Solubility Physical Equilibria

OUTCOME QUESTION(S): C12-4-11 PREDICTING EQUILIBRIUM Use the value of the reaction quotient, Q to explain how far a system at equilibrium has gone towards completion. Write solubility product (Ksp) expressions from balanced chemical equations for salts with low solubility and solve related problems. Vocabulary & Concepts Dissociation Ionization Saturate

These next 3 slides should be a review of grade 11… There are 3 actions that affect solubility: These next 3 slides should be a review of grade 11… 1. Nature of the solute / solvent “like dissolves like” Polar / ionic (charged) solutes dissolve in polar solvents Non-polar (uncharged) solutes dissolve in non-polar solvents 2. Temperature Solids placed in liquids: ↑ temperature - ↑ solubility Gases placed in liquids: ↑ in temperature - ↓ solubility 3. Pressure Does not affect the solubility of (s)/(l) For gases (g): ↑ pressure ↑ solubility

Aqueous (aq) phase means “dissolved in water” Dissolving: whole particles separate in solution Any solute that is soluble, will dissolve to some degree Following “like dissolves like” rule C6H12O6(s) C6H12O6(aq) Aqueous (aq) phase means “dissolved in water” SOLVENT Dissociating: particles disperse as individual ions Only ionic and highly polar compounds Ions (charged particles) create electrolyte solutions NaCl(s) Na+ (aq) + Cl-(aq) – + – + – + + – + – – + – + + – + – SOLUTE HCl(l) H+ (aq) + Cl-(aq)

AaBb aA+(aq) + bB¯(aq) AgCl (s) Mg(OH)2 (s) Ag+(aq) + Cl¯(aq) This is a general ionization/dissociation equation: AaBb aA+(aq) + bB¯(aq) silver chloride: AgCl (s) Ag+(aq) + Cl¯(aq) If you can build the compound, reverse engineer it - what are the parts of the compound? Break it up as ions magnesium hydroxide Mg(OH)2 (s) Mg2+(aq) + 2 OH¯(aq) lead (II) phosphate Pb3(PO4)2 (s) 3 Pb2+(aq) + 2 PO43-(aq)

Ksp Ksp = [A+]a[B-]b = AaBb(s) aA+(aq) + bB¯(aq) Equilibrium will be reached for a saturated solution at a constant temperature: AaBb(s) aA+(aq) + bB¯(aq) Remember: solids and liquids are incorporated into the constant value Ksp = [A+]a[B-]b {AaBb} Ksp = [A+]a[B-]b Ksp, called the solubility product constant.

Remember to look out for volume values and unit irregularities At equilibrium, the [Ag+] = 1.3 x 10-5 M and the [Cl-] = 1.3 x 10-5 M, find the Ksp of silver chloride? AgCl (s) Ag+(aq) + Cl-(aq) Remember to look out for volume values and unit irregularities Ksp = [Ag+][Cl-] Ksp =(1.3 x 10-5)(1.3 x 10-5) Think about why these two concentrations are the same (remember ICE tables) Ksp = 1.7 x 10-10

Ksp = [Pb+2][Cl -]2 Ksp = [1.62 x 10 -2][ 3. 24 x 10 -2]2 Calculate Ksp of lead (II) chloride if a 1.0 L saturated solution has of lead ions. 1.62 x 10-2 1.62 x 10-2 M PbCl2 (s) Pb2+ (aq) + 2 Cl1- (aq) 1 1 [I] (solid - doesn’t matter) X [C] - x + x + 2x [E] 2(1.62 x 10-2) In a saturated solution ALL solute has dissolved – in all cases – making this more of a Type III problem Ksp = [Pb+2][Cl -]2 Ksp = [1.62 x 10 -2][ 3. 24 x 10 -2]2 Remember: the coefficient 2 creates the (2x) and also is the power ([ ]2) Ksp = 1.70 x 10-5

This is a “reverse” question Ksp of magnesium hydroxide is 8.9 x 10-12. What are the [equilibrium] of ions in saturated solution? Mg(OH)2 (s) Mg2+(aq) + 2 OH¯(aq) [I] x 0 0 -x +x +2x 0 x 2x [C] [E] 1.3 x 10-4 2.6 x 10-4 Ksp = [Mg2+][OH-]2 This is a “reverse” question 8.9 x 10-12 = 4x3 8.9 x 10-12 = [x][2x]2 4 4 8.9 x 10-12 = [x]4x2 3√ 3√ 2.23 x 10-12 = x3 8.9 x 10-12 = 4x3 1.3 x 10-4 M = x

First change into mol/L – need dimensional analysis and the molar mass The solubility of PbF2 is . What is the value of the solubility product constant? 0.466 g/L PbF2(s) Pb2+(aq) + 2 F¯(aq) 207 + 2 (19) = 245g/mol Ksp = [Pb2+][F-]2 First change into mol/L – need dimensional analysis and the molar mass 0.466 g 1 mol = 1.90 x 10-3 M PbF2 1 L 245.2 g

PbF2(s) Pb2+(aq) + 2 F¯(aq) [I] 1.9 x 10-3 M [C] - x + x + 2x [E] 1.9 x 10-3 M 3.8 x 10-3 M In a saturated solution ALL solute has dissolved ~ The original concentration is the final concentration (for 1:1) Ksp = [Pb2+][F-]2 Ksp = (1.90 x 10-3)(3.80 x 10-3)2 Ksp = 2.74 x 10-8

Predicting Precipitation

Compare Q value with given Ksp to determine if solution is saturated / unsaturated / supersaturated 1. Q = Keq Solution is at equilibrium. Qc = [A+]a[B¯]b Solution is saturated 2. Q < Keq Solution is NOT at equilibrium. Qc = [A+]a[B¯]b Solution is unsaturated More solute can dissolve 3. Q > Keq Solution is NOT at equilibrium. Qc = [A+]a[B¯]b Solution is supersaturated Some solute left undissolved

CAN YOU / HAVE YOU? C12-4-11 PREDICTING EQUILIBRIUM Use the value of the reaction quotient, Q to explain how far a system at equilibrium has gone towards completion. Write solubility product (Ksp) expressions from balanced chemical equations for salts with low solubility and solve related problems. Vocabulary & Concepts Dissociation Ionization Saturated