E 12 Water and Soil

Solve problems relating to removal of heavy –metal ions

and phosphates by chemical precipitation

http://www.chem.purdue.edu/gchelp/howtosolveit/equilibrium/solubility_products.htm

Chemical Precipitation

• Many metal ions form insoluble / sparingly soluble salts

• Chemical precipitation of heavy metals and phosphates

Chemical Precipitation -1

• Use of Hydrogen sulfide• Removes sulfides of Hg, Cd, Pb,

Zn• Hg 2+ (aq) + H2S(aq)

HgS(s)• Pb 2+ (aq) + H2S(aq)

PbS(s)

Chemical Precipitation -2

• Use of Hydroxide to remove • copper• Cobalt• Iron• Cu 2+ (aq) + 2 OH- (aq) • Cu(OH)2 (s)

Solubility Rules

Quantitative Aspect of Solubility Equilibria

Solubility Products

Consider the equilibrium that exists in a saturated solution of BaSO4 in water:

BaSO4(s) Ba2+(aq) + SO42−(aq)

Solubility Products

The equilibrium constant expression for this equilibrium is

Ksp = [Ba2+] [SO42−]

= 1.0 x 10 -10 mol 2 dm -6 at 25○Cwhere the equilibrium constant, Ksp, is called the solubility product constant

Solubility Products• Ksp is not the same as solubility.• Solubility is generally expressed as the mass of solute dissolved in • 1 L (g/L) or • 100 mL (g/mL) of solution, • or in mol/L (M).• Temperature dependent

Solubility Product Constant

• Concentration of solid is not in the expression

• Solution is saturated• Precipitation depends on Ksp

Will a Precipitate Form?

• In a solution,– If Q = Ksp, the system is at equilibrium

and the solution is saturated.– If Q < Ksp, more solid will dissolve until Q = Ksp.

– If Q > Ksp, the salt will precipitate until Q = Ksp.

Solubility Product

Example 1

Ksp = [Ba2+] [SO42−]

= 1.0 x 10 -10 mol 2 dm -6 at 25○CX = 1.0 x 10 -5 mol dm -3

Factors Affecting Solubility

• The Common-Ion Effect– If one of the ions in a solution

equilibrium is already dissolved in the solution, the equilibrium will shift to the left and the solubility of the salt will decrease.

BaSO4(s) Ba2+(aq) + SO42−(aq)

Example 2

• Consider solubility of BaSO4 in 0.10 M sodium sulfate, strong electrolyte

• Calculate the con. Of Ba 2+ ions

BaSO4 Ba 2+ SO42-

Solubility of Barium sulfate = Y mol / dm3

Ksp = [Ba2+] [SO42−]

= 1.0 x 10 -10 mol 2 dm -6 at 25○C= 0.10

Application of Precipitation in Soil and Water Chemistry

• PPT of calcium ions as CaSO4

• by addition of sodium sulfate

Ksp = [Ca2+] [SO42−]

= 3.0 x 10 -5 mol 2 dm -6 at 25○C

Lead as Lead Chloride

• Pb 2+ (aq) + 2 Cl- (aq) • Pb(Cl)2 (s)

Ksp = [Pb2+] [Cl−]2

= 1.73 x 10 -7 mol 3 dm -9 at 25○C

Lead as Lead Sulfate

• Pb 2+ (aq) + SO42- (aq)

PbSO4 (s)

Ksp = [Pb2+] [SO42-]

= 6.3 x 10 -7 mol 2 dm -6 at 25○C

Calcium as calcium phosphate

• 3 Ca 2+ (aq) + 2 PO43- (aq)

Ca3(PO4)2 (s)

Ksp = [Ca2+] 3 [PO43-]2

= 6.3 x 10 -7 mol 2 dm -6 at 25○C

Phosphate as Aluminum or Iron or Calcium

• Fe 3+ (aq) + PO43- (aq)

FePO4 (s)

Arsonate, AsO4 3-

• Removal by addition of :• Aluminum nitrate• Iron(III)chloride• Iron(III)sulfate

• Al 3+ (aq) + AsO4 3- (aq) AlAsO4(s)

Examples 1

• Write the solubility product constant expression for the following compounds

• and indicate Ksp value in terms of X if X is its molar solubility

Example

• A) AgCl• B) PbCl2• C) As2S3

• D) Ca3(PO4)2

• E) Fe(OH)3

Silver chloride

• AgCl (s) Ag + (aq) + Cl- (aq)

x x

Ksp = [ Ag +][Cl-]

= x 2

Calcium Phosphate

• Ca3(PO4)2 (s)

• 3 Ca 2+ (aq) + 2 PO43- (aq)

• 3x 2x• Ksp = [Ca 2+] 3 [PO4

3- ] 2

• (3x)3 (2x)2

= 108 x 5

Example 2

• A) Calcium ions present in hard water are ppted by adding sulfate ions.

