ofb chapter 9 lecture notes -...

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6/17/2003 OFB Chapter 9 1 Chapter 9 Dissolution and Precipitation Equilibria 9-1 The Nature of Solubility Equilibria 9-2 The Solubility of Ionic Solids 9-3 Precipitation and the Solubility Product 9-4 The Effects of pH on Solubility 9-5 Complex Ions and Solubility 9-6 Controlling Solubility in Qualitative Analysis

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6/17/2003 OFB Chapter 9 1

Chapter 9Dissolution and Precipitation

Equilibria

9-1 The Nature of Solubility Equilibria

9-2 The Solubility of Ionic Solids9-3 Precipitation and the

Solubility Product9-4 The Effects of pH on

Solubility9-5 Complex Ions and Solubility9-6 Controlling Solubility in

Qualitative Analysis

6/17/2003 OFB Chapter 9 2

The Nature of Solubility Equilibria

Recrystallization:

Saturated solution:

Dissolution and precipitation are reverse of each other.

A solution begins to deposit a compound when it is brought to the point of saturation with respect to that compound.

A solution in which a dissolution-precipitation (solubility) equilibrium exists between the solid substance and its dissolved form.

Solvate: In dissolution, the solute-solvent attractions I2(CCl4) replace the solute-solute interactions I2(s)formerly present in the solid, and the molecules go off into the bulk of the solvent.

I2(s) ↔I2(CCl4)

6/17/2003 OFB Chapter 9 3

Saturated Solution: a solution in equilibrium with excess solute

e.g., NaCl solubility in grams per 100 grams water is approximately 36.0 grams = saturated solution

Unsaturated Solution: contains less than the equilibrium concentration of the soluteSupersaturated Solution: a solution that temporarily contains more of a solute than the equilibrium quantity

6/17/2003 OFB Chapter 9 4

AgF has two such changes

AgF·4H20

AgF·2H20

AgF

6/17/2003 OFB Chapter 9 5

The Nature of Solubility Equilibria

[I2]CCl4= K

The molarity of a saturated solution is a constant.

Le Chatelier’s principle:

I2(s) ↔I2(CCl4)

6/17/2003 OFB Chapter 9 6

9-2 Solubility of Salts

This chapter considers only salts which are sparingly soluble or insoluble for which concentrations of saturated salts are [salt] = 0.1 Mol L-1

or less

6/17/2003 OFB Chapter 9 7

Solubility Product Ksp

Describes a chemical equilibrium in which an excess solid salt is in equilibrium with a saturated aqueous solution of its separated ions.

General equation

AB (s) ↔ A+ (aq) + B- (aq)

6/17/2003 OFB Chapter 9 8

The Solubility of Ionic Solids

The Solubility Product

AgCl(s) ↔Ag+ (aq) + Cl-(aq)

= 1.6 × 10-10 at 25oC

=Ksp =

Ksp

The solid AgCl, which is in excess, is understood to have a concentration of

1 mole per liter.

6/17/2003 OFB Chapter 9 9

The Solubility of Ionic Solids

The Solubility Product

Ag2SO4(s) ↔2Ag+(aq) + SO42-(aq)

Ksp =

Fe(OH)3(s) ↔Fe+3(aq) + 3OH-1(aq)

Ksp =

6/17/2003 OFB Chapter 9 10

The Solubility of Ionic SolidsThe Solubility Product

Exercise 9-1

Write the Ksp equation for the dissolution of aluminum hydroxide (Al(OH)3) in water.

Al(OH)3(s) ↔Al3+(aq) + 3 OH-(aq)

6/17/2003 OFB Chapter 9 11

The Solubility of Ionic SolidsThe Solubility Product

TABLE 9-1contains Ksp values at 25C

6/17/2003 OFB Chapter 9 12

6/17/2003 OFB Chapter 9 13

6/17/2003 OFB Chapter 9 14

6/17/2003 OFB Chapter 9 15

The Solubility of SaltsSolubility and Ksp

Exercise 9-2Determine the mass of lead(II) iodate dissolved in 2.50 L of a saturated aqueous solution of Pb(IO3)2 at 25oC. The Ksp of Pb(IO3)2 is 2.6 × 10-13.

Gram solubility of

Lead (II) iodate

Pb(IO3)2(s) ↔Pb2+(aq) + 2 IO3-(aq)

[y] [y] [2y][Pb2+][IO3

-]2 = Ksp

y = 4.0 × 10-5∴ [Pb(IO3)2] = [Pb2+] = y = 4.0 × 10-5 mol L-1

∴ [IO3-] = 2y = 8.0 × 10-5 mol L-1

= (4.0 × 10-5 mol L-1) × (557 g mol-1)

= 0.0223 g L-1 × 2.50 L

Pb=207.2

I=126.9 O=16

Pb(IO3)2 = 557g per mole

6/17/2003 OFB Chapter 9 16

The Solubility of SaltsSolubility and Ksp

Exercise 9-3

Compute the Ksp of silver sulfate (Ag2SO4) at 25oC if its mass solubility is 8.3 g L-1.

