4.8 Solubility Product, Ksp
The solubility product constant, Ksp, is an equilibrium constant that describes the solubility of sparingly soluble compounds. As with other equilibrium constants, Ksp can be determined from the Gibbs free energy of reaction.
4.8.1 Aqueous solubility of AgCl(s)
Consider the silver(I) chloride salt, an insoluble salt according to the solubility rules. While this description is qualitative, we can quantify the aqueous solubility of AgCl(s). We will do this at three different temperatures: 298.15 K (25 °C), 323.15 K (50 °C), and 373.15 K (100 °C).
\[\mathrm{AgCl}(s) \rightleftharpoons \mathrm{Ag^+}(aq) + \mathrm{Cl^-}(aq)\]
Let us pull the thermodynamic values for each substance in the reaction from the Standard Thermodynamic Values for Select Substances Table (all values given in kJ mol–1).
| Species | ΔHf° | S° |
|---|---|---|
AgCl(s) |
-127 |
0.0963 |
Ag+(aq) |
105.6 |
0.07268 |
Cl–(aq) |
-167.2 |
0.0565 |
Next, we find the enthalpy (ΔH) and entropy (ΔS) of reaction
\[\begin{align*} \Delta H_{\mathrm{rxn}}^{\circ} &= (105.6~\mathrm{kJ~mol^{-1}} + -167.2~\mathrm{kJ~mol^{-1}}) - (-127~\mathrm{kJ~mol^{-1}}) \\[1.25ex] &= 65.4~\mathrm{kJ~mol^{-1}} \\[2.00ex] \Delta S_{\mathrm{rxn}}^{\circ} &= (0.07268~\mathrm{kJ~mol^{-1}~K^{-1}} + 0.0565~\mathrm{kJ~mol^{-1}~K^{-1}}) - (0.0963~\mathrm{kJ~mol^{-1}~K^{-1}}) \\[1.25ex] &= 0.03288~\mathrm{kJ~mol^{-1}~K^{-1}} \end{align*}\]
Notice that the dissolution process is endothermic (ΔH > 0) and increases in entropy (ΔS > 0). We can now calculate the Gibbs free energy of reaction at the three different temperatures.
Gibbs free energy, ΔG°, at 298.15, 323.15, and 373.15 K
ΔG° at 298.15 K
\[\begin{align*} \Delta G^{\circ} &= \Delta H^{\circ} - T\Delta{S}^{\circ} \\ &= 65.4~\mathrm{kJ~mol^{-1}} - (298.15~\mathrm{K})(0.03288~\mathrm{kJ~mol^{-1}~K^{-1}}) \\ &= 55.60~\mathrm{kJ~mol^{-1}} \end{align*}\]
ΔG° at 323.15 K
\[\begin{align*} \Delta G^{\circ} &= \Delta H^{\circ} - T\Delta{S}^{\circ} \\ &= 65.4~\mathrm{kJ~mol^{-1}} - (323.15~\mathrm{K})(0.03288~\mathrm{kJ~mol^{-1}~K^{-1}}) \\ &= 54.77~\mathrm{kJ~mol^{-1}} \end{align*}\]
ΔG° at 373.15 K
\[\begin{align*} \Delta G^{\circ} &= \Delta H^{\circ} - T\Delta{S}^{\circ} \\ &= 65.4~\mathrm{kJ~mol^{-1}} - (373.15~\mathrm{K})(0.03288~\mathrm{kJ~mol^{-1}~K^{-1}}) \\ &= 53.13~\mathrm{kJ~mol^{-1}} \end{align*}\]
We note that ΔG is positive (thermodynamically unfavorable process) and only slightly decreases with increasing temperature indicating that AgCl will always be insoluble in water regardless of T.
Plot
Figure 4.2: Gibbs free energy vs. temperature
Next, we determine the solubility product, Ksp, at three temperatures.
Equilibrium constant, Ksp, at 298.15, 323.15, and 373.15 K
Ksp at 298.15 K
\[\begin{align*} K_{\mathrm{sp}} &= e^{\frac{-\Delta G^{\circ}}{RT}} \\ &= e^{\frac{-\left (55.6~\mathrm{kJ~mol^{-1}}\right ) \left (\frac{10^3~\mathrm{J}}{\mathrm{kJ}}\right )}{\left (8.314~\mathrm{J~mol^{-1}~K^{-1}}\right ) \left (500~\mathrm{K} \right)}} \\ &= 1.82\times 10^{-10} \end{align*}\]
Ksp at 323.15 K
\[\begin{align*} K_{\mathrm{sp}} &= e^{\frac{-\Delta G^{\circ}}{RT}} \\ &= e^{\frac{-\left (54.77~\mathrm{kJ~mol^{-1}}\right ) \left (\frac{10^3~\mathrm{J}}{\mathrm{kJ}}\right )}{\left (8.314~\mathrm{J~mol^{-1}~K^{-1}}\right ) \left (323.15~\mathrm{K} \right)}} \\ &= 1.40\times 10^{-9} \end{align*}\]
Ksp at 373.15 K
\[\begin{align*} K_{\mathrm{sp}} &= e^{\frac{-\Delta G^{\circ}}{RT}} \\ &= e^{\frac{-\left (53.13~\mathrm{kJ~mol^{-1}}\right ) \left (\frac{10^3~\mathrm{J}}{\mathrm{kJ}}\right )}{\left (8.314~\mathrm{J~mol^{-1}~K^{-1}}\right ) \left (373.15~\mathrm{K} \right)}} \\ &= 3.65\times 10^{-8} \end{align*}\]
As expected, the solubility product constants are very small for a thermodynamically unfavored process.
