Lecture Presentation Unit 9 AcidBase Equilibria Day 2 2015 Pearson Education, Inc. James F. Kirby Quinnipiac University Hamden, CT Warm Up 1. Orange Juice has a H+ concentration of 5.0 x 10-4 M. What is the pH?
Acids and Bases 2015 Pearson Education, Inc. Warm Up 1. Orange Juice has a H+ concentration of 5.0 x 10-4 M. What is the pH? pH = -log [5.0 x 10-4] = 3.3 This means 10-3.3 = 5.0 x 10-4 Acids and Bases 2015 Pearson Education, Inc.
Agenda Lecture: Acids and Bases Practice: Acids and Bases Due Friday: Lab Notebook Set Up (Acid and Base Titration) Acids and Bases 2015 Pearson Education, Inc. Strong Acids You will recall that the seven strong acids are ________________________________ ___________and ________. These are, by definition, strong electrolytes and
exist totally as ions in aqueous solution; e.g., _________________________ So, for the monoprotic strong acids, [H3O+] = [acid] Acids and Bases 2015 Pearson Education, Inc. Strong Acids You will recall that the seven strong acids are HCl, HBr, HI, HNO3, H2SO4, HClO3, and HClO4. These are, by definition, strong electrolytes and
exist totally as ions in aqueous solution; e.g., HA + H2O H3O+ + A So, for the monoprotic strong acids, [H3O+] = [acid] 2015 Pearson Education, Inc. Acids and Bases Strong Bases Strong bases are the soluble hydroxides, which are the ________________and ___________________hydroxides (Ca 2+, Sr2+, and Ba2+). Again, these substances dissociate completely in aqueous solution; e.g.,
MOH(aq) M+(aq) + OH(aq) or M(OH)2(aq) M2+(aq) + 2 OH(aq) Acids and Bases 2015 Pearson Education, Inc. Strong Bases Strong bases are the soluble hydroxides, which are the alkali metal and alkaline earth metal hydroxides (Ca2+, Sr2+, and Ba2+). Again, these substances dissociate completely in aqueous solution; e.g., MOH(aq) M+(aq) + OH(aq) or M(OH)2(aq) M2+(aq) + 2 OH(aq) Acids
and Bases 2015 Pearson Education, Inc. Weak Acids For a weak acid, the equation for its dissociation is HA(aq) + H2O(l) H H3O+(aq) + A(aq) Since it is an equilibrium, there is an equilibrium constant related to it, called the __________________ _______________ The __________ the ______________ value of Ka, the ________________ __________ is the acid. Acids
and Bases 2015 Pearson Education, Inc. Weak Acids For a weak acid, the equation for its dissociation is HA(aq) + H2O(l) H H3O+(aq) + A(aq) Since it is an equilibrium, there is an equilibrium constant related to it, called the acid-dissociation constant, The greater the value Ka = [H3O+][A] of K , the stronger is [HA] a the acid.
Acids and Bases 2015 Pearson Education, Inc. Comparing Strong and Weak Acids What is present in solution for a strong acid versus a weak acid? Strong acids _____________dissociate to ions. Weak acids only _________ dissociate to ions. Acids and Bases 2015 Pearson Education, Inc. Comparing Strong and Weak Acids
What is present in solution for a strong acid versus a weak acid? Strong acids completely dissociate to ions. Weak acids only partially dissociate to ions. Acids and Bases 2015 Pearson Education, Inc. Calculating Ka from the pH The pH of a 0.10 M solution of formic acid, HCOOH, at 25 C is 2.38. Calculate Ka for formic acid at this temperature. We know that [H3O+][HCOO] Ka = [HCOOH]
To calculate Ka, we need the equilibrium concentrations of all three things. We can find [H3O+], which is the same as [HCOO], from the pH. _____________________________________ __________________________________ Acids and Bases 2015 Pearson Education, Inc. Calculating Ka from the pH The pH of a 0.10 M solution of formic acid, HCOOH, at 25 C is 2.38. Calculate Ka for formic acid at this temperature. We know that [H3O+][HCOO] Ka = [HCOOH]
To calculate Ka, we need the equilibrium concentrations of all three things. We can find [H3O+], which is the same as [HCOO], from the pH. [H3O+] = [HCOO] = 102.38 = 4.2 103 Acids and Bases 2015 Pearson Education, Inc. Calculating Ka from pH Now we can set up a table for equilibrium concentrations. We know initial HCOOH (_________) and ion concentrations (_______________); we found equilibrium ion concentrations (________________); so we calculate the change, then the equilibrium
HCOOH concentration. Acids and Bases 2015 Pearson Education, Inc. Calculating Ka from pH Now we can set up a table for equilibrium concentrations. We know initial HCOOH (0.10 M) and ion concentrations (0 M); we found equilibrium ion concentrations (4.2 103 M); so we calculate the change, then the equilibrium HCOOH concentration. Acids
and Bases 2015 Pearson Education, Inc. Calculating Ka from pH Reaction [HCOOH], M [H3O+], M [HCOO], M Initially Change At
equilibrium Acids and Bases 2015 Pearson Education, Inc. Calculating Ka from pH Reaction Initially [HCOOH], M [H3O+], M [HCOO], M 0.10
0 0 4.2 103 4.2 103 Change At equilibrium Acids and Bases 2015 Pearson Education, Inc.
