃) with a strong acid (e.g., HCl).

Examples of Buffer Systems and Appropriateness

• Will form a buffer:

  • (a) HNO₂ and NaNO₂ (YES)

  • (d) CH₃NH₃Cl and CH₃NH₂ (YES)

• Will not form a buffer:

  • (b) HCl and KCl (NO)

  • (c) HNO₃ and NaOH (NO)

  • (e) NH₃ and HNO₃ (YES, if NH₃ is in excess)

Henderson-Hasselbalch Equation

• The equation relates the pH of a buffer to the concentration of acid and conjugate base.

• $\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-$

• The equation is:
  $pH = pK_a + \log\frac{[A^-]}{[HA]}$

• Assumption: If $x << [HA]$ and $x << [A^-]$, initial concentrations can be used:

  • Where: $pK_a = -\log K_a$

Special Case: Equal Concentrations

• When $[HA] = [A^-]$, it simplifies to:
  $pH = pK_a$

Calculating pH of Buffers

Example: Sodium Benzoate Buffer

• Given:

  • Sodium benzoate $\text{NaC}_6\text{H}_5\text{COO}$ (molar mass = 144 g/mol)

  • Amount = 2.88 g

  • Volume of benzoic acid $C_6H_5COOH$ = 100 mL with concentration = 0.120 M,

  • $K_a = 6.3 \times 10^{-5}$

Calculation Steps:

1. Calculate concentrations:

   • Concentration of Benzoic Acid: $0.120 \, M$

   • Base concentration after adding sodium benzoate: $\text{C}_6\text{H}_5\text{COO}^- = \frac{(2.88 \, g)}{(144 \, g/mol \times 0.1 \, L)} = 0.200 \, M$

2. Calculate pK_a:

   • $pK_a = -\log(6.3 \times 10^{-5}) = 4.20$

3. Using Henderson-Hasselbalch Equation:

   • $pH = 4.20 + \log\frac{0.200}{0.120} = 4.42$

Buffer Response to Strong Acids and Bases

Strong Acid Addition

• Strong acids will react with the conjugate base of the buffer, lowering the pH.

Strong Base Addition

• Strong bases will react with the weak acid of the buffer, increasing the pH.

Buffer Capacity

• Defined as the amount of acid or base a buffer can neutralize before significant pH change occurs.

• Buffer capacity is determined by the concentrations of the weak acid and conjugate base (higher concentration = greater capacity).

Titrations

Section 17.3: Strong Acid - Strong Base Titrations

• General Reaction:
  $\text{HCl(aq) + NaOH(aq)} \rightarrow \text{H}_2\text{O(l) + NaCl(aq)}$

Titration Curve Analysis

• The equivalence point occurs at pH 7 when moles of acid equal moles of base.

• Prior to reaching equivalence, pH is less than 7; after equivalence, pH exceeds 7.

Example of Titration

• Consider 50 mL of 0.100 M HCl being titrated with 0.100 M NaOH.

1. Before Titration (0 mL NaOH):

   • $pH = -\log{[H_3O^+]} = -\log(0.100) = 1.00$

2. At 49 mL NaOH:

   • Moles of HCl = 0.100 M × 0.050 L = 0.0050 mol

   • Moles of NaOH = 0.100 M × 0.049 L = 0.0049 mol

   • After reaction:

   • $[H_3O^+] = \frac{0.0001 \, mol}{0.050 + 0.049 \, L} = 1.0 \times 10^{-3} \, M$

   • $pH = -\log(1.0 \times 10^{-3}) = 3.00$

3. At 50 mL NaOH (Equivalence Point):

   • Acid is fully neutralized.

   • Resulting solution is NaCl, so $pH = 7.0$.

4. At 51 mL NaOH:

   • Excess NaOH will determine concentration and pH will rise above 7:

   • Calculate remaining OH⁻ from excess NaOH accordingly.

Weak Acid - Strong Base Titrations

Weak Acid - Strong Base Analysis

• Involve equilibria between the acid and base during titration process.

• The pH at equivalence point for a weak acid is greater than 7 due to the production of a conjugate base.

Example: Titrating Acetic Acid with NaOH

• A 100.0 mL sample of 0.250 M acetic acid will require calculating pH:

  1. Initial pH Calculation.

  2. After adding NaOH, consider equivalence point (125.0 mL total NaOH).

  3. Post equivalence point analysis.

Conclusion

• Understanding the properties and calculations surrounding acids, bases, buffers, and their interactions during titration processes is crucial for mastery in this subject.