Chapter 17: Solubility & Complex Ion Equilibria ◆ The goal of this chapter is to understand the equilibria that exist between ionic solids and their ions in solution, and factors that affect that equilibrium. ◆ write (heterogeneous) equilibrium equations & K expressions calculate and interpret Ksp Ksp is the solubility product constant using Ksp, calculate solubility of salts mol/L (molar solubility), and g/L ◆ factors that affect solubility common ion effect pH formation of complex ions ◆ calculations to determine whether precipitation of a solid will occur when 2 solutions are combined Overview of the Chapter Solubility Equilibria ◆ equilibrium between a solid and its ions in solution ex. for calcium phosphate Ca 3 (PO 4 ) 2 (s) ⇄ 3 Ca 2+ (aq) + 2 PO 4 3– (aq) ◆ this is a heterogeneous equilibrium ◆ equilibrium constant: K sp solubility product constant for calcium phosphate: K sp = [Ca 2+ ] 3 [PO 4 3– ] 2 ◆ we can now refine our understanding of solubility rules from Chapter 4 solids that we classified as “insoluble” typically have small K sp ’s and low molar solubilities “slightly soluble” or “sparingly soluble”
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Chapter 17:Solubility & Complex
Ion Equilibria
◆ The goal of this chapter is to understand the equilibria that exist between ionic solids and their ions in solution, and factors that affect that equilibrium.
◆ write (heterogeneous) equilibrium equations & K expressionscalculate and interpret Ksp
Ksp is the solubility product constantusing Ksp, calculate solubility of salts
mol/L (molar solubility), and g/L
◆ factors that affect solubilitycommon ion effectpHformation of complex ions
◆ calculations to determine whether precipitation of a solid will occur when 2 solutions are combined
Overview of the Chapter
Solubility Equilibria◆ equilibrium between a solid and its ions in solution
ex. for calcium phosphateCa3(PO4)2 (s) ⇄ 3 Ca2+ (aq) + 2 PO43– (aq)
Calculation of Solubility After Complex Ion Formation
example (see example 17.11 in text):
Calculate the molar solubility of AgBr in 2.25 M Na2S2O3 (aq).
For AgBr, Ksp = 5.0 x 10–13; for the complex ion [Ag(S2O3)2]3–, Kf = 2.9 x 1013.
Calculation of Solubility After Complex Ion Formation
example (see example 17.11 in text):
Calculate the molar solubility of CuI in 0.88 M KCN (aq).
For CuI, Ksp = 1.1 x 10–12; for the complex ion [Cu(CN)2]–, Kf = 1.0 x 1016.
Precipitation of Ionic Solids
◆ For an ionic solid, MX the solubility equilibrium is given by:
MX (s) ⇄ Mn+ (aq) + Xn– (aq)
◆ When 2 solutions are combined – one sol’n containing Mn+, and one sol’n containing Xn– – will a precipitate of MX form?
◆ Is the system at equilibrium? If not, in what direction does the reaction proceed to reach equilibrium?
◆ Q vs K calculation
Precipitation of Ionic Solids
◆ determine concentrations of ions after solutions are combined:
[ion] = ––––––––––––––
◆ calculate the ion product, Q
◆ compare Q to Ksp:if Q = Ksp, the solution is saturated;
solution is at equilibrium
if Q > Ksp, the solution is supersaturated; precipitation of solid will occur
if Q < Ksp, the solution is unsaturated; precipitation will not occur - more solid will dissolve
mol iontotal sol’n volume
example:
250.0 mL of 0.0062 M AgNO3 (aq) and 250.0 mL of 0.00014 M Na2CO3 are combined. Will a precipitate of Ag2CO3 form? For Ag2CO3, Ksp = 8.1 x 10–12.
Ag2CO3 (s) ⇄ 2 Ag+ (aq) + CO32– (aq)
Q = [Ag+]2[CO32–]
example:
Determine the minimum concentration of carbonate ion required to cause the precipitation of silver carbonate from a 5.8 x 10–4 M solution of AgNO3. For Ag2CO3, Ksp = 8.1 x 10–12.
Ag2CO3 (s) ⇄ 2 Ag+ (aq) + CO32– (aq)
Ksp = [Ag+]2[CO32–]
◆ solve for [CO32–] present in a saturated solution of Ag2CO3;
[CO32–] at equilibrium OR when Q = Ksp
◆ any greater [CO32–] will result in Q > Ksp and precipitation of solid