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Concept 07 / 08Equilibrium

Buffer Solutions and the Henderson-Hasselbalch Equation

Acidic and basic buffers and how they resist pH change, the Henderson-Hasselbalch equation, buffer capacity and range, calculating pH after adding strong acid or base, and the weak acid-strong base titration curve with its buffer region and half-equivalence point.

Figure 1
A buffer solution resists changes in pH when small amounts of acid or base are added, or on dilution. Blood (pH 7.47.4, mainly H2CO3/HCO3−\mathrm{H_{2}CO_{3}/HCO_{3}^{-}}) is a natural buffer. Types. Acidic buffer: a weak acid and its salt with a strong base, for example CH3COOH+CH3COONa\mathrm{CH_{3}COOH + CH_{3}COONa}. Basic buffer: a weak base and its salt with a strong acid, for example NH4OH+NH4Cl\mathrm{NH_{4}OH + NH_{4}Cl}. How it works. In an acetate buffer, added H+\mathrm{H^{+}} is consumed by the large reserve of acetate ions (CH3COO−+H+→CH3COOH\mathrm{CH_{3}COO^{-} + H^{+} \rightarrow CH_{3}COOH}), and added OH−\mathrm{OH^{-}} is consumed by acetic acid (CH3COOH+OH−→CH3COO−+H2O\mathrm{CH_{3}COOH + OH^{-} \rightarrow CH_{3}COO^{-} + H_{2}O}). Only the ratio of the two components changes slightly, so pH changes little. Henderson-Hasselbalch equation. From Ka=[H+][A−][HA]K_{a} = \frac{[\mathrm{H^{+}}][\mathrm{A^{-}}]}{[\mathrm{HA}]}, with [A−]≈[\mathrm{A^{-}}] \approx salt and [HA]≈[\mathrm{HA}] \approx acid (ionisation suppressed by the common ion): pH=pKa+log⁡[salt][acid]\text{pH} = \text{p}K_{a} + \log\frac{[\text{salt}]}{[\text{acid}]}. For a basic buffer: pOH=pKb+log⁡[salt][base]\text{pOH} = \text{p}K_{b} + \log\frac{[\text{salt}]}{[\text{base}]}. Because only the ratio matters, moles can be used directly and dilution hardly changes the pH. Buffer capacity and range. A buffer is most effective when [salt]=[acid][\text{salt}] = [\text{acid}], where pH=pKa\text{pH} = \text{p}K_{a} (maximum buffer capacity). It is useful over roughly pKa±1\text{p}K_{a} \pm 1, where the ratio lies between 110\frac{1}{10} and 1010. To make a buffer of a desired pH, choose an acid with pKa\text{p}K_{a} close to it. Higher concentrations of both components give greater capacity. Titration curve of a weak acid (see figure). Titrating CH3COOH\mathrm{CH_{3}COOH} with NaOH: the initial pH is that of the weak acid (2.872.87 for 0.1 M0.1\,\mathrm{M}). As base is added, an acetate buffer forms and the curve is flat. At the half-equivalence point, [salt]=[acid][\text{salt}] = [\text{acid}] and pH=pKa\text{pH} = \text{p}K_{a}: a direct experimental way to measure pKa\text{p}K_{a}. At equivalence, the solution contains only sodium acetate, which is basic by hydrolysis (pH about 8.78.7), so phenolphthalein (range 8.28.2 to 1010) is a suitable indicator, not methyl orange. The strong acid curve (dashed) starts lower and rises sharply at pH 77. Key Takeaways: - pH=pKa+log⁡[salt][acid]\text{pH} = \text{p}K_{a} + \log\frac{[\text{salt}]}{[\text{acid}]}; pOH=pKb+log⁡[salt][base]\text{pOH} = \text{p}K_{b} + \log\frac{[\text{salt}]}{[\text{base}]}. - Maximum capacity at pH=pKa\text{pH} = \text{p}K_{a}; useful range pKa±1\text{p}K_{a} \pm 1. - Half-equivalence point of a weak acid titration: pH=pKa\text{pH} = \text{p}K_{a}. - Weak acid plus strong base: equivalence pH above 77; use phenolphthalein. - Pitfall: a weak acid partly neutralised by a strong base is a buffer; compute the moles after reaction before applying Henderson's equation. Brain-Teaser 1: A buffer is 0.1 M0.1\,\mathrm{M} in CH3COOH\mathrm{CH_{3}COOH} and 0.1 M0.1\,\mathrm{M} in CH3COONa\mathrm{CH_{3}COONa} (pKa=4.74\text{p}K_{a} = 4.74). Find the pH before and after adding 0.01 mol0.01\,\mathrm{mol} HCl to 1 L1\,\mathrm{L}. Solution: Before: pH =4.74= 4.74. HCl converts acetate to acid: acid 0.110.11, salt 0.090.09. pH =4.74+log⁡0.090.11=4.74−0.09=4.65= 4.74 + \log\frac{0.09}{0.11} = 4.74 - 0.09 = 4.65. The same acid added to 1 L1\,\mathrm{L} of pure water would drop the pH from 77 to 22. Brain-Teaser 2: 50 mL50\,\mathrm{mL} of 0.2 M0.2\,\mathrm{M} NH3\mathrm{NH_{3}} is mixed with 50 mL50\,\mathrm{mL} of 0.1 M0.1\,\mathrm{M} HCl. Find the pH (pKb=4.74\text{p}K_{b} = 4.74). Solution: NH3\mathrm{NH_{3}}: 10 mmol10\,\mathrm{mmol}; HCl: 5 mmol5\,\mathrm{mmol}, converting 5 mmol5\,\mathrm{mmol} NH3\mathrm{NH_{3}} to NH4+\mathrm{NH_{4}^{+}}. Left: 5 mmol5\,\mathrm{mmol} base and 5 mmol5\,\mathrm{mmol} salt, a basic buffer. pOH =4.74+log⁡1=4.74= 4.74 + \log 1 = 4.74, pH =9.26= 9.26. Brain-Teaser 3: In what ratio should sodium formate and formic acid (pKa=3.74\text{p}K_{a} = 3.74) be mixed to make a buffer of pH 4.044.04? Solution: 4.04=3.74+log⁡[salt][acid]4.04 = 3.74 + \log\frac{[\text{salt}]}{[\text{acid}]}, so log⁡[salt][acid]=0.30\log\frac{[\text{salt}]}{[\text{acid}]} = 0.30 and the ratio is 2:12 : 1 (salt : acid).

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Buffer Solutions and the Henderson-Hasselbalch Equation — DPP

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