Frequently Asked Questions
What is the Henderson-Hasselbalch equation?
pH = pK<sub>a</sub> + log<sub>10</sub>([A<sup>-</sup>]/[HA]), where [A<sup>-</sup>] is the conjugate base concentration and [HA] is the weak acid concentration. It approximates the pH of a buffer solution from the pK<sub>a</sub> of the weak acid and the ratio of its ionized to un-ionized forms. The approximation works well when both concentrations are above ~0.001 M and pK<sub>a</sub> is between about 3 and 11.
What is the effective buffering range of a buffer?
A buffer works most effectively within one pH unit of its pK<sub>a</sub> (pK<sub>a</sub> − 1 to pK<sub>a</sub> + 1). Outside this range, the ratio of acid to conjugate base becomes extreme (>10:1 or <1:10), and the buffer has little capacity to resist pH changes. For example, an acetate buffer (pK<sub>a</sub> 4.76) is effective from about pH 3.8 to 5.8.
How do I prepare a buffer at a specific pH?
Choose a weak acid whose pK<sub>a</sub> is within 1 unit of your target pH. Use the Henderson-Hasselbalch equation to find the required [A<sup>-</sup>]/[HA] ratio: ratio = 10<sup>(pH − pK<sub>a</sub>)</sup>. Mix the weak acid and its conjugate base (or its salt) in that ratio at your desired total concentration. For pH 7.4 with phosphate buffer (pK<sub>a</sub> 7.20), the ratio is 10<sup>0.20</sup> ≈ 1.58 base to acid.
How does Henderson-Hasselbalch differ from an exact ICE-table calculation?
Henderson-Hasselbalch assumes the equilibrium concentrations of the weak acid and conjugate base equal the amounts you mixed in, ignoring the small shift from dissociation and from water's own H<sup>+</sup> and OH<sup>-</sup>. An exact ICE-table (initial, change, equilibrium) treatment solves the full equilibrium and water balance, so it accounts for that shift. The two agree closely for typical buffers but diverge in very dilute solutions (below about 0.001 M), when the buffer ratio is extreme, or when the target pH is far from the pK<sub>a</sub>, because the neglected dissociation and water contribution then become a large fraction of the total. For those cases prefer an exact solver.
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