• Write the net ionic equation for the reaction

2

• Ca 2+(aq) + SO4

2-(aq) CaSO4 (s)

• B) Given the Ksp, = 3.0 x 10 -5 mol 2 dm-6 at 25 deg. C; calculate its molar solubility in water at 25 deg. C

Ex B

• CaSO4 (s) Ca 2+(aq) + SO4

2-(aq)

• Ksp = [Ca 2+ ][SO42- ]

• x x• = x2 = 3.0 x 10 -5 mol 2 dm-6 • x = 5.5 x 10 -3 mol dm-3

Ex C

• Determine if a ppt will form when its ion concentrations are

• [Ca 2+ ] = 1.0 x 10 -3 mol dm-3 • [SO4

2-] = 1.0 x 10 -2 mol dm-3

• B) Given the Ksp, = 3.0 x 10 -5 mol 2 dm-6 at 25 deg. C;

Ex 1 C

• Ksp = [Ca 2+] [SO42- ]

• (x) (x)

= x 2 = (1.0 x 10 -3 mol dm-3 ) (1.0 x 10 -2 mol dm-3)= 1.0 x 10 -5 < KspPpt will not form

Ex1D

• Calculate minimum ion of sulfate required to ppt.

• [Ca 2+ ] = 0.001 M• [Sulfate] = x = ksp• Calculate for X

Ex3

Ex4

Ex5

Applications of Ksp

4. A 200.0 mL sample of a saturated solution of Mg(OH)2 weighs 222.1210 g. When the beaker containing the solution is evaporated to dryness it weighs 22.1213 g. The mass of the empty beaker is 22.1200 g. Calculate the Ksp.

Mass of Beaker + Mg(OH)2 22.1213 g- Mass of Beaker - 22.1200 g

Mass of Mg(OH)2 0.0013 g note sig figs

s = 0.0013 g x 1 mole58.3 g

0.2000 L

= 1.1149 x 10-4 M

Mg(OH)2 ⇌ Mg2+ + 2OH-

s s 2sKsp = [Mg2+][OH-]2 = [s][2s]2 = 4s3

= 4(1.1149 x 10-4)3 = 5.5 x 10-12

Ksp

Solubility ProductSpecial KeqSaturated solutionsNo UnitsIncreasing Temperature increases the Ksp

4. A 200.0 mL sample of a saturated solution of Mg(OH)2 weighs 222.1210 g. When the beaker containing the solution is evaporated to dryness it weighs 22.1213 g. The mass of the empty beaker is 22.1200 g. Calculate the Ksp.

Mass of Beaker + Mg(OH)2 22.1213 g- Mass of Beaker - 22.1200 g

Mass of Mg(OH)2 0.0013 g note sig figs

s = 0.0013 g x 1 mole58.3 g

0.2000 L

= 1.1149 x 10-4 M

Mg(OH)2 ⇌ Mg2+ + 2OH-

s s 2sKsp = [Mg2+][OH-]2 = [s][2s]2 = 4s3

= 4(1.1149 x 10-4)3 = 5.5 x 10-12

The Molar Solubility is the molarity required to saturate of fill the solution at any given temperature.

1. The solubility (s) of BaCO3 is 5.1 x 10-5 M @ 250 C. Calculate the solubility product or Ksp.

BaCO3(s) ⇌ Ba2+ + CO32-

s s s

Ksp = [Ba2+][CO32-]

Ksp = [s][s]

Ksp = s2

Ksp = (5.1 x 10-5)2

Ba2+ CO32-

BaCO3(s)

The Molar Solubility is the molarity required to saturate of fill the solution at any given temperature.

1. The solubility (s) of BaCO3 is 5.1 x 10-5 M @ 250 C. Calculate the solubility product or Ksp.

BaCO3(s) ⇌ Ba2+ + CO32-

s s s

Ksp = [Ba2+][CO32-]

Ksp = [s][s]

Ksp = s2

Ksp = (5.1 x 10-5)2

Ksp = 2.6 x 10-9

Ba2+ CO32-

BaCO3(s)

3. If 0.00243 g of Fe2(CO3)3 is required to saturate 100.0 mL of solution. What is the solubility product?

Fe2(CO3)3 ⇌ 2Fe3+ + 3CO32-

s 2s 3s

s = 0.00243 g x 1 mole 291.6 g0.100 L

= 8.333 x 10-5 M

Ksp = [Fe3+]2[CO32-]3

Ksp = [2s]2[3s]3

Ksp = 108s5

Ksp = 108(8.333 x 10-5)5

Ksp = 4.34 x 10-19

Factors Affecting Solubility• pH

– If a substance has a basic anion, it will be more soluble in an acidic solution.

– Substances with acidic cations are more soluble in basic solutions.

Factors Affecting Solubility

• Complex Ions– Metal ions can act as Lewis acids and

form complex ions with Lewis bases in the solvent.

Factors Affecting Solubility

• Complex Ions– The

formation of these complex ions increases the solubility of these salts.

Factors Affecting Solubility

• Amphoterism– Amphoteric metal

oxides and hydroxides are soluble in strong acid or base, because they can act either as acids or bases.

– Examples of such cations are Al3+, Zn2+, and Sn2+.

Amphoteric–Amphoteric metal oxides and

hydroxides are soluble in strong acid or base, because they can act either as acids or bases.

–Examples of such cations are Al3+, Zn2+, and Sn2+.

Selective Precipitation of Ions

Precipitation

One can use differences in solubility's of salts to separate ions in a mixture.