The Equilibrium of a Buffer
In physical chemistry, weak acid buffers are modeled by the dissociation equilibrium of the weak acid $\text{HA}$:
Because $\text{HA}$ is a weak acid, it only partially dissociates, meaning there is a high concentration of undissociated $\text{HA}$ molecules. The salt (e.g. $\text{NaA}$) completely dissociates, providing a high concentration of the conjugate base ions $\text{A}^-$.
Le Chatelier's Buffer Mechanism
The solution resists pH changes by shifting the position of equilibrium in accordance with Le Chatelier's Principle:
On adding small amounts of acid ($\text{H}^+$)
The added $\text{H}^+$ ions react with the high concentration of conjugate base $\text{A}^-$ to form undissociated $\text{HA}$ molecules: $$\text{H}^+(aq) + \text{A}^-(aq) \rightarrow \text{HA}(aq)$$ The position of equilibrium shifts to the left. The added $\text{H}^+$ is consumed, and pH remains stable.
On adding small amounts of alkali ($\text{OH}^-$)
The added $\text{OH}^-$ ions react with the $\text{H}^+$ ions to form water: $$\text{OH}^-(aq) + \text{H}^+(aq) \rightarrow \text{H}_2\text{O}(l)$$ This decreases the $\text{H}^+$ concentration, causing the weak acid $\text{HA}$ to dissociate further to replace them: $$\text{HA}(aq) \rightarrow \text{H}^+(aq) + \text{A}^-(aq)$$ The position of equilibrium shifts to the right, and the pH remains stable.
Calculating Buffer pH
To calculate the pH of an acidic buffer, rearrange the acid dissociation constant ($K_a$) expression:
Then convert $[\text{H}^+]$ to pH using the logarithmic expression: $\text{pH} = -\log_{10}[\text{H}^+]$. Or directly use the Henderson-Hasselbalch equation:
Worked Exam Example
Worked A-Level Example
Question: A buffer solution is made by mixing $25.0\text{cm}^3$ of $0.100\text{mol/dm}^3$ ethanoic acid ($\text{CH}_3\text{COOH}$) and $25.0\text{cm}^3$ of $0.200\text{mol/dm}^3$ sodium ethanoate ($\text{CH}_3\text{COONa}$).
Given that $K_a$ of ethanoic acid is $1.74 \times 10^{-5}\text{ mol/dm}^3$, calculate the pH of the resulting buffer solution at $298\text{K}$.
Solution:
- Determine the moles of weak acid ($\text{HA}$) and conjugate base ($\text{A}^-$) in the mixture: $$\text{Moles of } \text{CH}_3\text{COOH} = \frac{25.0}{1000} \times 0.100 = 0.0025\text{ mol}$$ $$\text{Moles of } \text{CH}_3\text{COO}^- = \frac{25.0}{1000} \times 0.200 = 0.0050\text{ mol}$$
- Note that since the volumes are equal, the concentration ratio is equal to the mole ratio.
- Substitute values into the rearranged $K_a$ equation to calculate $[\text{H}^+]$: $$[\text{H}^+] = K_a \times \frac{[\text{HA}]}{[\text{A}^-]} = (1.74 \times 10^{-5}) \times \frac{0.0025}{0.0050}$$ $$[\text{H}^+] = 8.70 \times 10^{-6}\text{ mol/dm}^3$$
- Calculate pH: $$\text{pH} = -\log_{10}(8.70 \times 10^{-6}) = 5.06$$
Answer: pH of buffer = $5.06$.
Frequently Asked Questions
If $x$ moles of strong acid ($\text{H}^+$) are added, the moles of conjugate base $\text{A}^-$ decrease by $x$, and the moles of weak acid $\text{HA}$ increase by $x$: $$[\text{H}^+] = K_a \times \frac{n(\text{HA}) + x}{n(\text{A}^-) - x}$$ Substitute the new values into the equation to calculate the adjusted pH.
A basic buffer resists pH changes at alkaline values. It consists of a weak base (e.g. ammonia, $\text{NH}_3$) and its conjugate acid salt (e.g. ammonium chloride, $\text{NH}_4\text{Cl}$).
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