Quiz 3 - Solubility and Precipitation Reactions | Solubility and Precipitation Reactions

Calcite and aragonite are two crystalline forms of calcium carbonate. The solubility product constant (K_sp​) reflects the solubility of each form, where a lower K_sp​ indicates lower solubility and greater thermodynamic stability. Calcite has a K_sp = 3.4 x 10⁻⁹, while aragonite has a K_sp = 6.0 x 10⁻⁹. Because calcite’s K_sp​ is lower, it dissolves less in water, making it more stable thermodynamically than aragonite.

• Step 1: Write the Dissociation Equation and Expression for K_sp
  PbF₂ (s) ⇌ Pb²⁺ (aq) + 2 F⁻ (aq)
  The K_sp​ expression for PbF₂​ is: Ksp = [Pb²⁺][F⁻]²

• Step 2: Set Up the Molar Solubility in Terms of s
  [Pb²⁺] = s
  [F⁻] = 2s (since each mole of PbF₂​ produces 2 moles of F⁻)
  Ksp = [Pb²⁺][F⁻]² = 4s³

• Step 3: Solve s
  s = $\sqrt[3]{\frac{{\mathrm{K}}_{\mathrm{sp}}}{4}}$ = $\sqrt[3]{\frac{4.0 \times {10}^{-8}}{4}}$ = 2.15 x 10^-3 M

• Step 4: Convert Molar Solubility to Mass:
  Molar mass of PbF₂​ = 207.2 + (2 x 19.0) = 245.2g/mol
  Mass = s x molar mass = 2.15 x 10^-3 x 245.2 = 0.53 g
   

The solubility of a compound in water is related to its K_sp​ value (the solubility product constant). A higher K_sp​ value indicates greater solubility.

The solubility of PbF_2​ increases in strong acids because F⁻ ions, as the conjugate base of the weak acid HF, react with H⁺ to form HF, removing F⁻ from the solution and shifting equilibrium.

In contrast, PbCl₂​, PbBr_2​, and PbI₂​ contain halide ions (Cl⁻, Br⁻, I⁻) that are conjugate bases of strong acids (HCl, HBr, HI), making them very weak bases that do not react significantly with H⁺. Therefore, their solubility remains largely unaffected by the presence of strong acids.

• Calculate the number of moles of AgCl dissolved:
  n (AgCl) = $\frac{1.9 \times {10}^{-4} \mathrm{g}}{143.4 \mathrm{g}.{\mathrm{mol}}^{-1}}$ = 1.32 x 10^-6 mol
• Calculate the molar solubility in mol/L (concentration of dissolved Ag⁺ and Cl⁻):
  s (AgCl) = $\frac{1.32 \times {10}^{-6} \mathrm{mol}}{0.1 \mathrm{L}}$ = 1.32 x 10^-5 M
• The dissociation of AgCl in water is: AgCl (s) ⇌ Ag⁺ (aq) + Cl⁻ (aq)
  Since AgCl dissociates into one Ag⁺ ion and one Cl⁻ ion, the concentration of each ion in solution will be equal to the molar solubility: 
  [Ag⁺] =[Cl⁻] = 1.32 x 10⁻⁵ M
• The solubility product constant, K_sp, for AgCl is:
  K_sp = [Ag⁺][Cl^-] = (1.32 x 10^-5)² = 1.8 x 10^-10

• Calcium hydroxide dissociates in water as: Ca(OH)₂ (s) ⇌ Ca²⁺ (aq) + 2 OH⁻ (aq)

• Let s be the molar solubility of Ca(OH)₂ in mol.L^-1. Then, in terms of s:
  [Ca²⁺] = s
  [OH^-] = 2s (since each Ca(OH)₂ produces 2 OH⁻ ions)

• The K_sp​ expression for Ca(OH)₂​ is:
  K_sp = [Ca²⁺][OH^-]² = 4s³
  ⇒ s = $\sqrt[3]{\frac{{\mathrm{K}}_{\mathrm{sp}}}{4}}$ = $\sqrt[3]{\frac{4.0 \times {10}^{-6}}{4}}$ = 1.0 x 10^-2 M

To find the concentration of Mg²⁺ in a solution that is 0.10 M in NaF and is saturated with MgF₂​, we can use the K_sp​ and account for the common ion effect from the F⁻ ions provided by NaF.

• The dissociation of MgF₂​ in water is: MgF₂​ (s) ⇌ Mg²⁺ (aq) + 2 F⁻ (aq)
• The solubility product expression K_sp​ for MgF₂​ is: K_sp = [Mg²⁺][F^-]²
  with K_sp = 6.4 x 10^-9 and [F^-] = 0.10 M (from NaF addition)
• Let s be the concentration of Mg²⁺ in the saturated solution. Then: K_sp = [Mg²⁺][F^-]²
  ⇒ [Mg²⁺] = s = $\frac{{\mathrm{K}}_{\mathrm{sp}}}{{\left[{\mathrm{F}}^{-}\right]}^2}$ = $\frac{6.4 \times {10}^{-9}}{0.{10}^2}$ =6.4 x 10^-7 M

To determine if a precipitate of PbBr₂​ will form in each mixture, we need to calculate the ion product Q for each mixture and compare it with K_sp. For a precipitate to form, Q must be greater than K_sp​.

• Mixture I:

[Pb²⁺] = $\frac{0.006 M \times 100 mL}{150 mL}$ = 0.004 M

[Br^-] = $\frac{0.003 M \times 50 mL}{150 mL}$] = 0.001 M

Q = [Pb²⁺][Br^-]² = (0.004)(0.001)² = 4 x 10^-9 < K_sp ⇒ no precipitation

• Mixture II:

[Pb²⁺] = $\frac{0.008 M \times 100 mL}{200 mL}$ = 0.004 M

[Br^-] = $\frac{0.006 M \times 100 mL}{200 mL}$] = 0.003 M

Q = [Pb²⁺][Br^-]² = (0.004)(0.003)² = 3.6 x 10^-8 < K_sp ⇒ no precipitation