chapter 15
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Chapter 15. Applications of Aqueous Equilibria. Catalyst. Derive 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 - PowerPoint PPT PresentationTRANSCRIPT
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Chapter 15Applications of Aqueous Equilibria
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CatalystDerive the Henderson
Hasselbalch equation! DON’T LOOK AT YOUR NOTES
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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)
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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?
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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)
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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
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Buffered SolutionsWhat does it mean to have a
buffer?
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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!
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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)
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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!
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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:
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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.
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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.
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Example 15.4Calculate the pH of a solution
containing 0.75 M lactic acid and 0.25 M sodium lactate.
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Example 15.5A buffered solution contains 0.25
M ammonia and 0.40 M ammonium chloride. Calculate the pH of the solution
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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)
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SO HOW DO WE KNOW WHEN WE’VE MET THE EQUIVALENCE POINT IN A TITRATION?
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2 ways to tell equivalence point1. Use a pH meter2. Use an indicator that changes
color at the end point (equivalence point).
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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?
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When is the color visible for acidic solution?
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Choosing the Appropriate indicatorEx. 15.11)
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Use H-H equation to determine what pH will allow the indicator to change color!
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When is the color visible for BASIC solution?
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Figure 15.8!!!!All indicator ranges!
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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
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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.
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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.
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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.
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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.
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CatalystTurn in Prelab questionsAnswer the following:
◦What is the Kinetic Molecular Theory?
◦Write the Solubility Rules◦Write the strong acids◦Write the strong bases
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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
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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?
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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).
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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.
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Complete Qualitative Analysis on your own!This is how Selective Precipitation is used in the lab!
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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
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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