Conductometric Titrations and Solubility Products

1Titrating without an indicator

How do you find the end point of a titration when the solution is dark brown, cloudy with a precipitate, or so dilute that no indicator changes sharply? You follow the conductivity instead. As titrant is added, one kind of ion is replaced by another that moves faster or slower, and the conductivity changes in straight lines whose slopes change at the end point. The same measurements, made on a saturated solution, can even give the solubility product of a salt as insoluble as silver chloride.

2Shapes of conductometric titration curves

Figure 12.4: Conductometric titration curves calculated from limiting ionic conductivities, for 50.0 cm³ of 0.0100 mol dm⁻³ analyte titrated with 0.100 mol dm⁻³ titrant (a concentrated titrant keeps the dilution small, so the lines stay straight).

• Strong acid with strong base. ${\mathrm{H}}^{+}+\mathrm{O}{\mathrm{H}}^{-}\to {\mathrm{H}}_2\mathrm{O}\backslash mathrm\{H^\{+\}\}\; +\; \backslash mathrm\{OH^\{-\}\}\; \backslash rightarrow\; \backslash mathrm\{H\_2O\}$: each fast ${\mathrm{H}}^{+}\backslash mathrm\{H^\{+\}\}$ (349.7) is replaced by a slow $\mathrm{N}{\mathrm{a}}^{+}\backslash mathrm\{Na^\{+\}\}$ (50.1), so κ falls steeply. After the end point, extra $\mathrm{N}{\mathrm{a}}^{+}\backslash mathrm\{Na^\{+\}\}$ and fast $\mathrm{O}{\mathrm{H}}^{-}\backslash mathrm\{OH^\{-\}\}$ (198.0) make κ rise steeply. A sharp V.
• Weak acid with strong base. Ethanoic acid is only slightly ionised, so κ starts low. It may dip a little at first, as the ethanoate formed suppresses the acid's own ionisation; then it rises gently as $\mathrm{N}{\mathrm{a}}^{+}\backslash mathrm\{Na^\{+\}\}$ and $\mathrm{C}{\mathrm{H}}_3\mathrm{C}\mathrm{O}{\mathrm{O}}^{-}\backslash mathrm\{CH\_3COO^\{-\}\}$ build up. After the end point it rises more steeply ($\mathrm{O}{\mathrm{H}}^{-}\backslash mathrm\{OH^\{-\}\}$).
• Precipitation. $\mathrm{A}{\mathrm{g}}^{+}+\mathrm{C}{\mathrm{l}}^{-}\to \mathrm{A}\mathrm{g}\mathrm{C}\mathrm{l}(\mathrm{s})\backslash mathrm\{Ag^\{+\}\; +\; Cl^\{-\}\; \backslash rightarrow\; AgCl(s)\}$: $\mathrm{A}{\mathrm{g}}^{+}\backslash mathrm\{Ag^\{+\}\}$ (61.9) is replaced by ${\mathrm{K}}^{+}\backslash mathrm\{K^\{+\}\}$ (73.5), a slightly faster ion, so κ hardly changes (the small gain is cancelled by the small dilution). After the end point, both ${\mathrm{K}}^{+}\backslash mathrm\{K^\{+\}\}$ and $\mathrm{C}{\mathrm{l}}^{-}\backslash mathrm\{Cl^\{-\}\}$ accumulate and κ rises faster.
• Strong acid with weak base. κ falls as ${\mathrm{H}}^{+}\backslash mathrm\{H^\{+\}\}$ is replaced by $\mathrm{N}{\mathrm{H}}_4^{+}\backslash mathrm\{NH\_4^\{+\}\}$. After the end point the extra ammonia hardly ionises (its ionisation is suppressed by the $\mathrm{N}{\mathrm{H}}_4^{+}\backslash mathrm\{NH\_4^\{+\}\}$ already present), so the line is almost flat.

Advantages over indicator titrations: they work with coloured or turbid solutions, very dilute solutions, weak acids with weak bases, and precipitation reactions; and the end point comes from many readings, not one colour change.

3Solubility and $K_{sp}K\_\{sp\}$ from conductivity

A saturated solution of a sparingly soluble salt such as AgCl is so dilute that it is effectively at infinite dilution, so ${\mathrm{\Lambda }}_m\approx {\mathrm{\Lambda }}_m^{\circ }\backslash Lambda\_m\; \backslash approx\; \backslash Lambda^\{\backslash circ\}\_m$, which can be found from Kohlrausch's law. Then:

$$s=\frac{1000{\kappa }_{\text{salt}}}{{\mathrm{\Lambda }}_m^{\circ }}{\kappa }_{\text{salt}}={\kappa }_{\text{solution}}-{\kappa }_{\text{water}}s\; =\; \backslash frac\{1000\backslash ,\backslash kappa\_\{\backslash text\{salt\}\}\}\{\backslash Lambda^\{\backslash circ\}\_m\}\; \backslash qquad\; \backslash kappa\_\{\backslash text\{salt\}\}\; =\; \backslash kappa\_\{\backslash text\{solution\}\}\; -\; \backslash kappa\_\{\backslash text\{water\}\}$$

The conductivity of the water itself must be subtracted, because at such low concentrations it is no longer negligible.