Weak-acid equilibria, pH measurement and titration curves

What you’ll learn

• Calculate an acid dissociation constant from pH or half-equivalence data.
• Use titration-curve evidence to choose measurements and an indicator.

Understand the idea

Core practical 11 determines the acid dissociation constant of a weak acid. For a solution containing only a monoprotic weak acid HA and water, let the hydrogen-ion concentration be $x$. If water's contribution is negligible, the equilibrium concentrations are $[\mathrm{H^+}]=[\mathrm{A^-}]=x$ and $[\mathrm{HA}]=c-x$. The exact concentration expression is therefore $K_a=x^2/(c-x)$. The shortcut $x^2/c$ is justified only when the fraction dissociated is small.

A second route uses the weak-acid–strong-base titration curve. At half the equivalence volume, neutralisation has converted half the acid to its conjugate base. The acid and conjugate-base concentrations are approximately equal, so $\mathrm{pH}=\mathrm{p}K_a$. At equivalence the solution contains the conjugate base and is usually alkaline; do not assume that every equivalence point has pH 7. Choose an indicator whose transition interval lies within the steep pH rise.

A pH meter should be calibrated with suitable buffer standards at the measurement temperature. Rinsing between samples and gently removing excess rinse water reduce carry-over and dilution. In a titration-curve plan, closer volume intervals around the steep region locate equivalence more clearly, while repeated stable readings help identify drift. Temperature matters because equilibrium constants and electrode response both depend on it.

Key formulae

$$K_a=\frac{[\mathrm{H^+}][\mathrm{A^-}]}{[\mathrm{HA}]}=\frac{x^2}{c-x}$$

$$[\mathrm{H^+}]=10^{-\mathrm{pH}}$$

$$\mathrm{pH}=\mathrm{p}K_a+\log_{10}\frac{[\mathrm{A^-}]}{[\mathrm{HA}]}$$

$$K_a=10^{-\mathrm{p}K_a}$$

Worked example

A $0.100\ \mathrm{mol\,dm^{-3}}$ solution of a monoprotic weak acid has pH $2.90$. Neglect the contribution from water. Calculate $K_a$ using the equilibrium acid concentration rather than assuming it equals the initial concentration.

1. The hydrogen-ion concentration is $x=10^{-2.90}=1.2589\times10^{-3}\ \mathrm{mol\,dm^{-3}}$.
2. The conjugate-base concentration is $x$, while $[\mathrm{HA}]=0.100-x=0.098741\ \mathrm{mol\,dm^{-3}}$.
3. Hence $K_a=x^2/(0.100-x)=1.605\times10^{-5}\ \mathrm{mol\,dm^{-3}}$.

Answer: $K_a\approx1.61\times10^{-5}\ \mathrm{mol\,dm^{-3}}$.