Quiz 2 - Solubility and Precipitation Reactions | Solubility and Precipitation Reactions

Solubility-product constant K_sp: equilibrium-constant of a dissolution reaction.
Dissolution of tin(II) hydroxide: Sn(OH)₂ (s) $\rightleftharpoons$ Sn²⁺ (aq) + 2 HO^- (aq)

Dissolution of BaCO₃: BaCO₃ (s) $\rightleftharpoons$ Ba²⁺ (aq) + CO₃^2- (aq)
According to the reaction stoichiometry, molar solubility s of BaCO₃ (s) = [Ba²⁺] = [CO₃^2-]
K_sp = [Ba²⁺] [CO₃^2-] = s²    ⇒   s = K_sp^1/2

Dissolution of Ca₃(PO₄)₂: Ca₃(PO₄)₂ (s) $\rightleftharpoons$ 3 Ca²⁺ (aq) + 2 PO₄^3- (aq)
According to the reaction stoichiometry, molar solubility s of Ca₃(PO₄)₂ = $\frac{\left[{\mathrm{Ca}}^{2+}\right]}{3}$ = $\frac{\left[{{\mathrm{PO}}_4}^{2-}\right]}{2}$
K_sp = [Ca²⁺]³ [PO₄^3-]² = (3s)³ x (2s)² = 27s³ x 4s² =108s⁵   ⇒ s =  ${\left(\frac{K_{sp}}{108}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$ 

AgNO₃ and Ag₃PO₄ have a common ion: Ag⁺
Common ion effect: the solubility of an ionic solid decreases when a common ion is present in solution

The solubility of an ionic solid is increased by the formation of a soluble complex ion.

Al(OH)₃ is an amphoteric metal hydroxides:
Neutral solution: insoluble
Basic solution: Al(OH)₃ (s) + HO^- (aq) ⇌ [Al(OH)₄]^- (aq)     [formation of a soluble hydroxy complex ion]
Acidic solution: Al(OH)₃ (s) + 3 H₃O⁺ (aq) ⇌ Al³⁺ (aq) + 6 H₂O (l)     [acid-base neutralization]

Adding HCl will increase the solubility of BaSO_4​ because the H⁺ ions react with SO₄²⁻​ ions to form HSO₄⁻, reducing the concentration of SO₄²⁻ ions and shifting the equilibrium to dissolve more BaSO₄.

Since AgI has a lower K_sp​ than AgCl, I^- ions will precipitate as AgI before Cl^- ions precipitate as AgCl.

Group 5 cations, such as Na⁺, K⁺, and Mg²⁺, do not precipitate under the standard conditions used for Groups 1-4. These ions stay in solution and are often identified by flame tests or other methods.

For PbI₂​, the dissociation is PbI₂ ⇌ Pb²⁺ + 2 I⁻

Let s be the molar solubility of PbI₂​ in water. Then: K_sp = [Pb²⁺][I⁻]² = s (2s)² = 4s³ 

⇒ s = $\sqrt[3]{\frac{{\mathrm{K}}_{\mathrm{sp}}}{4}}$ = 1.2 x 10^-3 M