Solubility Equilibria Solubility Product Constant, K sp Precipitation Complex Ions.
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Transcript of Solubility Equilibria Solubility Product Constant, K sp Precipitation Complex Ions.
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Solubility EquilibriaSolubility Product Constant, Ksp
Precipitation
Complex Ions
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Solubility RulesConsider an ionic solid that completely dissociates into hydrated cations and anions:
NaF (s) Na+ (aq) + F- (aq)
Recall the “Solubility Rules” for substances that are 100% ionized.
Strong Electrolytes ( 100% ionized) A. Strong Acids: HCl, HBr, HI, H2SO4, HNO3, HClO4, HClO3 B. Strong Bases: Hydroxides of group IA and II A, Except Be and Mg C. Soluble Salts ( ionic compounds: metal/nonmetal) Always Soluble if these are in a compound Except with NO3
-, Group IA, NH4+, CH3COO-, ClO4
-, ClO3- No Exceptions
Cl-, Br-, I- Pb, Ag, Hg2
2+ SO4
2- Ag, Pb, Hg22+
Ca, Sr, Ba
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Solubility Product Constant Salts that are not listed on your solubility rules chart, are sparingly soluble to the
degree indicated by the solubility product constant, or Ksp.
With more concentration of ions and more collisions, the ions recombine to form the salt again:
CaF2 (s) Ca2+ (aq) + 2F- (aq) Ksp = [Ca2+][ F-]2
[CaF2](s)
Since the solid is not aqueous, the concentration designation is meaningless, and it is not included in the equilibrium expression.
Therefore, for Ksp problems, the equilibrium expression is simply one in this form:
Ksp = [Ca2+][ F-]2
Ksp is >> 1 for all the substances that are 100% ionized, or listed on the solubility rules.
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Solubility of Solids Increasing surface area (crushing, etc.) will increase the rate that
equilibrium occurs, but not the equilibrium position: More surface area increases rate of dissolving, but also rate of re-formation.
Solubility product = a constant at a given temperature
Solubility = an equilibrium position At the saturation point, which is controlled by the Ksp
expression, it does not matter how much solid you add; no more can dissolve.
If the Ksp expression is not exceeded, no precipitation will occur.
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Concentration Affects Recombination
CaF2 (s) Ca2+ (aq) + 2F- (aq)
If concentration of either ion is increased, the equilibrium shifts back toward the reactants, causing more solid, or precipitant to be formed.
OHH
-+
+-+
+
-+
+
-+
+-+
+
-++
Cu
CuF
-+ +
- ++
F
-++
-++
-+
+
-+
+
-++
-++
Cu
-+ +
- ++
F
-++
-++
-+
+
-++
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Ksp Example One liter of saturated barium sulfate solution contains 0.0025 g of the solute. What is the
molar solubility of BaSO4? Find the solubility product constant for BaSO4.
BaSO4 (s) Ba2+ (aq) + SO42- (aq) Ksp = [Ba2+][SO4
2-]
Molar solubility = mol : 0.0025 g BaSO4 |1.0 mol BaSO4 = 1.1 x 10-5 mol BaSO4
L 1.0 L |233 g BaSO4 L
One mole of each ion is formed by one mole of BaSO4:
1.1 x 10-5 M BaSO4, forms 1.1 x 10-5 mol/L Ba2+ and 1.1 x 10-5 mol/L SO4
2-
Ksp= [Ba2+][SO42-] = (1.1 x 10-5 ) (1.1 x 10-5 ) = 1.2 x 10-10
If the coefficient of either of the products is 2, [ion-] is squared.
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In Real Life… Calcium CarbonateCaCO3 (s) Ca2+ (aq) + CO3 2- (aq) Ksp = [Ca2+][CO3 2-] = 5.0 x 10-10
ΔH is negative
Sea shells are mainly made of calcium carbonate (reacts with acid but does not dissolve appreciably in water).
When the sea creatures die and the remains sink to great depth in the oceans, the shells dissolve because of changes in the position of two equilibria:
1. At great depths the pressure is much higher and the temperature lowerBoth factors favor more CO2 dissolvingThe forward reaction is promoted: CaCO3 (aq) + CO2(aq) + H2O (l) Ca2+ (aq) + 2HCO3 -(aq)
2. Lower temperature favors a higher Ksp value because the dissolving process is slightly exothermic, so more of the calcium carbonate will dissolve.
There are no shells lying at the bottom of the oceans! The limestone cliffs you see in the landscape were formed in warm shallow tropical seas.
http://www.docbrown.info/page07/equilibria4a.htm