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Buffer solutions resist large changes in pH when small amounts of acid or base are added. This cheat sheet covers how buffers work, how to identify buffer pairs, and how to calculate buffer pH using the Henderson-Hasselbalch equation. Students need these tools to solve equilibrium problems in chemistry, biology, and laboratory titrations. The most important idea is that a buffer contains a weak acid and its conjugate base, or a weak base and its conjugate acid. The Henderson-Hasselbalch equation connects pH, pKapK_a, and the concentration ratio [A][HA]\frac{[A^-]}{[HA]}. When the concentrations of acid and conjugate base are equal, pH=pKa\text{pH} = pK_a. A buffer works best when pH\text{pH} is within about 11 unit of pKapK_a.

Key Facts

  • A buffer made from a weak acid and its conjugate base follows pH=pKa+log([A][HA])\text{pH} = pK_a + \log\left(\frac{[A^-]}{[HA]}\right).
  • A buffer made from a weak base and its conjugate acid can be analyzed using pOH=pKb+log([BH+][B])\text{pOH} = pK_b + \log\left(\frac{[BH^+]}{[B]}\right), then pH=14.00pOH\text{pH} = 14.00 - \text{pOH} at 25C25^\circ\text{C}.
  • The acid dissociation constant is related to acid strength by pKa=log(Ka)pK_a = -\log(K_a).
  • When [A]=[HA][A^-] = [HA], the ratio is 11, so log(1)=0\log(1) = 0 and pH=pKa\text{pH} = pK_a.
  • If [A]>[HA][A^-] > [HA], then pH>pKa\text{pH} > pK_a because the buffer has more conjugate base than acid.
  • If [A]<[HA][A^-] < [HA], then pH<pKa\text{pH} < pK_a because the buffer has more weak acid than conjugate base.
  • A useful buffer range is approximately pKa1<pH<pKa+1pK_a - 1 < \text{pH} < pK_a + 1.
  • Buffer capacity increases when the total concentration [HA]+[A][HA] + [A^-] increases, even if the ratio [A][HA]\frac{[A^-]}{[HA]} stays the same.

Vocabulary

Buffer
A solution that resists major pH changes when small amounts of acid or base are added.
Weak acid
An acid that only partially ionizes in water and has an equilibrium described by KaK_a.
Conjugate base
The particle formed when an acid loses a proton, such as AA^- from HAHA.
Henderson-Hasselbalch equation
An equation that calculates buffer pH using pH=pKa+log([A][HA])\text{pH} = pK_a + \log\left(\frac{[A^-]}{[HA]}\right).
Buffer capacity
The amount of acid or base a buffer can neutralize before its pH changes significantly.
Equivalence point
The point in a titration where stoichiometrically equal amounts of acid and base have reacted.

Common Mistakes to Avoid

  • Reversing the ratio in the Henderson-Hasselbalch equation is wrong because the acid buffer equation uses [A][HA]\frac{[A^-]}{[HA]}, not [HA][A]\frac{[HA]}{[A^-]}.
  • Using a strong acid and strong base as a buffer is wrong because buffers require a weak acid with its conjugate base, or a weak base with its conjugate acid.
  • Substituting moles and concentrations inconsistently is wrong because the ratio must compare the same kind of quantity, such as moles over moles or molarity over molarity, in the same final volume.
  • Forgetting to account for neutralization before using Henderson-Hasselbalch is wrong because added strong acid or base changes the amounts of HAHA and AA^- first.
  • Assuming every weak acid solution is a buffer is wrong because a buffer needs significant amounts of both the weak acid and its conjugate base.

Practice Questions

  1. 1 Calculate the pH of a buffer with [HA]=0.20M[HA] = 0.20\,\text{M}, [A]=0.30M[A^-] = 0.30\,\text{M}, and pKa=4.76pK_a = 4.76.
  2. 2 A buffer contains 0.50mol0.50\,\text{mol} of HAHA and 0.25mol0.25\,\text{mol} of AA^- in the same solution. If pKa=5.10pK_a = 5.10, find the pH.
  3. 3 What ratio [A][HA]\frac{[A^-]}{[HA]} is needed to make a buffer with pH=7.40\text{pH} = 7.40 using an acid with pKa=7.20pK_a = 7.20?
  4. 4 Explain why a solution containing only 0.10M0.10\,\text{M} HCl and 0.10M0.10\,\text{M} NaCl is not a buffer, even though it contains an acid and a salt.