Chapter 18: Chemical Equilibrium

Chapter 18: Chemical Equilibrium

Table of ContentsTable of Contents

18.3: Using equilibrium constants

Equilibrium Expressions and Constants

Chemical Equilibrium: Basic ConceptsChemical Equilibrium: Basic ConceptsChemical Equilibrium: Basic ConceptsChemical Equilibrium: Basic Concepts

• What is Keq?

• Why do we use Keq?

• Write Keq for the following reaction:

Chemical Equilibrium: Additional ConceptsChemical Equilibrium: Additional Concepts Chemical Equilibrium: Additional ConceptsChemical Equilibrium: Additional Concepts

Solubility equilibria• The solubility product constant (Ksp) is

used for dissolving a sparingly soluble ionic compound in water.

Equilibrium of a dissolved substance

Tells us if we have more solid or more ions


Solubility equilibria

When you know Ksp, you can…

1. calculate the molar solubility of a sparingly soluble ionic compound

2. calculate moles per liter of a saturated substance.

3. calculate ion concentrations in a saturated solution.

These all use the same process!

1. Calculating Molar Solubility

ex1)What is the molar solubility of copper II hydroxide?

1. Write the eq. expression

2. Look up Ksp 2.2 x 10-20 = [Cu2+][OH-]2

3. Replace Molarites with x 2.2 x 10-20 = [x][2x]2

4. Solve for x 2.2 x 10-20 = [x][2x]2 = x4x2 = 4x3

[x]3 = 5.5 x 10-21

[x] = 1.8 x 10-7 M

The solubility of Cu(OH)2 is = 1.8 x 10-7 mol/L

1. Calculating Molar Solubility

ex1)What is the molar solubility of copper II hydroxide?

The moles per liter of the saturated substance is 1.8x10-7 mol/L

Ion concentrations x = [Cu2+]; 2x = [OH-]

[OH-] = 2x = 2(1.8 x 10-7 M)

[OH-] = 3.6 x 10-7 M

WE ALSO KNOW…

[Cu2+] = 1.8x10-7 M

PracticePracticeEx2) How many moles per liter of silver chloride will

be in a saturated solution of AgCl? Ksp = 1.8x10-10

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

Ex3) Calculate the molar solubility of strontium chromate (SrCrO4)in water if Ksp = 3.7x10-5

1.3x10-5 mol/L

0.00608 mol/L

1.3e-05 mol/L


Predicting precipitates

• The ion product (Q)

• Ksp = Qsp at only at equilibrium

• Q can predict the shift in equilibrium

• Calculated the same way

• Ksp = initial concentrations

• Qsp = instantaneous concentrations


Predicting precipitates

• If Qsp < Ksp, shift to reactants, no precipitate forms

• If Qsp > Ksp, shift to products, precipitate will form

• If Qsp = Ksp, no change will occur

Q ex4) Predict whether a precipitate of PbCl2 will form if 1L of 0.01M NaCl is added to 1L of 0.2M Pb(NO3)2. Ksp = 1.7x10-5

How did they know which precipitate?Solubility rules!

Back of the periodic table

Q ex4) Predict whether a precipitate of PbCl2 will form if 1L of 0.01M NaCl is added to 1L of 0.2M Pb(NO3)2. Ksp = 1.7x10-5

Calculate Q

1)Q expression

2) Molarities

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

(aq)

Qsp = [Pb2+] [Cl-]2

Total volume

Q ex4) Predict whether a precipitate of PbCl2 will form if 1 L of 0.01 M NaCl is added to 1 L of 0.2 M Pb(NO3)2. Ksp = 1.7x10-5

Calculate Q

3) Solve!

4) Compare:

Qsp = [Pb2+] [Cl-]2

= [0.01M] [0.005M]2

= 2.5x10-7

Qsp (2.5x10-7) Ksp(1.7x10-5)So… shift to reactants

no precipitate will form

<

PracticePracticeEx5) Predict whether a precipitate of PbF2 will

form when equal volumes of 0.10M Pb(NO3)2 and 0.30M NaF are mixed. Ksp = 3.3x10-8

PbF2(s) Pb2+(aq) + 2F-

(aq)

Ex6) Predict whether a precipitate Ag2SO4 will form when equal volumes of .25 M K2SO4 and .01 AgNO3 are mixed. Ksp = 1.2x10-5

Shift to reactants, precipitate of PbF2

Shift to products, no precipitate


Common ion effect

• Common ion effect: The solubility of a substance is reduced when the substance is dissolved in a solution containing a common ion.

• For example, PbI2 is less soluble in an aqueous solution of NaI than in pure water.

• Because the common ion I– is already present in the NaI solution. It reduces the maximum possible concentration of Pb2+ and thus reduces the solubility of PbI2.

Chapter 18: Chemical Equilibrium

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