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 • Suppose we have a solution of HF and NaF (remember salts fully dissociate) • Step 1: Identify MAJOR SPECIES • HF, 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 Problem • The 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 Solutions • What does it mean to have a buffer?
Buffered Solutions • A 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 Solutions • A buffered solution is one that resists change in its pH when either OH- or H+ ions are added. • A buffered solution may contain aWEAK ACID and it’s SALT (HF and NaF) OR a WEAK BASE and it’s SALT (NH3 and NH4Cl)
Buffered Solutions • A 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!
By solving the next set of example problems, our goal is to answer the question: How does a buffered solution resist changes in pH when an acid or a base is added?
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 15.4 • Calculate the pH of a solution containing 0.75 M lactic acid and 0.25 M sodium lactate.
Example 15.5 • A 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 solution • Calculate 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)
2 ways to tell equivalence point • Use a pH meter • 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?
Choosing the Appropriate indicator • Ex. 15.11)
Use H-H equation to determine what pH will allow the indicator to change color!
Figure 15.8!!!! All indicator ranges!
Solubility Equilibria • What does it mean to be soluble? • If something is NOT soluble…what will you see in the solution? • Solubility product constant or solubility product = Ksp • Table 15.4
Ex 15.12) Calculating Kspfrom Solubility I pg. 718 • Copper (I) Bromide has a measured solubility of 2.0 x 10-4mol/L at 25 °C. Calculate its Ksp value.
Ex 15.13) Calculating Kspfrom Solubility II pg. 719 • Calculating Kspvalue for bismuth sulfide, which has a solubility of 1.0 x 10-15mol/L at 25 °C.
Ex 15.14) Calculating Solubility from Ksp pg.720 • The 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. 723 • Calculate the solubility of solid CaF2 (Ksp = 4.0 x 10-11) in a 0.025 M NaF solution.
Catalyst • Turn in Prelab questions • Answer the following: • What is the Kinetic Molecular Theory? • Write the Solubility Rules • Write the strong acids • Write the strong bases
Precipitation • Precipitation 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. 725 • A solution is prepared by adding 750.0 mL of 4.00 x 10-3M 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 15.17 • A 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 Precipitation • Ex. 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+ + NH3Ag(NH3)+ K1 = 2.1 x 103 Ag(NH3)++ NH3 Ag(NH3)2+ K2= 8.2 x 103
Complex Ion Equilibria Ex 15.19) Complex Ions Calculate the concentrations of Ag+, Ag(S2O3)-, and Ag(S2O3)23- 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+ + S2O32- Ag(S2O3)- K1= 7.4 x 108 • Ag(S2O3)- + S2O32- Ag(S2O3)23- K2= 3.9 x 104