Chapter 15Applications of Aqueous Equilibria

CatalystDerive the Henderson

Hasselbalch equation! DON’T LOOK AT YOUR NOTES

Solutions of Acids or Bases Containing a Common Ion

We will talk about solutions that contain HA AND it’s conjugate base NaA

Suppose we have a solution of HF and NaF (remember salts fully dissociate)◦Step 1: Identify MAJOR SPECIESHF, Na+, F-, H2O (F is the common ion)

Solutions of Acids or Bases Containing a Common Ion

Let’s Compare 2 solutions:◦0.1 M HF solution ◦0.1 M HF solution + 0.1 M NaF

◦Step 1: Identify MAJOR SPECIES◦Step II: Write out the equations

How will LeChatelier’s Principle apply?

Solutions of Acids or Bases Containing a Common Ion

Let’s Compare 2 solutions:◦0.1 M HF solution ◦0.1 M HF solution + 0.1 M NaF

Common Ion Effect: The equilibrium position of HF will shift because the F- is already in solution!…so the pH with NaF will be higher! (less acidic)

Example ProblemThe equilibrium concentration of

H+ in a 1.0 M HF solution is 2.7 x 10-2 M and the % dissociation is 2.7%.

Calculate the [H+] and the % dissociation of HF in a solution containing 1.0 M HF (Ka = 7.2 x 10-4) and 1.0 M NaF

Buffered SolutionsWhat does it mean to have a

buffer?

Buffered SolutionsA buffered solution is one

that resists change in its pH when either OH- or H+ ions are added.

Example: Our blood – it can absorb acids and bases produced in our bodily reactions – but it must maintain a balanced pH to keep our cells alive!

Buffered SolutionsA buffered solution is one

that resists change in its pH when either OH- or H+ ions are added.

A buffered solution may contain a WEAK ACID and it’s SALT (HF and NaF) OR a WEAK BASE and it’s SALT (NH3 and NH4Cl)

Buffered SolutionsA buffered solution is one

that resists change in its pH when either OH- or H+ ions are added.

By choosing the correct components, a solution can resist change at almost any pH!

How does a buffered solution resist

changes in pH when an acid or a base is

added?

By solving the next set of example problems, our goal is to answer the question:

REMEMBER your SYSTEMATIC approach!

A buffered solution contains 0.5 M acetic acid (HC2H3O2, Ka = 1.8 x 10-5) and 0.5 M Sodium Acetate. Calculate the pH of this solution.

REMEMBER your SYSTEMATIC approach!

Calculate the change in pH that occurs when 0.010 mol solid NaOH is added to 1.0 L of the buffered solution described in the previous example.

Compare the pH change with that which occurs when 0.010 mol solid NaOH is added to 1.0 L of water.

Example 15.4Calculate the pH of a solution

containing 0.75 M lactic acid and 0.25 M sodium lactate.

Example 15.5A buffered solution contains 0.25

M ammonia and 0.40 M ammonium chloride. Calculate the pH of the solution

Adding strong acid to a buffered solutionCalculate the pH of the solution

that results when 0.10 mol gaseous HCl is aded to 1.0 L of a buffered solution that contains 0.25 M ammonia and 0.40 M ammonium chloride (15.5)

SO HOW DO WE KNOW WHEN WE’VE MET THE EQUIVALENCE POINT IN A TITRATION?

2 ways to tell equivalence point1. Use a pH meter2. Use an indicator that changes

color at the end point (equivalence point).

What is an indicator?It is a weak acid (HIn) that changes

color when the H+ leaves, leaving an (In- ion)

Let’s try a problem to see how they function…◦Assume you have some hypothetical

indicator HIn, Ka = 1.0 x 10-8

◦Let’s write the equation…◦Write Ka expression…◦What if we add this indicator to a solution

with a pH of 1.0?◦What color will it be?◦What if we add OH?...eventually…what?

When is the color visible for acidic solution?

Choosing the Appropriate indicatorEx. 15.11)

Use H-H equation to determine what pH will allow the indicator to change color!

When is the color visible for BASIC solution?

Figure 15.8!!!!All indicator ranges!

Solubility EquilibriaWhat does it mean to be soluble?If something is NOT soluble…

what will you see in the solution?Solubility product constant or

solubility product = KspTable 15.4

Ex 15.12) Calculating Ksp from Solubility I pg. 718Copper (I) Bromide has a

measured solubility of 2.0 x 10-4 mol/L at 25 °C. Calculate its Ksp value.

Ex 15.13) Calculating Ksp from Solubility II pg. 719Calculating Ksp value for bismuth

sulfide, which has a solubility of 1.0 x 10-15 mol/L at 25 °C.

Ex 15.14) Calculating Solubility from Ksp pg.720The Ksp for copper (II) iodate,

Cu(IO3)2, is 1.4 x 10-7 at 25 °C. Calculate its solubility at 25 °C.

Ex 15.15) Solubility and Common Ions pg. 723Calculate the solubility of solid

CaF2 (Ksp = 4.0 x 10-11) in a 0.025 M NaF solution.

CatalystTurn in Prelab questionsAnswer the following:

◦What is the Kinetic Molecular Theory?

◦Write the Solubility Rules◦Write the strong acids◦Write the strong bases

PrecipitationPrecipitation and Qualitative Analysis

What is Q?How do we calculate it again?For precipitation predictions:

◦Q < K : no precipitation◦Q > K : precipitation will occur

Ex 15.16) Determining Precipitation Conditions pg. 725A solution is prepared by adding

750.0 mL of 4.00 x 10-3 M Ce(NO3)3 to 300.00 mL of 2.00 x 10-2 M KIO3. Will Ce(IO3)3 (Ksp = 1.9 x 10-10) precipitated from this solution?

Ex 15.17A solution is prepared by mixing

150.0 mL of 1.00 x 10-2 M Mn(NO3)2 and 250.0 mL of 1.0 x 10-1 M NaF. Calculate the concentrations of Mg2+ and F- at equilibrium with solid MgF2

(Ksp = 6.4 x 10-9).

Selective PrecipitationEx. 15.18) A solution contains 1.0

x 10-4 M Cu+ and 2.0 x 10-3 M Pb2+. If a source of I- is added gradually to this solution, will PbI2 (Ksp = 1.4 x 10-8) or CuI (Ksp = 5.3 x 10-12) precipitate first? Specify the concentration of I- necessary to begin precipitation of each salt.

Complete Qualitative Analysis on your own!This is how Selective Precipitation is used in the lab!

Complex Ions form Coordination Complex’s

These are metals surrounded by ligands (Lewis Base)

Common Ligands:◦H2O, NH3, Cl-, CN-

Metal Ions add ligands one at a time…in a stepwise fashion:

Ag+ + NH3 Ag(NH3)+ K1 = 2.1 x 103

Ag(NH3)+ + NH3 Ag(NH3)2+

K2 = 8.2 x 103

Complex Ion EquilibriaEx 15.19) Complex IonsCalculate the concentrations of Ag+, Ag(S2O3)-, and Ag(S2O3)2

3- in a solution prepared by mixing 150.0 mL of 1.00 x 10-3 M AgNO3 with 200.0 mL of 5.00 M Na2S2O3. The stepwise formation equilibria are:Ag+ + S2O3

2- Ag(S2O3)-

K1 = 7.4 x 108

Ag(S2O3)- + S2O32- Ag(S2O3)2

3-

K2 = 3.9 x 104