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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.

Understanding Buffer Solutions & Henderson-Hasselbalch

The protection comes from two fast chemical reactions. If a small amount of strong acid enters a solution containing acetate ions, the acetate ions take up the extra hydrogen ions and form acetic acid. The added acid is changed into a weak acid, which only partly ionizes in water.

If a small amount of strong base enters, acetic acid gives hydrogen ions to the hydroxide ions, making water. Its conjugate base remains behind. In each case, one member of the pair removes most of the added substance before the pH can shift very far.

The Henderson-Hasselbalch equation is most useful after the chemistry has been handled correctly. When strong acid or base is added, do not put the original buffer amounts straight into the equation. First use mole ratios from the reaction.

Added hydrogen ions reduce the amount of conjugate base by the same number of moles and increase the amount of weak acid. Added hydroxide ions reduce the weak acid amount and increase the conjugate base amount.

Then use the new amounts in the concentration ratio. If all substances are in the same final volume, the volume cancels, so a ratio of moles gives the same result.

A buffer can have the right pH yet still be easily overwhelmed. This is the difference between buffer pH and buffer capacity. The ratio of the two buffer components sets the pH.

The total amount present controls how much acid or base the solution can absorb. For example, two solutions may have the same acid to base ratio.

A solution made with ten times more of each component has the same starting pH, but it can neutralize about ten times more added acid or base before its pH changes greatly. Dilution usually leaves the ratio nearly unchanged, though it lowers capacity because fewer particles are available to react.

Students often meet buffers during titrations. Before the equivalence point of a weak acid titrated with a strong base, both the original weak acid and the conjugate base formed by the reaction are present. That region behaves as a buffer.

At the half equivalence point, exactly half of the original acid has been converted. The amounts of acid and conjugate base are equal at that moment, making it a useful way to determine the acid's pKa from a titration curve. Near the equivalence point, the buffer is used up and the pH can change sharply.

Buffers matter in living things because many enzymes work only over a narrow pH range. Blood contains a carbonic acid and bicarbonate system that helps limit pH changes. Cells, medicines, foods, and laboratory solutions use similar ideas.

When solving problems, identify which species reacts first with any added strong acid or base. Keep track of moles, not just labels on a bottle.

Check whether enough of both buffer components remain after the reaction. Finally, make sure the chosen weak acid has a pKa close to the target pH, since a very uneven ratio gives poor resistance to further change.