Chemistry 1011 Slot 51 Chemistry 1011 TOPIC Solubility Equilibrium TEXT REFERENCE Masterton and...

Chemistry 1011 Slot 5 1

Chemistry 1011

TOPICSolubility Equilibrium

TEXT REFERENCEMasterton and Hurley Chapter 16.1


16.1 Solubility Equilibrium

YOU ARE EXPECTED TO BE ABLE TO:

• Write an expression for the solubility product constant, Ksp, for a substance

• Calculate the concentration of ions at equilibrium, given Ksp

• Calculate the solubility product constant for a substance given its solubility and formula

• Predict whether a combination of ions will form a precipitate, given Ksp and ion concentrations

• Calculate the solubility of a substance in water, given Ksp

• Use Le Chatelier’s Principle to determine the effect of adding a common ion to a solution.

• Calculate the solubility of a substance in the presence of a common ion


Formation of Precipitates• A precipitate is formed when

– Two solutions are mixed, and– The cation from one solution combines with the anion

from the other solution to form an insoluble solid

NaCl(aq) + AgNO3(aq) NaNO3(aq) + AgCl(s)

Na+(aq) + Cl(aq) + Ag+

(aq) + NO3(aq) Na+

(aq) + NO3aq) + AgCl(s)

Ag+(aq) + Cl(aq) AgCl(s)

• An equilibrium is established between the solid and the corresponding ions in solution


Solubility Equilibrium

AgCl(s) Ag+(aq) + Cl(aq)

• An expression can be written for the equilibrium constant:

Ksp = [Ag+]x[Cl] • [solid] does not appear in the equilibrium

constant expression• Ksp is known as the solubility product constant• Solubility product data are normally measured at

25oC


Determining Ion ConcentrationsAg3PO4(s) 3Ag+

(aq) + PO43

(aq)

Ksp = [Ag+]3x[PO43] = 1 x 1016

PbCl2(s) Pb2+(aq) + 2Cl(aq)

Ksp = [Pb2+]x[Cl]2 = 1.7 x 105

Q: Calculate [Pb2+] and [Cl] in a solution of PbCl2 at 25oC

[Cl] = 2 x [Pb2+]

Ksp = [Pb2+] x [2Pb2+]2 = 1.7 x 105

Ksp = 4 [Pb2+]3 = 1.7 x 105

[Pb2+] = 1.6 x 102 mol/L [Cl] = 3.2 x 102 mol/L


Calculating Ksp

• The solubility of a salt can be determined by experiment• Ksp for the salt can be determined from these results

• Q: The solubility of magnesium hydroxide is found to be 8.4 x 104 g/100cm3 at 25oC. Find Ksp

Mg(OH)2(s) Mg2+(aq) + 2OH

(aq)

Ksp = [Mg2+]x[OH]2 = ??

Solubility = 8.4 x 104 g/100cm3 at 18oCSolubility = (8.4 x 104)g x 1000cm3/L = 1.44 x 104 mol/L 58.3 g/mol 100cm3

[Mg2+] = 1.44 x 104 mol/L; [OH] = 2.88 x 104 mol/L

Ksp = [Mg2+]x[OH]2 = 1.2 x 1011


Determining Precipitate Formation• To determine whether a precipitate will form

when two solutions are mixed:1. Determine the concentrations of the reacting ions in

the mixture

2. Calculate the ion product, P

3. Compare the ion product, P, with Ksp

4. If P > Ksp then precipitate will form

5. If P < Ksp then no precipitate

6. If P = then no precipitate – solution is saturated


Determining Precipitate FormationQ: Will a precipitate form when 5.0mL of 1.0 x 103 mol/L silver

nitrate is added to 5.0mL of 1.0 x 105 mol/L potassium chromate? Ksp Ag2CrO4 = 1.0 x 1012

2AgNO3(aq) + K2CrO4(aq) 2KNO3(aq) + Ag2CrO4(s)

2Ag+(aq) + CrO4

2(aq) Ag2CrO4(s)

Ksp = [Ag+]2 x [CrO42] = 1.0 x 1012

[Ag+] = 5.0 x 104 mol/L

[CrO42] = 5.0 x 106 mol/L

Ion Product, P = [Ag+]2 x [CrO42] = (5.0 x 104 )2 x (5.0 x 106 )

P = 1.25 x 1012

P > Ksp A precipitate will form


Determining Precipitate Formation

Q: 1.0 mL of 1.0 mol/L barium chloride is added to 10.0 mL of a solution containing a small amount of magnesium sulfate. Determine the minimum concentration of magnesium sulfate that will cause a precipitate to form.