1 Ag2SO4(s) ↔2 Ag+(aq) + 1 SO42-(aq)

[y] [2y] [y]

6/17/2003 OFB Chapter 9 17

A common problem is to calculate if a precipitate will form at equilibrium

(Chapter 7) Reaction quotient before mixing occurs:

Q(init) = [A+](init)[B-](init)

If Q(init) < Ksp, no solid AB can appear.

If Q(init) > Ksp, solid AB precipitates until

Q = Ksp

AB (s) ↔ A+ + B-

6/17/2003 OFB Chapter 9 18

AgCl(s) ↔Ag+ (aq) + Cl-(aq)

Ksp = [Ag+][Cl-]If Q > Ksp then the solid precipitates

Q (init) = [Ag+] (init) [Cl-] (init)= Reaction quotient

Precipitation from Solution: Does a solid ppt form?

6/17/2003 OFB Chapter 9 19

Precipitation and the Solubility Product

Precipitation from SolutionExercise 9-4:The Ksp of thallium (I) iodate is 3.1 × 10-6 at 25oC. Suppose that 555 mL of a 0.0022 M solution of TlNO3is mixed with 445 mL of a 0.0022 M solution of NaIO3. Does TlIO3 precipitate at equilibrium?

Evaluate : Reaction quotient before mixing occurs:

Q(init) = [Tl+](init)[IO3-](init)

If Q(init) < Ksp, no solid TlIO3 can appear.

If Q(init) > Ksp, solid TlIO3 precipitates until Q = Ksp

[Tl+]

[IO3-]

Q > Ksp

Solid ppt

Q < Ksp

No ppt

6/17/2003 OFB Chapter 9 20

Exercise 9-4The Ksp of thallium(I) iodate is 3.1 × 10-6 at 25oC. Suppose that 555 mL of a 0.0022 M solution of TlNO3is mixed with 445 mL of a 0.0022 M solution of NaIO3. Does TlIO3 precipitate at equilibrium?

6/17/2003 OFB Chapter 9 21

Precipitation and the Solubility Product

The Common Ion Effect

If a solution and a solid salt to be dissolved in it have an ion in common, then the solubility of the salt is depressed.

6/17/2003 OFB Chapter 9 22

The Common Ion Effect

Exercise 9-6

The Ksp of thallium(I) iodate (TlO3) is 3.1 × 10-6 at 25oC. Determine the molar solubility of TlIO3 in 0.050 mol L-1 KIO3 at 25oC.

[Tl+] (mol L-1) [IO3-] (mol L-1)

Initial concentration

Equilibrium concentration

Change in concentration

[Tl+][IO3-] = Ksp

6/17/2003 OFB Chapter 9 23

The Effects of pH on Solubility

Solubility of Hydroxides

Zn(OH)2(s) ↔Zn2+(aq) + 2 OH-(aq)

[Zn2+][OH-]2 = Ksp = 4.5 × 10-17

Many solids dissolve more readily in more acidic solutions

If pH decreases (or made more acidic), the [OH-] decreases. In order to maintain Ksp the [Zn2+] must increase and consequently more solid Zn(OH)2 dissolves.

6/17/2003 OFB Chapter 9 24

The Effects of pH on SolubilitySolubility of Hydroxides

Exercise 9-7Estimate the molar solubility of Fe(OH)3 in a solution that is buffered to a pH of 2.9.

In pure water:

[OH-] = 3y = 1.3 × 10-9 mol L-1

pOH = 8.87 (and pH = 5.13)

[Fe3+] = y [OH-] = 3yy(3y)3 = 27y4 = Ksp = 1.1 × 10-36

y = 4.5 × 10-10 mol L-1 = [Fe3+] = [Fe(OH)3]=

In pure water, Fe(OH)3 is 5 x 10 6 less soluble than at pH = 2.9

6/17/2003 OFB Chapter 9 25

9-7 The Effects of pH on Solubility

• Solubilities of Hydroxides• Solubility of Salts and Weak Bases• Selective Precipitation of Ions• Metal Sulfides

But as before solubility of Metal Sulfides increase as pH decreases

Ksp = [M2+][OH-][HS-]As pH decreases (or made more acidic), the [OH-] decreases. In order to maintain Ksp the [M2+] must increase and consequently more solid Metal Sulfide dissolves.

Somewhat more complicated due to other competing reactions. E.g.,

MS + H2O ↔ M2+ + OH- + HS-

(Metal Sulfide)

6/17/2003 OFB Chapter 9 26

• Examples / Exercises– 9-1, Ksp calculations– 9-2, Ksp calculations– 9-3, Ksp calculations– 9-4, ppt Q ? Ksp– 9-5, Equilibrium concentrations– 9-6, Common Ion effect– 9-7 Effect of pH of on solubility

• Problems– 16, 19, 23, 30, 39, 41, 42