Solubility product constant (Ksp) Knowledge-Based Study

Study notes

1. Dynamic Equilibrium in Saturated Solutions

   When a sparingly soluble ionic compound is placed in water, dissolution begins as water molecules hydrate and separate the surface ions. Simultaneously, dissolved ions collide with the solid surface and recrystallize. Eventually, the rate of dissolution equals the rate of precipitation, establishing a heterogeneous dynamic equilibrium:

   $$\text{M}_x\text{A}_y(s) \rightleftharpoons x\,\text{M}^{y+}(aq) + y\,\text{A}^{x-}(aq)$$

   Because the concentration of a pure solid is constant, its activity is defined as 1 and it does not appear in the equilibrium expression. The equilibrium constant for this dissolution process is known as the solubility product constant ($K_{sp}$).

   The value of $K_{sp}$ is temperature-dependent and serves as a quantitative measure of how much an ionic substance dissolves to reach saturation at a given temperature. A smaller $K_{sp}$ generally reflects lower solubility for compounds with similar ion ratios.

   Writing Equilibrium Expressions for Sparingly Soluble Salts

   The solubility product constant expression is the product of the equilibrium molar concentrations of the constituent ions, each raised to a power equal to its stoichiometric coefficient in the balanced dissociation equation.

   • 1:1 Electrolytes:
     For silver chloride:
     $$\text{AgCl}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Cl}^-(aq)$$
     $$K_{sp} = [\text{Ag}^+][\text{Cl}^-]$$
   • 1:2 Electrolytes:
     For lead(II) iodide:
     $$\text{PbI}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\,\text{I}^-(aq)$$
     $$K_{sp} = [\text{Pb}^{2+}][\text{I}^-]^2$$
   • Complex Stoichiometry (3:2):
     For calcium phosphate:
     $$\text{Ca}_3(\text{PO}_4)_2(s) \rightleftharpoons 3\,\text{Ca}^{2+}(aq) + 2\,\text{PO}_4^{3-}(aq)$$
     $$K_{sp} = [\text{Ca}^{2+}]^3[\text{PO}_4^{3-}]^2$$

   Important Note: Always ensure the stoichiometric coefficient is applied both as a multiplier to the ion concentration (if derived from salt dissolution) and as the exponent in the $K_{sp}$ expression.

   Calculating Solubility from Equilibrium Constants

   Molar solubility ($s$) is the number of moles of solute that dissolve per liter of solution to produce a saturated solution (expressed in $\text{mol/L}$). We can express $K_{sp}$ directly in terms of $s$:

   Case 1: 1:1 Salt (e.g., $\text{BaSO}_4$)

   Dissociation yields $s$ moles of each ion per liter:

   $$K_{sp} = [s][s] = s^2 \implies s = \sqrt{K_{sp}}$$

   Case 2: 1:2 Salt (e.g., $\text{Mg(OH)}_2$)

   Dissociation produces $s$ of $\text{Mg}^{2+}$ and $2s$ of $\text{OH}^-$:

   $$K_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2 = (s)(2s)^2 = 4s^3 \implies s = \sqrt[3]{\frac{K_{sp}}{4}}$$

   Comparing Solubilities: You can directly compare $K_{sp}$ values to determine relative solubilities only if the salts yield the same total number of ions upon dissolution. If the ion ratios differ (e.g., comparing a 1:1 salt with a 1:2 salt), you must calculate their actual molar solubilities ($s$) to compare them accurately.

   The Ion Product Quotient ($Q_{sp}$)

   To predict whether a precipitate will form when solutions containing potential reactant ions are combined, we calculate the reaction quotient ($Q_{sp}$) using initial ion concentrations:

   $$Q_{sp} = [\text{M}^{y+}]_0^x [\text{A}^{x-}]_0^y$$

   Comparing $Q_{sp}$ to $K_{sp}$ gives three possible outcomes:

   1. $Q_{sp} < K_{sp}$: The solution is unsaturated. No precipitate forms, and more solute can dissolve until dynamic equilibrium is reached.
   2. $Q_{sp} = K_{sp}$: The solution is saturated and at dynamic equilibrium. No visible change occurs.
   3. $Q_{sp} > K_{sp}$: The solution is temporarily supersaturated. Precipitation occurs spontaneously, removing excess ions from solution until $Q_{sp} = K_{sp}$.

   Calculation Tip: When two solutions are mixed, remember to calculate the new diluted concentrations of each ion before computing $Q_{sp}$.