Plot
Figure 4.3: Solubility product constant vs. temperature
We now turn our attention to determining the equilibrium concentrations of a saturated AgCl(s) aqueous solution.
At 25 °C
| AgCl(s) | ⇌ | Ag+(aq) | + | Cl–(aq) | |
|---|---|---|---|---|---|
I |
0 |
0 | |||
C |
+x |
+x | |||
E |
x |
x |
Note that the first column of the ICE table is ignored since AgCl is a solid.
Next we solve for the equilibrium concentrations.
\[\begin{align*} \mathrm{[Ag^+][Cl^-]} &= K \\ (x)(x) &= 1.82\times 10^{-10} \\ x^2 &= 1.82\times 10^{-10} \\ x &= 1.35\times 10^{-5}~M \\[2ex] [\mathrm{Ag^+}]_{\mathrm{eq}} &= [\mathrm{Cl^-}]_{\mathrm{eq}} = 1.35\times 10^{-5}~M \end{align*}\]
As expected, the concentrations of the dissolved AgCl salt is very low.
Practice
What are the equilibrium concentrations of a saturated AgCl(s) aqueous solution at 50 °C and 100 °C?
We can now determine the mass of AgCl in a saturated solution. Let us assume a 1 L solution.
\[\begin{align*} m_{\mathrm{AgCl}} &= \left ( \dfrac{1.35\times 10^{-5}~\mathrm{mol~Ag^+}}{\mathrm{L~solution}}\right ) \left ( \dfrac{1~\mathrm{mol~AgCl}}{\mathrm{1~mol~Ag^+}}\right ) \left ( \dfrac{143.43~\mathrm{AgCl}}{\mathrm{mol}} \right ) \left ( 1~\mathrm{L~solution}\right ) \\[1.25ex] &= 1.94\times 10^{-3}~\mathrm{g~AgCl} \end{align*}\]
Finally, we can determine the aqueous solubility of AgCl in g/100g. Given that the saturated solution is nearly all water with very little AgCl solute, we will approximate the density of the solution to be that of pure water (at 25 °C).
\[\begin{align*} d_{\mathrm{solution}} \approx d_{\mathrm{water}} = 0.997~\mathrm{g~mL^{-1}}~(\mathrm{at~25~^{\circ}C}) \end{align*}\]
Let us find the mass of solution using the density and volume of the solution
\[\begin{align*} d_{\mathrm{soln}} &= \dfrac{m_{\mathrm{soln}}}{V_{\mathrm{soln}}} ~\rightarrow \\[1.25ex] m_{\mathrm{soln}} &= d_{\mathrm{soln}}V_{\mathrm{soln}} \\[1.25ex] &= \left (\dfrac{0.997~\mathrm{g}}{\mathrm{mL}^{-1}} \right ) \left (1~\mathrm{L}\right ) \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}}\right ) \\[1.25ex] &= 997~\mathrm{g} \end{align*}\]
and the mass of solvent.
\[\begin{align*} m_{\mathrm{solvent}} &= m_{\mathrm{solution}} - m_{\mathrm{solute}} \\[1.25ex] &= 997~\mathrm{g~solution} - 1.94\times 10^{-3}~\mathrm{g~solute} \\[1.25ex] &= 996.998~\mathrm{g} \end{align*}\]
We can now find the aqueous solubility of AgCl in g/100g.
\[\begin{align*} \dfrac{1.94\times 10^{-3}~\mathrm{g~AgCl}}{996.998~\mathrm{g~H_2O}} \times 100~\mathrm{g~H_2O} &= 1.94\times 10^{-4}~\mathrm{g/100g} \end{align*}\]
How do our values stack up to authoritative data? Below is a table that compares the solubility constant and solubility of AgCl to values published in the CRC Handbook of Chemistry and Physics 97th ed. (p. 4-84 and 5-178, respectively)4. Notice the excellent agreement.
| Ksp | g/100g | |
|---|---|---|
Ours |
1.82e-10 |
0.000194 |
CRC |
1.77e-10 |
0.00019 |