Lead (II) chromate, PbCrO4, is a slightly soluble salt. Write a net ionic equation for the dissolution of solid PbCrO4 in water. Write an equilibrium constant expression for this reaction. K = What might one add to increase the solubility of the PbCrO4? Write equations to describe how this added reagent works to increase the PbCrO4 solubility. What reagent could be added to decrease the concentration of Pb2+ in solution
The Correct Answer and Explanation is:
Net Ionic Equation and Equilibrium Expression
Lead(II) chromate (PbCrO4\text{PbCrO}_4PbCrO4) is a slightly soluble salt, and its dissolution in water can be represented by the following net ionic equation: PbCrO4(s)⇌Pb2+(aq)+CrO42−(aq)\text{PbCrO}_4(s) \rightleftharpoons \text{Pb}^{2+}(aq) + \text{CrO}_4^{2-}(aq)PbCrO4(s)⇌Pb2+(aq)+CrO42−(aq)
The corresponding equilibrium constant expression, known as the solubility product (Ksp_\text{sp}sp), is: Ksp=[Pb2+][CrO42−]K_{sp} = [\text{Pb}^{2+}][\text{CrO}_4^{2-}]Ksp=[Pb2+][CrO42−]
This expression quantifies how much PbCrO4\text{PbCrO}_4PbCrO4 dissolves in water at equilibrium.
Increasing Solubility of PbCrO₄
To increase the solubility of PbCrO4\text{PbCrO}_4PbCrO4, one can add a substance that removes either Pb2+\text{Pb}^{2+}Pb2+ or CrO42−\text{CrO}_4^{2-}CrO42− from solution, thereby shifting the equilibrium to the right (according to Le Châtelier’s Principle).
Example: Adding Hydrochloric Acid (HCl)
Adding HCl introduces H+\text{H}^+H+ ions, which react with CrO42−\text{CrO}_4^{2-}CrO42− to form HCrO4−\text{HCrO}_4^-HCrO4−: CrO42−(aq)+H+(aq)⇌HCrO4−(aq)\text{CrO}_4^{2-}(aq) + \text{H}^+(aq) \rightleftharpoons \text{HCrO}_4^-(aq)CrO42−(aq)+H+(aq)⇌HCrO4−(aq)
This reaction decreases the concentration of CrO42−\text{CrO}_4^{2-}CrO42−, driving the dissolution equilibrium to the right and increasing the solubility of PbCrO4\text{PbCrO}_4PbCrO4.
Decreasing Pb²⁺ Concentration
To decrease the concentration of Pb2+\text{Pb}^{2+}Pb2+ in solution, a reagent that forms an insoluble compound with lead can be added.
Example: Adding Sodium Sulfate (Na₂SO₄)
Sulfate ions react with lead(II) ions to form insoluble lead(II) sulfate: Pb2+(aq)+SO42−(aq)→PbSO4(s)\text{Pb}^{2+}(aq) + \text{SO}_4^{2-}(aq) \rightarrow \text{PbSO}_4(s)Pb2+(aq)+SO42−(aq)→PbSO4(s)
This reaction removes Pb2+\text{Pb}^{2+}Pb2+ from solution by forming a precipitate, lowering its concentration.
Summary
- Dissolution reaction: PbCrO4(s)⇌Pb2+(aq)+CrO42−(aq)\text{PbCrO}_4(s) \rightleftharpoons \text{Pb}^{2+}(aq) + \text{CrO}_4^{2-}(aq)PbCrO4(s)⇌Pb2+(aq)+CrO42−(aq)
- Equilibrium: Ksp=[Pb2+][CrO42−]K_{sp} = [\text{Pb}^{2+}][\text{CrO}_4^{2-}]Ksp=[Pb2+][CrO42−]
- To increase solubility: Add HCl (removes CrO42−\text{CrO}_4^{2-}CrO42−)
- To decrease Pb2+\text{Pb}^{2+}Pb2+: Add Na₂SO₄ (forms PbSO4\text{PbSO}_4PbSO4 precipitate)
These strategies exploit chemical equilibria to control solubility and ion concentrations in solution
