AP Chem Get Equilibrium Practice Stamped Today: Solubility Equilibrium (Ksp) Unit 6 Test: Th 1/31 (MC) and Fri 2/1(FRQ)

Solubility & Equilibrium Initially, when a solid is placed in water: XY(s)  X+ (aq) + Y- (aq) As time continues, more of the solid dissolves and the [ions] begins to increase.

Solubility & Equilibrium When the concentration of ions increases, eventually there is a greater the chance that they will collide and reform the solid XY(s)  X+ (aq) + Y- (aq) When it becomes saturated it reaches equilibrium. XY(s) ⇌ X+ (aq) + Y- (aq)

XY(s) ⇌ X+ (aq) + Y- (aq) Equilibrium Constant for Solubility-Ksp Ksp = [X+]x [Y-]y The concentrations of the ions at equilibrium is known as the solubility or molar solubility of the substance.

Meaning of Ksp The value of Ksp tells you how soluble a substance is (how much it will dissolve and dissociate into ions). A really small Ksp means the substance is relatively not that soluble in water (not much will dissolve and dissociate into ions) We can still use ICE tables to solve Ksp problems.

AgI(s) ⇆ Ag+(aq)+ I-(aq) Write the Ksp expression for this solubility equilibrium. Ksp= [Ag+][I-]

AgI(s) ⇆ Ag+(aq)+ I-(aq) C E Consider the solubility equilibrium: AgI (s) ⇌ Ag+(aq) + I-(aq) (Ksp = 8.52 x 10-17). Calculate the solubility of AgI in pure water. +x +x x x Ksp= [Ag+][I-] 8.52 x 10-17 = x2 x = 9 x 10-9 M  molar solubility of AgI in pure water

Common Ion Effect Instead of adding your solute to water, you’re adding it to a solution that contains a common ion as your solute This means the starting concentration of one of your ions will NOT be 0.

AgI (s) ⇆ Ag+(aq) + I-(aq) C E 0.001 Then calculate the new solubility of AgI in a solution containing 1.00 x 10-3 M NaI. +x +x x 0.001 + x Ksp= [Ag+][I-] 8.52 x 10-17 = x(.001+x) x = 8.52 x 10-14 M  solubility in a solution containing NaI