Buffers

Homogeneous vs. Heterogeneous Equilibrium

Application of Le Chatelier's principle.
Qualitative understanding.
Example: Acetic acid solution with added sodium acetate.
- The common ion is acetate.
- Equilibrium system: CH3COOH \rightleftharpoons H^+ + CH3COO^-
- Adding sodium acetate increases acetate concentration, shifting the equilibrium to the left.
- This reduces H+ concentration, increasing pH.

Used to quantify the common ion effect.
Derivation:
- Start with the weak acid dissociation constant: K_a = \frac{[H^+][A^-]}{[HA]}
- Take the negative log of both sides: -log(K_a) = -log(\frac{[H^+][A^-]}{[HA]})
- Define pKa = -log(Ka)
- Using logarithm properties:
  - pK_a = -log[H^+] - log[A^-] + log[HA]
  - pK_a = pH - log[A^-] + log[HA]
- Rearrange to get the Henderson-Hasselbalch equation:
  - pH = pK_a + log(\frac{[A^-]}{[HA]})
Assumptions:
- Equilibrium concentrations are approximately equal to initial concentrations due to small K_a values.

Definition: A solution that resists changes in pH upon addition of small amounts of acid or base.
Buffers contain both an acid and a base to neutralize excess base or acid, respectively.
Key: Use a weak acid and its conjugate base (or a weak base and its conjugate acid) to avoid neutralization.
- Example: Acetic acid and sodium acetate.
- If H^+ is neutralized by acetate, it regenerates acetic acid.
- If acetate hydrolyzes to form hydroxide, acetic acid neutralizes it back to acetate.
Adding excess acid (e.g., HCl) to a buffer:
- Acetate neutralizes the acid.
- Small pH change occurs.
Adding excess base (e.g., NaOH) to a buffer:
- Acetic acid neutralizes the base.
- Slight pH change occurs.
Demonstration in Lab:
- Compare pH changes in buffer vs. distilled water upon adding strong acid/base.

Buffer solutions often contain equal amounts of weak acid and conjugate base (e.g., acetic acid and sodium acetate).
- Adding excess hydronium shifts equilibrium left, consuming added acid and restoring pH.
- Adding excess base removes hydronium, shifting equilibrium right to restore original concentration.
- Equilibrium shifts until Q = K.
Buffer Capacity:
- Adding enough excess acid consumes all acetate, crashing the buffer.
- Adding enough excess hydroxide consumes all acetic acid, also crashing the buffer.
- Exceeding buffer capacity means the solution can no longer resist pH changes.
- Stoichiometry matters.

Determine if pairs of solutions can function as a buffer by checking for a weak acid and its conjugate base or a weak base and its conjugate acid.
Examples:
- Nitric acid and sodium nitrate:
  - Nitric acid is a strong acid, so no buffer.
- HF and KF:
  - HF is a weak acid, and F- is its conjugate base, so it's a buffer.
- Ammonia and NH4Cl:
  - Ammonia is a weak base, and NH4+ is its conjugate acid, so it's a buffer.
- Two weak acids cannot form a buffer.
- Strong base and neutral salt cannot form a buffer.

Equation: pH = pK_a + log(\frac{[A^-]}{[HA]})
Where:
- [A^-] is the concentration of the conjugate base.
- [HA] is the concentration of the weak acid.
Assumptions:
- Initial concentrations are approximately equal to equilibrium concentrations.
Tenfold change in the ratio of conjugate base to weak acid leads to a pH change of one unit.
To prepare a buffer of a specific pH, choose an acid whose pK_a is close to the desired pH.

Distilled Water Control:
- Adding HCl or NaOH results in significant pH changes.
- Using M1V1 = M2V2 to calculate concentration of acid/base.
- pH = -log[H^+] or pH = 14 + log[OH^-]
Buffer Calculations:
- Using Henderson-Hasselbalch equation:
  - pH = pK_a + log(\frac{[A^-]}{[HA]})
- Initial Concentrations: Calculate using stoichiometry and dilution.
- Example with acetic acid and sodium acetate:
  - Moles of sodium acetate = grams / molecular weight.
  - Moles of acetic acid = volume * molarity.
- pKa = -log(Ka) = 4.7
- If initial concentrations are equal, pH = pKa.
Adding HCl to Buffer:
- HCl neutralizes acetate.
- Changes in moles of acetate and acetic acid.
- Use Henderson-Hasselbalch to find new pH.
Adding NaOH to Buffer:
- NaOH neutralizes acetic acid.
- Changes in moles of acetate and acetic acid.
- Use Henderson-Hasselbalch to find new pH.

Problem: Calculate pH of a buffer with KF and HF.
Use Henderson-Hasselbalch equation: pH = pK_a + log(\frac{[F^-]}{[HF]})
[F-] = [KF] because KF dissociates completely.
Use grams of KF and its molecular weight to find moles of KF, and then concentration of F-.
Use M1V1 = M2V2 to find [HF] after dilution.
Plug in values to get pH.