Precipitation, Reaction Quotient (Q), and Selective Precipitation in Aqueous Solutions

Precipitation and the Reaction Quotient (Q)

Understanding Precipitation in Aqueous Solutions

Precipitation occurs when the concentration of ions in solution exceeds the solubility of an ionic compound, resulting in the formation of a solid. The extent of solubility is quantified by the solubility product constant (Ksp), and the prediction of precipitation is made using the reaction quotient (Q).

• Ksp is the equilibrium constant for the dissolution of a sparingly soluble salt.

• Q is calculated using the actual (not necessarily equilibrium) concentrations of ions at any moment.

• For a salt such as CaF2: $Q = [Ca^{2+}][F^-]^2$

To determine the state of the solution, compare Q to Ksp:

• Q < Ksp: Solution is unsaturated; more solid can dissolve, no precipitation occurs.

• Q = Ksp: Solution is saturated; at equilibrium, no net change.

• Q > Ksp: Solution is supersaturated; precipitation occurs until Q is reduced to Ksp.

Condition	Solution State	What Happens?	Example
Q < Ksp	Unsaturated	More solid dissolves; no precipitation	Drop solid into water → it dissolves
Q = Ksp	Saturated (equilibrium)	No net change; solid in equilibrium with ions	Excess solid in contact with solution
Q > Ksp	Supersaturated	Solid precipitates until Q = Ksp	Mix two solutions → precipitation

Supersaturated Solutions

Supersaturated solutions contain more dissolved solute than allowed by equilibrium. Precipitation may not occur immediately due to kinetic hindrance, but the addition of a seed crystal can trigger rapid crystallization.

• Example: Supersaturated sodium acetate solution (Q > Ksp).

• Crystallization is thermodynamically favored but kinetically hindered until a nucleation site is provided.

Predicting Precipitation: Step-by-Step Method

How to Determine if a Precipitate Will Form

1. Identify cross products: When mixing solutions, determine all possible ion pairs. Only those with low Ksp (insoluble) can precipitate.

2. Write the Q expression: For the potentially insoluble product, write $Q = [cation]^x[anion]^y$ using the Ksp stoichiometry.

3. Calculate ion concentrations after mixing: Account for dilution. If equal volumes are mixed, new [ion] = old [ion] / 2. Otherwise, use $C_1V_1 = C_2V_2$.

4. Calculate Q: Substitute the post-mixing concentrations into the Q expression.

5. Compare Q to Ksp: If Q > Ksp, a precipitate forms. If Q < Ksp, no precipitate forms. If Q = Ksp, the solution is at equilibrium (rare).

Example: Predicting Precipitation

Problem: A solution containing lead(II) nitrate is mixed with one containing sodium bromide to form a solution that is 0.0150 M in Pb(NO3)2 and 0.00350 M in NaBr. Does a precipitate form? (Ksp for PbBr2 = $4.67 \times 10^{-6}$)

• Possible cross product: PbBr2 (potentially insoluble)

• Q = [Pb2+][Br-]2 = (0.0150)(0.00350)2 = $1.84 \times 10^{-7}$

• Since Q < Ksp, no precipitate forms; the solution remains unsaturated.

Practice Problem: Will AgI Precipitate?

Stepwise Solution

• Given: [AgNO3] = 0.010 M, [KI] = 0.015 M, Ksp (AgI) = $8.51 \times 10^{-17}$

• Q = [Ag+][I-] = (0.010)(0.015) = $1.5 \times 10^{-4}$

• Q >> Ksp → AgI will precipitate

• Limiting ion: Ag+ (smaller moles); after precipitation, Ag+ is nearly zero, excess I- remains.

Selective Precipitation: Separating Mixed Metal Cations

Principles and Strategy

Selective precipitation is a technique to separate ions in a mixture by adding a reagent that precipitates one ion but not others. The key is to choose a precipitating anion so that:

• Q > Ksp for the less soluble compound (target) → it precipitates first.

