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Ksp
November 2014
Review Keq
• Keq is an expression relating concentrations of products to concentrations of reactant at equilibrium. • Solids are never included.
Solids are never included
• Pure liquids are never included. Saturated Solutions
• Solution that has dissolved the maximum amount of solute.
• Different solids have a high or low solubility in water despite not seeming like it would
water, despite not seeming like it would. • E.g. plastics 1
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Dissociation
• Dissociation can occur therefore an equilibrium is established between solid and ions. • Example of solubility equilibrium equation: Example of solubility equilibrium equation:
• BaSO4 (s) ÅÆ Ba2+ (aq) + SO42‐ (aq)
Ksp
• BaSO4 (s) ÅÆ Ba2+ (aq) + SO42‐ (aq)
• The Keq expression depends only on the molar concentrations of the of species in solution (no solids). • For ionic compounds dissolving in water, the Keq is given a special name: "solubility product constant” or Ksp; it expresses the degree to which the solid is soluble in water.
Ksp example
• BaSO4 (s) ÅÆ Ba2+ (aq) + SO42‐ (aq)
• Ksp=[Ba2+][SO42−] • The value of Ksp for BaSO4 is 1.1 x 10‐10, a very small number, indicating that only a small amount of solid will dissolve in water. 2
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Ksp
• Ksp represents the level at which a solute dissolves in solution. • Ksp is proportional to the [ions] • LARGE Ksp
G
= more soluble (lots of ions) l bl (l
fi )
• SMALL Ksp = less soluble (fewer ions) Solids
• Solids are not included when calculating Keq, because their change in concentration are very small/insignificant, therefore omitted. Rules for Ksp calculations
1. Leave out the solid. (Include the aqueous ions only)
2. Change coefficients in the balanced equation to exponents in the Ksp expression. to exponents in the Ksp
expression
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Example
• Write the solubility equilibrium equation for lead (II) chloride dissolving. • PbCl2(s) ÅÆ Pb2+(aq) + 2Cl‐(aq) • Write the Ksp expression. • Ksp = [Pb2+] [Cl‐]2
Solubility Differentiation
• Solubility ‐ The quantity of substance that dissolves to form a saturated solution (g solute/L solution) • Molar Solubility –
Molar Solubility mol/L mol/L
• Ksp – equilibrium constant for the equilibrium between an ionic solid and its saturated solution. Factors affecting the Ksp
• Ksp can be affected by: – pH – Concentrations of (other) ions – Temperature Temperature
• Ksp has only one value for a given solute at a given temperature 4
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Example
• The solubility of PbS in water is 1.84 x 10‐14 mol/L
• Calculate the Ksp value.
Answer
•
•
•
•
•
PbS (s) ÅÆ Pb2+(aq)+S2‐(aq)
Ksp= [Pb2+][S2‐]
Ksp= [s][s]
Ksp= (1.84 x 10‐14 )(1.84 x 10‐14 )
Ksp= 3.38x 10‐28 Example
• The net‐ionic equation for the substance, AgCl
dissolving in water is: – AgCl(s) ÅÆAg+(aq) + Cl‐(aq)
‐10
• Ksp for AgCl
for AgCl = 1.8 x
1 8 x 10‐10 – Find the molar solubility.
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Answer
Ksp = [Ag+] [Cl‐]
Ksp= (s)2
1.8 x 10‐10 = (s)2
s= 1.34164 x 10‐5 M Where "s"
Where s is the moles/L of AgCl
is the moles/L of AgCl which which
dissolve • We can say, the molar solubility of AgCl is 1.34164 x 10‐5 moles/L •
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•
•
•
Try on your own!
• The solubility of Ag2CrO4 in water is 1.31 x 10‐4
moles/L. Calculate the value of Ksp. Answer
• Ag2CrO4 (s) ÅÆ2Ag+(aq) + CrO42‐(aq)
• The solubility of Ag2CrO4 in water is 1.31 x 10‐4
moles/L
• Once the solution is saturated (reaches O
h
l i i
d(
h
equilibrium): • [Ag+] = 2.62 x 10‐4 moles/L • [CrO42‐] = 1.31 x 10‐4 moles/L 6
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Answer
• The Ksp expression is:
Ksp = [Ag+]2 [CrO42‐] • Plugging in the concentrations directly we get: • Ksp = (2.62 x
(2 62 10
0‐44)2 (1.31 x
( 3 10
0‐44) )
‐12 • Ksp = 8.99 x 10
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