Aqueous Ionic Equilibria: Buffer Solutions and Their Properties

Chapter 18: Aqueous Ionic Equilibria

Buffer Solutions: Definition, Components, and Properties

Buffer solutions are essential in maintaining a relatively constant pH in chemical and biological systems. They are composed of a weak acid and its conjugate base, or a weak base and its conjugate acid, often provided by soluble salts.

• Acidic Buffer: Mixture of a weak acid (HA) and its conjugate base (A–), e.g., acetic acid and sodium acetate (HC2H3O2/NaC2H3O2).

• Basic Buffer: Mixture of a weak base (B) and its conjugate acid (HB+), e.g., ammonia and ammonium chloride (NH3/NH4Cl).

• Soluble Salts: Used to provide the conjugate acid or base component.

Examples of Buffer Systems:

• HC2H3O2/C2H3O2– (acetic acid/acetate)

• NH3/NH4+ (ammonia/ammonium)

• H2CO3/HCO3– (carbonic acid/bicarbonate)

• HCO3–/CO32– (bicarbonate/carbonate)

Naturally Occurring Buffers:

• Blood: Maintains pH ~7.45 using buffers such as H2CO3/HCO3–.

• Natural Waters: Buffer capacity against acid rain is determined by H2CO3/HCO3– concentration.

Understanding Buffer Action

The fundamental property of a buffer is its ability to resist changes in pH upon addition of small amounts of strong acid or base, or upon dilution.

• Acid Addition: H3O+ from the acid reacts with A– to form HA, increasing [HA] and decreasing [A–].

• Base Addition: OH– from the base reacts with HA to form A–, decreasing [HA] and increasing [A–].

• Result: The ratio [A–]/[HA] changes only slightly, so pH remains nearly constant.

Key Equilibrium:

For a weak acid buffer:

$\mathrm{HA(aq) + H_2O(\ell) \rightleftharpoons H_3O^+(aq) + A^-(aq)}$

Acid dissociation constant:

$K_a = \frac{[H_3O^+][A^-]}{[HA]}$

Thus, $[H_3O^+]$ (and pH) depends on the ratio $[A^-]/[HA]$.

Henderson-Hasselbalch Equation and pH Calculation

The Henderson-Hasselbalch equation provides a convenient way to calculate the pH of buffer solutions using the concentrations of the acid and base components.

Derivation:

Starting from the acid dissociation equilibrium:

$K_a = \frac{[H_3O^+][A^-]}{[HA]}$

Solving for $[H_3O^+]$ and taking the negative logarithm:

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

Generalized Form:

$pH = pK_a + \log \left( \frac{\text{base}}{\text{acid}} \right)$

• For weak acid/conjugate base: base = A–, acid = HA

• For weak base/conjugate acid: base = B, acid = HB+

Key Features:

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

• Each tenfold change in $[A^-]/[HA]$ changes pH by 1 unit.

• Dilution does not affect pH, as the ratio remains constant.

Diprotic Acid Buffer Systems:

• For diprotic acids (H2A), two buffer systems exist:

First dissociation:

$H_2A(aq) + H_2O(\ell) \rightleftharpoons H_3O^+(aq) + HA^-(aq)$

$pH = pK_{a1} + \log \left( \frac{[HA^-]}{[H_2A]} \right)$

Second dissociation:

$HA^-(aq) + H_2O(\ell) \rightleftharpoons H_3O^+(aq) + A^{2-}(aq)$

$pH = pK_{a2} + \log \left( \frac{[A^{2-}]}{[HA^-]} \right)$

Example: Oxalic Acid Buffers

• H2C2O4/HC2O4– (Ka1 = $6.5 \times 10^{-2}$)

• HC2O4–/C2O42– (Ka2 = $6.1 \times 10^{-5}$)

Sample Calculations

• Acidic Buffer Example: 0.30 M acetic acid and 0.20 M sodium acetate, $K_a = 1.8 \times 10^{-5}$

$pH = -\log(1.8 \times 10^{-5}) + \log\left(\frac{0.20}{0.30}\right) = 4.74 + (-0.176) = 4.57$

• Basic Buffer Example: 0.20 M NH3 and 0.50 M NH4Cl, $K_b = 1.8 \times 10^{-5}$

First, find $K_a$ for NH4+:

$K_a = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.6 \times 10^{-10}$

$pH = -\log(5.6 \times 10^{-10}) + \log\left(\frac{0.20}{0.50}\right) = 9.25 + (-0.398) = 8.85$

Limitations of the Henderson-Hasselbalch Equation

• Not valid if $K_a > 10^{-3}$ or if concentrations are less than $10^{-3}$ M.

• In such cases, a quadratic equation must be used for accurate pH calculation.

Preparation of Buffer Solutions

Buffer solutions are prepared in the laboratory by mixing calculated amounts of acid and conjugate base (or base and conjugate acid) to achieve a desired pH and total concentration.

1. Identify the acid dissociation equation relevant to the buffer system (monoprotic, diprotic, or triprotic).

2. Select the acid/conjugate base pair and obtain $K_a$ and molar masses from reference data.

3. Calculate the required masses of each component using the Henderson-Hasselbalch equation and the total buffer concentration.

4. Weigh or measure the components accurately. For liquids, use density to convert mass to volume.

5. Adjust the pH as needed by adding small amounts of strong acid or base, monitoring with a pH meter.

Example: Preparation of a Sodium Bioxalate/Sodium Oxalate Buffer (pH 4.77)

Species	Ka	Molar Mass (g/mol)	Compound
HC2O4–	5.1 × 10–5	130.03	NaHC2O4·H2O
C2O42–	—	134.04	Na2C2O4

Given: [HC2O4–] + [C2O42–] = 0.20 M

Let x = [HC2O4–], then [C2O42–] = 0.20 – x

Using Henderson-Hasselbalch:

$pH = pK_{a2} + \log \left( \frac{[C_2O_4^{2-}]}{[HC_2O_4^-]} \right)$

Given pH = 4.77, $pK_{a2} = 4.29$:

$4.77 = 4.29 + \log \left( \frac{0.20 - x}{x} \right)$

$0.48 = \log \left( \frac{0.20 - x}{x} \right)$

$10^{0.48} = \frac{0.20 - x}{x} \implies x = 0.050\,\mathrm{M},\ [C_2O_4^{2-}] = 0.15\,\mathrm{M}$

For 100.0 mL solution:

• Mass NaHC2O4 = $0.050 \times 0.100 \times 130.03 = 0.650$ g

• Mass Na2C2O4 = $0.15 \times 0.100 \times 134.04 = 2.01$ g

Practical Considerations

• Measured pH may differ from calculated due to uncertainties in $K_a$ values and calculation simplifications.

• Final pH adjustment is done by adding strong acid or base dropwise while monitoring with a pH meter.

Summary Table: Buffer Preparation Steps

Step	Description
1	Identify relevant acid/base pair and dissociation equation
2	Obtain $K_a$ and molar masses
3	Calculate required concentrations and masses
4	Weigh/measure components accurately
5	Mix and adjust pH as needed

Additional info: The notes emphasize the importance of buffer solutions in biological and environmental systems, and provide practical laboratory guidance for buffer preparation and adjustment.