chapter 9 dissolution and precipitation equilibria

04/19/23 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


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 solute

Supersaturated Solution: a solution that temporarily contains more of a solute than the equilibrium quantity


AgF has two such changes

AgF·4H20

AgF·2H20

AgF

Endothermic – heat is added to a system

Exothermic – heat is removed from a system

Sharp changes in slope occur if water of crystallization is lost or gained by the solid that is in contact with the solution


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


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)


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.


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 =


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)


The Solubility of Ionic SolidsThe Solubility Product

TABLE 9-1contains Ksp values at 25C


The Solubility of SaltsSolubility and Ksp

Exercise 9-2

Determine 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



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]

MassMolar x molesmassMassMolar

massmoles

spK Find 4.

solubilitymolar in expressed

expression mequilibriu Find 3.

solubilitymolar Find 2.

solubility mass Given 1.



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]

[Ag+]2[SO42-] = Ksp

[y] = (8.3 g Ag2SO4 L-1) (1 mol Ag2SO4/311.8 g)

[y] = [2.66 10-2 ] mol Ag2SO4 L-1


Review

The Nature of Solubility Equilibria

Dissolution and precipitation are reverse of each other.

General reaction

X3Y2 (s) ↔ 3X+2 (aq) + 2Y-3 (aq)

Ksp =

Dissolution (Solubility)

[X+2]3 [Y-3]2

[s] [3s] [2s]

= (3s)3 (2s)2

s = molar solubility expressed in moles per liter


Review

The Nature of Solubility Equilibria

Dissolution and precipitation are reverse of each other.

General reaction

X3Y2 (s) ↔ 3X+2 (aq) + 2Y-3 (aq)

Ksp =

Precipation

[X+2]3 [Y-3]2

[Y-3][X+2]Mix [X+2] and [Y-3]

Does a ppt of X3Y2 form?

Reaction quotient before mixing occurs:

Q(init) = [X+2]3(init)[Y-3]2

(init) Q > K ?


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?


Precipitation and the Solubility Product

Precipitation from Solution

Exercise 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 TlNO3 is 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


[Tl+](init) = (0.0022 mol L-1)(555 mL/1000 mL)

= 0.0012 mol L-1

[IO3-](init) = (0.0022 mol L-1)(445 mL/1000 mL)

= 0.00098 mol L-1

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

= (0.0012)(0.00098) = 1.17 10-6

Because Q(init) < Ksp, solid TlIO3 does NOT precipitate!

Exercise 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 TlNO3 is mixed with 445 mL of a 0.0022 M solution of NaIO3. Does TlIO3 precipitate at equilibrium?

Q(init) ? Ksp = 1.17 10-6 < 3.1 10-6

Tl(IO3) (s) ↔Tl+(aq) + IO3-(aq)


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.


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 Assume s is small

s = [TlIO3]= 6.2 × 10-5 mol L-1 which is depressed 28 times relative to the 1.76 x 10-3 conc. without the common ion

TlIO3(s) ↔ Tl+(aq) + IO3-(aq)


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.


The Effects of pH on SolubilitySolubility of Hydroxides

Exercise 9-7

Estimate 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-] = 3y

y(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


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)


• 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

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