Chapter 18

Chapter 18

The Solubility Product Constant

Write the net ionic equation

• Solutions of sodium chromate and barium chloride are mixed.

Ba2+ + CrO42- → BaCrO4

Ba2+ + CrO42- ↔ BaCrO4

BaCrO4 ↔ Ba2+ + CrO42-

• Since the reaction is reversible when can flip it.• Write the equilibrium expression for this reaction.


K = [Ba2+] [CrO42-]


Ksp = [Ba2+] [CrO42-]

• This is the solubility product expression• It is used to determine the concentrations of

ions in a saturated solution of a slightly soluble salt and determine whether a precipitate will form within a solution.

Write the solubility product expression for AgCl

Ksp = [Ag+] [Cl-]

Write the solubility product expression for MgF2

Ksp = [Mg2+] [F-]2

Write the solubility product expression for Ag3PO4

Ksp = [Ag+]3 [PO43-]

Molar Solubility

• Molar solubility is the maximum solubility of a substance expressed in moles per liter.

• Example: The molar solubility of silver chromate (Ag2CrO4) is 1.3 x 10-4 mol/L

The molar solubility of Hg2Cl2 is 6.5 x 10-7 mol/L. Find the solubility product.

Ksp = 1.1 x 10-18

The molar solubility of Ag2CrO4 is 1.3 x 10-4 mol/L.

Find the solubility product.

Ksp = 8.8 x 10-12

The solubility of PbCrO4 is 4.30 x 10-5 g/L. Find the solubility product.

What’s the difference here?


Ksp = 1.8 x 10-14


Ksp = 1.8 x 10-14

See Appendix D (Pages A6 & A7)

What is the molar solubility PbCrO4?Ksp of PbCrO4 = 1.8 x 10-14

Molar solubility = 1.3 x 10-7 mol PbCrO4/L[Pb2+] = [CrO4

2-] = 1.3 x 10-7 M

What are the ion concentrations?

What is the molar solubility Ag2CrO4?Ksp of Ag2CrO4 = 8.8 x 10-12

Molar solubility = 1.3 x 10-4 mol Ag2CrO4/L[Ag+] = 2.6 x 10-4 M [CrO4

2-] = 1.3 x 10-4 M


What is the molar solubility Ag3PO4?Ksp of Ag3PO4 = 1.8 x 10-18

Molar solubility = 1.6 x 10-5 mol Ag3PO4/L[Ag+] = 4.8 x 10-5 M [PO4

3-] = 1.6 x 10-5 M


What is the molar solubility Pb3(PO4)2?Ksp of Pb3(PO4)2 = 3.0 x 10-44

Molar solubility = 7.7 x 10-10 mol Pb3(PO4)/L[Pb2+] = 2.3 x 10-9M [PO4

3-] = 1.5 x 10-9 M


I place a large scoopful of Ag3PO4 (a white crystalline solid) in a beaker of water and stir. I come back tomorrow.

What do I see in the beaker?

Did any of the Ag3PO4 dissolve? What are the concentrations in the solution?

What is the molar solubility Ag3PO4?Ksp of Ag3PO4 = 1.8 x 10-18

Molar solubility = 1.6 x 10-5 mol Ag3PO4/L [Ag+] = 4.8 x 10-5 mol Ag+/L [PO4

3-] = 1.6 x 10-5 mol PO43-/L


Did any of the Ag3PO4 dissolve? If any does dissolve what are the concentrations in the solution?

[Ag+] = 4.8 x 10-5M [PO4

3-] = 1.6 x 10-5M

How would you describe this solution?

Saturated Solutions and Solubility

Crystallization: opposite of the solution process (solute becomes reattached to surface of crystal)

Saturated solution: no more solid will dissolve and a dynamic equilibrium exists between the solution and the undissolved solid.

Solute + Solvent Solution

ecrystalliz

dissolve

What would we observe if we added sodium phosphate solution to a beaker saturated

solution of Ag3PO4?

What would we observe if we added sodium phosphate solution to the beaker containing a

saturated solution of Ag3PO4?

Ksp of Ag3PO4 = 1.8 x 10-18

[Ag+] = 4.8 x 10-5 mol Ag+/L[PO4

3-] = 1.6 x 10-5 mol PO43-/L

[Ag+]3[PO43-]

= [4.8 x 10-5]3 [1.6 x 10-5] = ?

Why does Ag3PO4 precipitate?

Reaction Quotient

If Q = Ksp, the system is at equilibrium and the solution is saturated

If Q < Ksp, the solution is not saturated precipitate does not form

If Q > Ksp, the solution exceeds saturation and a precipitate forms

Q vs Ksp

Does AgCl precipitate when equal volumes of 1.5 x 10-5M solutions of AgNO3 and NaCl mix?

??


NO

200.0ml of 0.0015M MgCl2 is mixed with 100.0ml of 0.0015M NaOH. Does Mg(OH)2 precipitate?

