8.9.A.1  Buffer pH and the Henderson-Hasselbalch Equation:

1. Weak Acid Equilibrium and pKa:

i. Weak Acid Dissociation:

A weak acid partially dissociates only in water, and comes into equilibrium:

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

* HA = weak acid
* H⁺ = hydrogen ion (or H₃O⁺ in aqueous solution)
* A⁻ = conjugate base

ii. Acid Dissociation Constant (Ka):

This balance is quantified by the acid dissociation constant, Ka:

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

* A larger Ka indicates a stronger acid (more dissociation).
* A smaller Ka indicates a weaker acid (less dissociation).

iii. What is pKa?

Since Ka values are frequently small, we employ the logarithmic form:

$$\text{p}{K}_a=-\log (K_a)$$

* Low pKa → Stronger acid
* High pKa → Weaker acid

Key Point:

* pKa informs you about how easily a weak acid will donate a proton.
* It also tells you what pH the acid/base pair is best at buffering to.

iv. Relationship to pH and Buffering:

At pH = pKa, the concentrations of the acid and its conjugate base are equal:

$$[\text{HA}]=[{\text{A}}^{-}]$$

This is the ideal buffering point, and it’s central to the Henderson-Hasselbalch equation:

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

v. Example: Acetic Acid:

• Formula: CH₃COOH ⇌ CH₃COO⁻ + H⁺

• $K_a \approx 1.8 \times 10^{-5}$

• $\text{p}K_a \approx 4.74$

Thus, acetic acid buffers most effectively at pH 4.74.

vi. Summary Table

2. Henderson-Hasselbalch Equation:

The Henderson-Hasselbalch equation is a simplified way to estimate the pH of a buffer solution, based on the ratio of a weak acid and its conjugate base.

i. The Equation:

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

Where:

* pH = the acidity of the solution
* pKa = the acid dissociation constant (log scale)
* [A⁻] = concentration of the conjugate base
* [HA] = concentration of the weak acid

ii. What It Tells You:

* The formula connects pH with pKa and the ratio of base to acid.
* It is used to calculate the pH of a buffer and how it will react to added acid or base.

iii. Key Insights:

iv. Example Calculation:

Buffer: Acetic acid (pKa = 4.74), with

• $[CH₃COOH] = 0.10\ M$[CH3​COOH]=0.10 M

• $[CH₃COO⁻] = 0.20\ M$[CH3​COO−]=0.20 M

$$\text{pH}=4.74+\log \left(\frac{0.20}{0.10}\right)=4.74+\log (2)\approx 4.74+0.30=5.04$$

So, the buffer’s pH is 5.04.

v. When to Use It:

Use the Henderson-Hasselbalch equation when:

* You’re working with a buffer solution.
* You know the concentrations of both the weak acid and its conjugate base.
* You want to estimate how pH changes with additions of acid/base.

vi. Limitations:

* Doesn’t work well for very dilute solutions.
* Assumes that activity ≈ concentration (fair for moderate concentrations).
* Not for strong acids or bases — only for weak acid/base pairs.

3. Buffer Response to Acid/Base Addition:

A buffer resists pH changes when small quantities of acid or base are added. This stability is due to the buffer’s capacity to neutralize added H⁺ or OH⁻ ions with its two main constituents:

* A weak acid (HA) — donates H⁺ when base is added
* A conjugate base (A⁻) — accepts H⁺ when acid is added

i. When Acid (H⁺) is Added:

Conjugate base (A⁻) of the buffer neutralizes the H⁺:

$${\text{H}}^{+}+{\text{A}}^{-}\to \text{HA}$$

* The H⁺ added reacts with A⁻ to give HA.
* Concentration of free H⁺ doesn’t rise much → pH remains almost the same.

ii. When Base (OH⁻) is Added:

Weak acid (HA) releases a proton to neutralize OH⁻:

$$\text{OH}^- + \text{HA} \rightarrow \text{A}^- + \text{H}_2\text{O}$$

* The OH⁻ reacts with HA to give water and more A⁻.
* OH⁻ is taken away → no big increase in pH.

iii. Why pH Remains Almost the Same

* The equilibrium is shifted a bit to re-equilibrate:

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

* Small amounts of H⁺ or OH⁻ are sorbed up by the buffer pair.
* Since both members are available in relatively high concentrations, the impact on the concentration of H⁺ (and hence pH) is small.

iv. Only Works Within Buffer Capacity:

* Buffer will only operate to a certain point.
* After most HA or A⁻ is depleted, buffer will no longer be able to counteract added acid or base → sharp pH change.

v. Summary: