Buffers, Hydrolysis and Titration

Solutions That Push Back

A buffer resists pH change on addition of acid or base. It contains a weak acid together with its conjugate base in comparable amounts, so that added $\mathrm{H^+}$ is mopped up by the base and added $\mathrm{OH^-}$ by the acid. The Henderson-Hasselbalch equation gives the pH:

$$\mathrm{pH} = \mathrm{p}K_a + \log\frac{[\mathrm{A^-}]}{[\mathrm{HA}]}$$

Two consequences follow immediately. When the two are equal the log term vanishes and $\mathrm{pH} = \mathrm{p}K_a$, so a buffer is chosen by picking an acid whose $\mathrm{p}K_a$ is near the target pH. And because only the ratio appears, diluting a buffer barely changes its pH, though it does reduce its capacity, the amount of acid or base it can absorb before failing.

Salt hydrolysis explains why dissolving a salt often gives a non-neutral solution. The ions inherit the strength of their parents:

Sodium acetate is basic because acetate is the conjugate of a weak acid and takes protons from water; ammonium chloride is acidic for the mirror reason.

A titration adds a known reagent until the reaction is complete. The equivalence point is where stoichiometrically equal amounts have been mixed; the endpoint is where the indicator changes. These are ideally, but not automatically, the same, and choosing an indicator whose range brackets the equivalence pH is the analyst's job.

The equivalence pH is not always 7. Strong acid with strong base gives 7; weak acid with strong base gives above 7, because the conjugate base remains in solution. Halfway to equivalence, exactly half the weak acid is neutralised, the buffer ratio is 1, and $\mathrm{pH} = \mathrm{p}K_a$, the standard way of measuring $K_a$ from a curve.

Common pitfall: assuming every equivalence point is at pH 7. Only the strong-strong case is neutral. Titrating acetic acid with sodium hydroxide reaches equivalence near pH 8.7, so phenolphthalein is appropriate and methyl orange would change far too early.