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Explore the applications of aqueous equilibria in strong acid/strong base titration, weak acid/strong base titration, and weak base reactions with water. Learn about acid-base properties of salts, buffered solutions, and the Henderson-Hasselbalch equation.
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Strong Acid/Strong Base Titration Endpoint is at pH 7 A solution that is 0.10 M HCl is titrated with 0.10 M NaOH
Strong Acid/Strong Base Titration A solution that is 0.10 M NaOH is titrated with 0.10 M HCl Endpoint is at pH 7 It is important to recognize that titration curves are not always increasing from left to right.
Weak Acid/Strong Base Titration A solution that is 0.10 M CH3COOH is titrated with 0.10 M NaOH Endpoint is above pH 7
Strong Acid/Weak Base Titration A solution that is 0.10 M HCl is titrated with 0.10 M NH3 Endpoint is below pH 7
Reaction of Weak Bases with Water The base reacts with water, producing its conjugate acid and hydroxide ion: CH3NH2 + H2O CH3NH3+ + OH- Kb = 4.38 x 10-4
Kb for Some Common Weak Bases Many students struggle with identifying weak bases and their conjugate acids.What patterns do you see that may help you?
Reaction of Weak Bases with Water The generic reaction for a base reacting with water, producing its conjugate acid and hydroxide ion: B + H2O BH+ + OH- (Yes, all weak bases do this – DO NOT make this complicated!)
What is [H+] of a 0.5 M HF solution? (Ka=7.2x10-4) HF -> H+ + F- I C E
What is [H+] of a 0.5 M solution of NaF? (Ka=7.2x10-4) F- + H2O HF +OH- I C E
Acid Base Properties of Salts • Sometimes a salt such as NaF can have acid base properties.
Common Ion Effect • What if you have two solutions mixed together and they are both found in the Kb expression? (Common Ion Effect)
What is [H+] of a 0.5 M HF solution mixed with a 0.5 M solution of NaF? (Ka=7.2x10-4) HF -> H+ + F- I C E
Buffered Solutions • A solution that resists a change in pH when either hydroxide ionsorprotons are added. • Buffered solutions contain either: • A weak acid and its salt • A weak base and its salt
Acid/Salt Buffering Pairs The salt will contain the anion of the acid, and the cation of a strong base (NaOH, KOH)
Base/Salt Buffering Pairs The salt will contain the cation of the base, and the anion of a strong acid (HCl, HNO3)
Titration of an Unbuffered Solution A solution that is 0.10 M CH3COOH and 0.10 M NaCH3COO is titrated with 0.10 M NaOH A solution that is 0.10 M CH3COOH is titrated with 0.10 M NaOH
Titration of a Buffered Solution A solution that is 0.10 M CH3COOH and 0.10 M NaCH3COO is titrated with 0.10 M NaOH Buffered Unbuffered
Comparing Results Unbuffered Buffered • In what ways are the graphs different? • In what ways are the graphs similar?
Comparing Results Buffered Unbuffered
Adding HCl to a buffer solution • Suppose that 0.250 liters of a buffer solution that contains 0.225 M acetic acid and 0.225 M sodium acetate. What would be the pH change if 30.0 mL of 0.100 M HCl is added to this buffer? Assume volumes are additive. Ka for acetic acid is 1.8 x 10 -5. 4.74 to 4.70
Solving Solubility Problems For the salt AgI at 25C, Ksp = 1.5 x 10-16 What is the solubility of AgI? AgI(s) Ag+(aq) + I-(aq) O O +x +x x x 1.5 x 10-16 = x2 x = solubility of AgI in mol/L = 1.2 x 10-8 M
Solving Solubility Problems For the salt PbCl2 at 25C, Ksp = 1.6 x 10-5 PbCl2(s) Pb2+(aq) + 2Cl-(aq) O O +2x +x 2x x 1.6 x 10-5 = (x)(2x)2 = 4x3 x = solubility of PbCl2 in mol/L = 1.6 x 10-2 M
Solving Solubility with a Common Ion For the salt AgI at 25C, Ksp = 1.5 x 10-16 What is its solubility in 0.050 M NaI? AgI(s) Ag+(aq) + I-(aq) 0.050 O +x +x 0.050+x x 1.5 x 10-16 = (x)(0.050+x) (x)(0.050) x = solubility of AgI in mol/L = 3.0 x 10-15 M
Complex Ions A Complex ion is a charged species composed of: 1. A metallic cation 2. Ligands – Lewis bases that have a lone electron pair that can form a covalent bond with an empty orbital belonging to the metallic cation
Coordination Number • Coordination number refers to the number of ligands attached to the cation • 2, 4, and 6 are the most common coordination numbers
Complex Ions and Solubility AgCl(s) Ag+ + Cl- Ksp = 1.6 x 10-10 Ag+ + NH3 Ag(NH3)+ K1 = 1.6 x 10-10 Ag(NH3)+ NH3 Ag(NH3)2+ K2 = 1.6 x 10-10 K = KspK1K2 AgCl + 2NH3 Ag(NH3)2+ + Cl-