??


Yes

AgNO3 solution is added to a solution of Na2CrO4. What precipitate forms?

AgNO3 is added to a 0.0010M solution of Na2CrO4. What is the [Ag+] when the Ag2CrO4 precipitate begins to form?

Clothing washed in water that has a manganese concentration exceeding 0.1 mg L-1 (1.8 x 10-6 M) may be stained with manganese. A laundry wishes to add a base to precipitate manganese as the hydroxide Mn(OH)2 (Ksp = 4.5 x 10-14). At what pH is [Mn2+] equal to 1.8 x 10-6 M?

Clothing washed in water that has a manganese concentration exceeding 0.1 mg L-1 (1.8 x 10-6 M) may be

stained with manganese. A laundry wishes to add a base to precipitate manganese as the hydroxide Mn(OH)2 (Ksp = 4.5

x 10-14). At what pH is [Mn2+] equal to 1.8 x 10-6 M?

Fractional Precipitation

• The process by which two aqueous substances in a solution are separated through the addition of a common ion, taking advantage of their different concentration needs (Ksp values) in order to form a precipitate.

• The ion with the ________ Ksp will precipitate first.

smaller

A solution contains 0.010 mol of KI and 0.10 mol of KCl per liter. AgNO3 is gradually added to this solution.

Which precipitate forms first, AgCl or AgI?

• This is a multiple equilibria problem.

• Strategy: If there is more than one equilibrium involved write both and solve for what you can.

A solution contains 0.010 mol of KI and 0.10 mol of KCl per liter. AgNO3 is gradually added to this solution.

Which precipitate forms first, AgCl or AgI?

A solution contains 0.010 mol of KI and 0.10 mol of KCl per liter. AgNO3 is gradually added to this solution. What is the [I ̄ ] in the solution when AgCl starts to precipitate?

A solution contains 0.010 mol of KI and 0.10 mol of KCl per liter. AgNO3 is gradually added to this solution. What percent of the

original [I ̄ ] remains in solution when AgCl starts to precipitate?

NH4Cl is added to a 750mL solution that is 0.10M in Mg2+ and 0.10M in NH3. The solution also contains a Mg(OH)2 precipitate. Calculate the [NH4

+] that is required to dissolve the precipitate.

NH4Cl is added to a 750mL solution that is 0.10M in Mg2+ and 0.10M in NH3. The solution also contains a Mg(OH)2 precipitate.

How many grams of NH4Cl were added?

We can increase or decrease the solubility of “insoluble” (slightly soluble) substances by

applying LeChatlier’s Principle.

• The Common Ion Effect• pH

The Common-Ion Effect• A salt is less soluble in a solution that has

an ion in common with the salt.• Calcium Sulfate (CaSO4) is “insoluble”

(slightly soluble) in water. • Write the equation that represents the

solubility of calcium sulfate in water.CaSO4(s) ⇌ Ca2+(aq) + SO4

2−(aq)

The Common-Ion Effect

CaSO4(s)⇌Ca2+(aq)+SO42−(aq)

• Describe the equilibrium system of Calcium Sulfate in water.

•The solution is saturated with most of the calcium sulfate in solid form.

•The [SO42−] in the solution is very

low and is equal to the [Ca2+].

The Common-Ion Effect

CaSO4(s)⇌Ca2+(aq)+SO42−(aq)

• Describe what would happen if a solution of sodium sulfate were added to the beaker.

•The [SO42−] would increase

causing the reaction to shift to the left reducing the solubility of CaSO4 causing the it CaSO4 to precipitate out of the solution.

pH

• The pH can affect the solubility of a solute in two ways: – through the common ion effect.

Al(OH)3 ↔ Al3+ + 3OH ̄

• Is Al(OH)3 more soluble at a lower pH, higher pH, or does pH not affect the solubility of Al(OH)3 ?

Confused?• What happens to this equilibrium if a strong

base such as NaOH is added?• What happens to this equilibrium if a strong

acid such as HCl is added?

pH

• The pH can affect the solubility of a solute in two ways: – through the common ion effect.

or– when a salt contains a basic anion such as F–,

CH3CO2–, or CN– it is normally more soluble at

low pH’s.– When a salt contains an acidic ion such as Ag+,

Zn2+, or Al3+ it is often normally more soluble at high pH’s.

FeS ↔ Fe2+ + S2-

• What happens to this equilibrium if a strong acid such as HCl is added?

• A strong acid such as HCl will often dissolve an insoluble salt by lowering the concentration of the anion from the solution and forming a weak acid.

AgCl ↔ Ag+ + Cl ̄

• What happens to this equilibrium if ammonia or a strong base is added?

Adding Strong Bases or Ammonia to Insoluble Salts

• A strong base or ammonia will often dissolve an insoluble salt by lowering the concentration of the cation from the solution and forming a complex ion.

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