Solubility equilibrium and the common ion effect are key concepts in understanding how dissolved substances behave in solution. These principles explain why adding certain ions can decrease solubility, and how to predict when precipitation will occur.

Precipitation reactions and selective precipitation build on these ideas, allowing us to predict and control the formation of solid compounds from solutions. These techniques are crucial for separating and identifying ions in mixtures, with wide-ranging applications in chemistry and beyond.

Solubility Equilibrium and the Common Ion Effect

Common ion effect on solubility

• Common ion effect occurs when a solution contains an ion that is also part of the solute's dissociation equilibrium shifts the equilibrium to the left, towards the solid form of the solute, decreasing solubility ($NaCl$ and $AgCl$)
• Le Chatelier's principle explains the common ion effect adding a common ion (stress) causes the equilibrium to shift in the direction that reduces the concentration of the common ion, counteracting the stress ($NaCl$ added to $AgCl$)
• Solubility product constant ($K_{sp}$) remains unchanged, but the concentration of ions in the solution decreases due to the common ion effect ($AgCl$ in $NaCl$ solution)

Ion concentrations with common ions

• To calculate ion concentrations in the presence of a common ion:
  1. Write the balanced dissociation equation for the solute ($AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$)
  2. Set up an ICE table (Initial, Change, Equilibrium) considering the common ion's initial concentration ($[Cl^-]_i = 0.1$ M from $NaCl$)
  3. Use the $K_{sp}$ expression to solve for the equilibrium concentrations of the ions ($1.8 \times 10^{-10} = [Ag^+][Cl^-]$)
• Example: Calculate $[Ag^+]$ in a solution containing $0.1$ M $NaCl$ and excess $AgCl$ ($K_{sp} = 1.8 \times 10^{-10}$)
  • ICE table: $[Ag^+]_i = 0$, $[Cl^-]_i = 0.1$ M, $[Ag^+]_e = x$, $[Cl^-]_e = 0.1 + x$
  • $K_{sp}$ expression: $1.8 \times 10^{-10} = (x)(0.1 + x)$, solve for $x$ to find $[Ag^+]$ at equilibrium

Precipitation Reactions and Selective Precipitation

Precipitation reactions using Q vs Ksp

• Compare the reaction quotient ($Q$) to the solubility product constant ($K_{sp}$) to predict the direction of a precipitation reaction ($Q < K_{sp}$: precipitation, $Q > K_{sp}$: dissolution, $Q = K_{sp}$: equilibrium)
• Calculate $Q$ using the initial concentrations of the ions in the solution for the reaction $aA^+ + bB^- \rightleftharpoons A_aB_b(s)$, $Q = [A^+]^a[B^-]^b$ ($BaSO_4$: $Q = [Ba^{2+}][SO_4^{2-}]$)
• Compare $Q$ to $K_{sp}$ to determine the direction of the reaction if $Q < K_{sp}$, precipitation occurs; if $Q > K_{sp}$, solid dissolves; if $Q = K_{sp}$, system is at equilibrium

Selective precipitation in analysis

• Selective precipitation separates and identifies ions in a mixture based on differences in solubility among various compounds ($Ag^+$ and $Pb^{2+}$ separated using $HCl$ and $KI$)
• Precipitating agents are added in a specific order to selectively precipitate certain ions while leaving others in solution, based on the solubility of the compounds formed with the ions in the mixture ($HCl$ precipitates $AgCl$, leaving $Pb^{2+}$ in solution)
• Precipitates are separated from the solution by filtration, and further tests can be performed on the precipitates and the remaining solution to identify the ions present ($AgCl$ precipitate and $Pb^{2+}$ in solution)
• Selective precipitation allows for the systematic identification of ions in a complex mixture by separating them based on their solubility differences ($Ag^+$, $Pb^{2+}$, and $Ba^{2+}$ in a mixture)