Predicting precipitation Knowledge-Based Study

Study notes

1. When studying the solubility of slightly soluble salts, the solubility product constant, $K_{sp}$, describes the dynamic equilibrium established between an undissolved ionic solid and its dissolved ions in a saturated solution.

   However, when two separate aqueous solutions containing ions are combined, the system is not necessarily at equilibrium. To determine whether the solution is at equilibrium, unsaturated, or supersaturated, we calculate the ion product quotient, denoted as $Q_{sp}$ (or simply the reaction quotient $Q$).

   For a generic dissolution process:

   $$M_aX_b(s) \rightleftharpoons a\,M^{n+}(aq) + b\,X^{m-}(aq)$$

   The expression for $Q_{sp}$ takes the same mathematical form as $K_{sp}$:

   $$Q_{sp} = [M^{n+}]_0^a [X^{m-}]_0^b$$

   The key distinction is that while $K_{sp}$ uses equilibrium concentrations, $Q_{sp}$ uses the initial or instantaneous molar concentrations of ions present in the solution immediately after mixing, before any precipitation takes place.

   Comparing the calculated value of $Q_{sp}$ to the known value of $K_{sp}$ at a specific temperature allows us to predict whether a precipitate will form:

   • $Q_{sp} < K_{sp}$: The solution is unsaturated. The concentration of ions is below the saturation threshold. No precipitate forms, and any added solid solute will continue to dissolve until saturation is reached.
   • $Q_{sp} = K_{sp}$: The solution is exactly saturated. The dissolved ions are in dynamic equilibrium with the solid phase. No precipitate forms spontaneously, and no additional solid can dissolve.
   • $Q_{sp} > K_{sp}$: The solution is temporarily supersaturated. The ion concentrations exceed the equilibrium limit. According to Le Chatelier's principle, the system shifts toward the reactants (solid phase) to relieve this stress, and a precipitate forms until the remaining ion concentrations satisfy $Q_{sp} = K_{sp}$.

   A common source of error in predicting precipitation is using the stock concentrations of the solutions rather than their concentrations after mixing.

   When two aqueous solutions are combined, the total volume of the mixture increases. Assuming volumes are additive:

   $$V_{\text{total}} = V_1 + V_2$$

   Because the volume increases, each ion is diluted. The new molar concentration for each ion must be calculated using the dilution equation before computing $Q_{sp}$:

   $$M_{\text{initial}} V_{\text{initial}} = M_{\text{diluted}} V_{\text{total}}$$

   $$M_{\text{diluted}} = \frac{M_{\text{initial}} V_{\text{initial}}}{V_{\text{total}}}$$

   Only these diluted concentrations represent the instantaneous ion concentrations immediately following the mixing event.

   To predict whether a precipitate will form, follow this systematic procedure:

   1. Identify the potential precipitate: Use solubility rules to determine which combination of cation and anion could form an insoluble or slightly soluble salt.
   2. Write the dissolution equilibrium: State the dissociation equation for the salt to identify the stoichiometric coefficients and exponents for the expression.
   3. Calculate post-mixing concentrations: Account for total volume and determine the diluted concentration of each relevant ion.
   4. Calculate $Q_{sp}$: Substitute the diluted ion concentrations into the ion product expression.
   5. Compare $Q_{sp}$ to $K_{sp}$: Conclude whether precipitation occurs based on whether $Q_{sp} > K_{sp}$.

   Worked Example

   Suppose 100 mL of $2.0 \times 10^{-3}\text{ M}$ $BaCl_2$ is mixed with 100 mL of $2.0 \times 10^{-3}\text{ M}$ $K_2SO_4$. Given that $K_{sp}(BaSO_4) = 1.1 \times 10^{-10}$, will $BaSO_4$ precipitate?

   • Total volume: $100\text{ mL} + 100\text{ mL} = 200\text{ mL}$.
   • Diluted concentrations:
     $$[Ba^{2+}] = \frac{(2.0 \times 10^{-3}\text{ M})(100\text{ mL})}{200\text{ mL}} = 1.0 \times 10^{-3}\text{ M}$$
     $$[SO_4^{2-}] = \frac{(2.0 \times 10^{-3}\text{ M})(100\text{ mL})}{200\text{ mL}} = 1.0 \times 10^{-3}\text{ M}$$
   • Calculate $Q_{sp}$:
     $$Q_{sp} = [Ba^{2+}][SO_4^{2-}] = (1.0 \times 10^{-3})(1.0 \times 10^{-3}) = 1.0 \times 10^{-6}$$
   • Comparison: $1.0 \times 10^{-6} > 1.1 \times 10^{-10}$, so $Q_{sp} > K_{sp}$.

   Conclusion: A precipitate of barium sulfate, $BaSO_4(s)$, will form.