In many chemical and biological systems, maintaining a constant pH is critical. For instance, human blood must be kept within a very narrow pH range (around 7.35 to 7.45) for metabolic enzymes to function. To control pH, chemists use buffer solutions.
A solution that resists changes in pH when small amounts of acid (H+ ions) or base (OH- ions) are added to it.
A buffer solution that maintains a pH below 7. It is commonly prepared by mixing a weak acid (e.g. ethanoic acid) with a solution of its salt containing the conjugate base (e.g. sodium ethanoate).
A buffer solution that maintains a pH above 7. It is commonly prepared by mixing a weak base (e.g. ammonia) with a solution of its salt containing the conjugate acid (e.g. ammonium chloride).
Key Principle
A buffer solution works by providing large reservoirs of both a weak acid (to react with added base) and a conjugate base (to react with added acid). Because these reservoirs are large, the addition of small amounts of H+ or OH- causes only a negligible shift in the ratio of acid to conjugate base, leaving the overall pH virtually unchanged.
Mechanism of Action: How Acidic Buffers Work
Consider an acidic buffer containing a weak acid (HA) and its conjugate base (A⁻) from a fully soluble salt (MA). Two equilibria represent the system, though the salt dissociation is complete:
HA(aq) ⇌ H⁺(aq) + A⁻(aq) (Dissociation is very small)
MA(aq) → M⁺(aq) + A⁻(aq) (Dissociation is complete)
This results in high concentrations of undissociated weak acid (HA) and conjugate base anions (A⁻) co-existing in solution.
- When H+ ions are added: The added protons react with the large reservoir of conjugate base ions (A⁻) to form undissociated weak acid molecules:
A⁻(aq) + H⁺(aq) → HA(aq)
The equilibrium shifts to the left. The concentration of H+ is kept virtually the same. - When OH- ions are added: The added hydroxide ions react with the weak acid molecules to form water and conjugate base ions:
HA(aq) + OH⁻(aq) → A⁻(aq) + H₂O(l)
Alternatively, they react with the small amount of H+ present to form water. This causes the weak acid equilibrium to shift to the right to replace the lost H+ ions. Either description explains how the OH- is consumed without raising pH.
Calculating the pH of an Acidic Buffer
To calculate the pH of an acidic buffer, we start with the acid dissociation constant expression:
\[ \text{K}_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \]
Rearranging the equation to solve for [H⁺]:
\[ [\text{H}^+] = \text{K}_a \times \frac{[\text{HA}]}{[\text{A}^-]} \]
Unlike the weak acid calculations in the previous lesson, we can no longer assume [H⁺] = [A⁻] because we have added a large amount of the salt (A⁻). Instead, we make two alternative approximations:
1. [HA]equilibrium≈[HA]initial (assuming very little weak acid dissociates).
2. [A⁻]equilibrium≈[salt]initial (assuming all conjugate base comes from the fully dissociated salt, and the small amount from acid dissociation is negligible).
Using these assumptions, we get: \( [\text{H}^+] = \text{K}_a \times \frac{[\text{acid}]}{[\text{salt}]} \). In calculations, you can use either concentrations or the actual moles of acid and salt, as the volume terms cancel out!
Step 1: Calculate the moles of acid and salt in the mixture:
Since the volumes are equal, the overall volume is doubled, meaning each concentration is halved. However, using moles is safer and works for any mixing ratio.
\[ \text{moles of CH}_3\text{COOH} = \text{concentration} \times \text{volume in dm}^3 = 0.200 \times 0.250 = 0.0500\text{ mol} \]
\[ \text{moles of CH}_3\text{COO}^- = \text{concentration} \times \text{volume in dm}^3 = 0.100 \times 0.250 = 0.0250\text{ mol} \]
Step 2: Substitute moles directly into the rearranged Ka expression:
\[ [\text{H}^+] = \text{K}_a \times \frac{\text{moles of acid}}{\text{moles of salt}} \]
\[ [\text{H}^+] = (1.74 \times 10^{-5}) \times \frac{0.0500}{0.0250} = (1.74 \times 10^{-5}) \times 2 = 3.48 \times 10^{-5}\text{ mol dm}^{-3} \]
Step 3: Calculate the pH:
\[ \text{pH} = -\log_{10}(3.48 \times 10^{-5}) = 4.46 \]
The pH of the buffer solution is 4.46 (to 2 decimal places).
Step 1: Calculate initial moles of reactants:
\[ \text{initial moles of HCOOH} = 0.400 \times 0.100 = 0.0400\text{ mol} \]
\[ \text{initial moles of NaOH} = 0.250 \times 0.050 = 0.0125\text{ mol} \]
Step 2: Account for the neutralisation reaction:
The strong base NaOH reacts completely with the weak acid: HCOOH + NaOH → HCOONa + H₂O.
- Methanoic acid is in excess. Remaining moles of HCOOH = 0.0400 − 0.0125 = 0.0275 mol.
- All added NaOH reacts, forming an equivalent amount of salt: moles of HCOO- formed = 0.0125 mol.
Step 3: Calculate the hydrogen ion concentration:
\[ [\text{H}^+] = \text{K}_a \times \frac{\text{moles of HCOOH}}{\text{moles of HCOO}^-} \]
\[ [\text{H}^+] = (1.78 \times 10^{-4}) \times \frac{0.0275}{0.0125} = (1.78 \times 10^{-4}) \times 2.20 = 3.916 \times 10^{-4}\text{ mol dm}^{-3} \]
Step 4: Calculate pH:
\[ \text{pH} = -\log_{10}(3.916 \times 10^{-4}) = 3.41 \]
The pH of the resulting buffer solution is 3.41.
A classic exam question asks you to calculate the pH of a buffer solution after a small volume of strong acid or base is added. In this case:
- If a strong acid is added, add the moles of H+ to the moles of weak acid, and subtract the same moles of H+ from the conjugate base.
- If a strong base is added, subtract the moles of OH- from the moles of weak acid, and add the same moles of OH- to the conjugate base.
Then, recalculate the pH using the updated mole values. See the worked example below to master this process.
Step 1: Calculate the moles of H+ added:
\[ \text{moles of H}^+ \text{ added} = 1.00 \times 0.00500 = 0.00500\text{ mol} \]
Step 2: Adjust buffer mole values:
The added H+ reacts with CH3COO- to form CH3COOH:
- New moles of CH3COO- = 0.0250 − 0.00500 = 0.0200 mol (conjugate base decreased).
- New moles of CH3COOH = 0.0500 + 0.00500 = 0.0550 mol (weak acid increased).
Step 3: Recalculate H+ concentration and pH:
\[ [\text{H}^+] = (1.74 \times 10^{-5}) \times \frac{0.0550}{0.0200} = (1.74 \times 10^{-5}) \times 2.75 = 4.785 \times 10^{-5}\text{ mol dm}^{-3} \]
\[ \text{pH} = -\log_{10}(4.785 \times 10^{-5}) = 4.32 \]
The pH decreased only slightly from 4.46 to 4.32, demonstrating the buffer's effectiveness.
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