• Q < Ksp for the more soluble compound → it stays dissolved.

• Ksp values should differ by at least a factor of 103 for effective separation.

Example: Mg2+ and Ca2+ in seawater with KOH.

• Mg(OH)2: Ksp = $2.06 \times 10^{-13}$ (less soluble, precipitates first)

• Ca(OH)2: Ksp = $4.68 \times 10^{-6}$ (more soluble, stays dissolved)

Finding Minimum Reagent Concentration for Selective Precipitation

To find the minimum [OH-] to precipitate Mg2+ from seawater, set Q = Ksp and solve for [OH-]:

• Given [Mg2+] = 0.059 M

• Q = [Mg2+][OH-]2 = (0.059)[OH-]2

• Set Q = Ksp: $(0.059)[OH^-]^2 = 2.06 \times 10^{-13}$

• Solve for [OH-]: [OH-] = $1.9 \times 10^{-6}$ M triggers Mg(OH)2 precipitation

Finding Ion Concentration After Selective Precipitation

To find [Mg2+] remaining when Ca(OH)2 just begins to precipitate:

• Find [OH-] for Ca(OH)2 precipitation: $(0.011)[OH^-]^2 = 4.68 \times 10^{-6}$

• Calculate [Mg2+] at this [OH-] using Ksp of Mg(OH)2

• Result: [Mg2+] = $4.9 \times 10^{-10}$ M when Ca2+ begins to precipitate

Practice Problem: Selective Precipitation of Ba2+ and Ca2+

• Given: [Ba2+] = 0.025 M, [Ca2+] = 0.025 M, Ksp (BaSO4) = $1.07 \times 10^{-10}$, Ksp (CaSO4) = $7.10 \times 10^{-5}$

• (a) BaSO4 has the smaller Ksp → precipitates first

• (b) Minimum [SO42-] to begin precipitating Ba2+: $(0.025)[SO_4^{2-}] = 1.07 \times 10^{-10}$ → [SO42-] = $4.3 \times 10^{-9}$ M

• (c) [SO42-] when Ca2+ begins to precipitate: $(0.025)[SO_4^{2-}] = 7.10 \times 10^{-5}$ → [SO42-] = $2.84 \times 10^{-3}$ M

• [Ba2+] remaining: $[Ba^{2+}] = \frac{1.07 \times 10^{-10}}{2.84 \times 10^{-3}} = 3.77 \times 10^{-8}$ M

• Conclusion: Ba2+ is essentially completely removed before Ca2+ begins to precipitate.

Practice Problem: Precipitation of PbCl2

• Mix 50.0 mL of 0.0200 M Pb(NO3)2 with 50.0 mL of 0.0300 M NaCl.

• (a) Ksp expression: PbCl2(s) ⇌ Pb2+(aq) + 2 Cl-(aq); $K_{sp} = [Pb^{2+}][Cl^-]^2$

• (b) After mixing, concentrations are halved: [Pb2+] = 0.0100 M, [Cl-] = 0.0150 M

• Q = (0.0100)(0.0150)2 = $2.25 \times 10^{-6}$

• (c) Q < Ksp ($1.17 \times 10^{-5}$) → no precipitate forms

• (d) No limiting reagent applies since no precipitation occurs.

Key Takeaways

• Q is calculated like Ksp but with actual (not equilibrium) concentrations at any moment.

• Q < Ksp: unsaturated (no ppt); Q = Ksp: saturated (equilibrium); Q > Ksp: precipitate forms.

• When solutions mix, concentrations are diluted—always recalculate concentrations before finding Q.

• Selective precipitation: choose a reagent anion to precipitate the less-soluble compound first.

• Selective precipitation requires Ksp values to differ by ≥ 103 for practical separation.

• Find minimum reagent concentration: set Q = Ksp for target compound and solve for reagent ion.