Ksp BaSO4 = 1.1 x 1010


Determining Precipitate Formation• 1.0 mL of 1.0 mol/L barium chloride is added to 10.0 mL of a solution containing a small

amount of magnesium sulfate. Determine the minimum concentration of magnesium sulfate that will cause a precipitate to form. Ksp BaSO4 = 1.1 x 1010

BaCl2(aq) + MgSO4(aq) MgCl2(aq) + BaSO4(s)

Ba2+(aq) + SO4

2(aq) BaSO4(s)

Ksp = [Ba2+]x[SO42] = 1.1 x 1010

[Ba2+] = 1.0 x 103L x 1.0 mol/L 1.00 x 102L = 1.0 x 101 mol/L

[SO42] = x mol/L

Ksp = [Ba2+]x[SO42] = (1.0 x 101 ) x (x) = 1.1 x 1010

x = [SO42] = minimum [MgSO4] =1.1 x 1010 mol/L


Selective Precipitation• Suppose that a solution contains two different cations,

for example Ba2+ and Ca2+

• Each forms an insoluble sulfate, BaSO4 and CaSO4

Ksp BaSO4 = 1.1 x 1010

Ksp CaSO4 = 7.1 x 105

• If sulfate ions are added to a solution containing equal amounts of Ba2+ and Ca2+, then the BaSO4 will precipitate first

• Only when the Ba2+ ion concentration becomes very small will the SO4

2ion concentration rise to the point that CaSO4 will be precipitated


Determining Solubility• The solubility, s, of a salt can be determined from Ksp data

Q: Determine the solubility of lead chloride in water at 25oC. Ksp = 1.7 x 105

Let solubility of lead chloride = s mol/L• For every mole of PbCl2 that dissolves,

1 mole of Pb2+(aq) and 2 moles of Cl(aq) are formed

PbCl2(s) Pb2+(aq) + 2Cl(aq)

[Pb2+] = s mol/L[Cl] = 2 x [Pb2+] = 2s mol/L

Ksp = [Pb2+]x[Cl]2 = (s) x(2s)2 = 1.7 x 105

4s3 = 1.7 x 105

s = 1.6 x 102 mol/L (Can also be expressed in grams/Litre)


The Common Ion Effect

• The presence of a common ion will reduce the solubility of an ionic salt (Le Chatelier)

• If a common ion is added to a saturated solution of a salt, then the salt will be precipitated (Le Chatelier)

For example, CaCO3 is less soluble in a solution containing CO3

2 ions than in pure water

CaCO3(s) Ca2+(aq) + CO3

2(aq)

Ksp = [Ca2+] x [CO32] = 4.9 x 109


The Common Ion EffectCaCO3(s) Ca2+

(aq) + CO32

(aq)

Ksp = [Ca2+] x [CO32] = 4.9 x 109

• Solubility of CaCO3 = [Ca2+]

• In pure water

[Ca2+] = [CO32] = (4.9 x 109) = 7.0 x 105 mol/L

Solubility of CaCO3 in 1.0 x 101 sodium carbonate solution = ???

• [CO32] = 1.0 x 101 mol/L (ignore CO3

2 from CaCO3 )

• [Ca2+] = Ksp = 4.9 x 109 = 4.9 x 108 mol/L

[CO32] 1.0 x 101


One More Example• Ksp for manganese II hydroxide is 1.2 x 1011 • Solid sodium hydroxide is added slowly to a 0.10 mol/L

solution of manganese II chloride. What will be the pH when a precipitate forms?

Mn(OH)2(s) Mn2+(aq) + 2OH

(aq)

Ksp = [Mn2+] x [OH]2 = 1.2 x 1011

• [Mn2+] = 0.10 mol/L • [OH] = Ksp [Mn2+] = 1.2 x 1011 0.10 = 1.1 x 105 mol/L• pH = 9.0

Chemistry 1011 Slot 51 Chemistry 1011 TOPIC Solubility Equilibrium TEXT REFERENCE Masterton and...

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Transcript of Chemistry 1011 Slot 51 Chemistry 1011 TOPIC Solubility Equilibrium TEXT REFERENCE Masterton and...