Calculating Ka from pH Reaction [HCOOH], M [H3O+], M [HCOO], M Initially 0.10 0 0
Change 4.2 103 At equilibrium +4.2 103 +4.2 103 4.2 103 4.2 103 Acids and Bases 2015 Pearson Education, Inc.
Calculating Ka from pH Reaction [HCOOH], M [H3O+], M [HCOO], M Initially 0.10 0 0 Change
4.2 103 At equilibrium 0.10 4.2 103 = 0.0958 = 0.10 +4.2 103 +4.2 103 4.2 103 4.2 103 Acids and Bases
2015 Pearson Education, Inc. Calculating Ka from pH This allows us to calculate Ka by putting in the equilibrium concentrations. Ka = = ____________ Acids and Bases 2015 Pearson Education, Inc. Calculating Ka from pH This allows us to calculate Ka by putting in the equilibrium concentrations. [4.2 103][4.2 103] Ka = [0.10]
= 1.8 104 Acids and Bases 2015 Pearson Education, Inc. Calculating Ka from the pH The pH of a 0.20 M solution of acetic acid, CH3COOH, at 25 C is 2.52. Calculate Ka for acetic acid at this temperature. We know that [H3O+][CH3COO] Ka = [CH3OOH] Acids and
Bases 2015 Pearson Education, Inc. Calculating Ka from pH Reaction [CH3COOH], M [H3O+], M [CH3COO-], M Initially Change At
equilibrium Acids and Bases 2015 Pearson Education, Inc. Calculating Percent Ionization [H3O+]eq Percent ionization = 100 [HA]initial In this example, [H3O+]eq = 4.2 103 M From pH = 2.38 [HCOOH]initial = 0.10 M 4.2 103 Percent ionization = 100 0.10
= 4.2% 2015 Pearson Education, Inc. Acids and Bases Calculating Percent Ionization [H3O+]eq Percent ionization = 100 [HA]initial In this example, [H3O+]eq = 4.2 103 M From pH = 2.38 [HCOOH]initial = 0.10 M 4.2 103 Percent ionization = 100
0.10 = 4.2% 2015 Pearson Education, Inc. Acids and Bases Your turn. By yourself, how would you attack this problem? When you have done as much as you can, turn to neighbor and tell them what you did. Question: A 0.30 M acid solution has a pH of 1.85. What is the % ionization? a. b.
c. d. 2015 Pearson Education, Inc. 21 % 16 % 6.2 % 4.7 % Acids and Bases A 0.30 M acid solution has a pH of 1.85. What is the % ionization? a.
b. c. d. 21 % 16 % 6.2 % 4.7 % Acids and Bases 2015 Pearson Education, Inc. Method to Follow to Calculate pH Using Ka 1) Write the _________________ for the ionization
equilibrium. 2) Write the __________________expression. 3) Set up a table for ____________________Concentration to determine equilibrium concentrations as a function of change (x). 4) Substitute equilibrium concentrations into the equilibrium constant expression and _________________________________. Acids and (Make assumptions if you can!) Bases 2015 Pearson Education, Inc. Method to Follow to Calculate pH Using Given Ka 1) Write the chemical equation for the ionization equilibrium.
2) Write the equilibrium constant expression. 3) Set up a table for Initial/Change in/Equilibrium Concentration to determine equilibrium concentrations as a function of change (x). 4) Substitute equilibrium concentrations into the equilibrium constant expression and solve for x. (Make assumptions if you can!) Acids and Bases 2015 Pearson Education, Inc. Example Calculate the pH of a 0.30 M solution of acetic acid, HC2H3O2, at 25 C. 1) HC2H3O2 + H2O H H3O+ + C2H3O2 2) Ka = _______________= _________ 3)
Reaction CH3COOH (M) H3O+ (M) CH3COO (M) Initial Change Equilibrium 2015 Pearson Education, Inc. Acids and Bases
Example Calculate the pH of a 0.30 M solution of acetic acid, HC2H3O2, at 25 C. 1) HC2H3O2 + H2O H H3O+ + C2H3O2 2) Ka = [H3O+][C2H3O2] / [HC2H3O2]= 1.8 X 10-5 3) Reaction Initial Change Equilibrium 2015 Pearson Education, Inc. CH3COOH (M)
H3O+ (M) CH3COO (M) 0.30 0 0 x +x +x 0.30 x
x x Acids and Bases Example (concluded) 4) Ka = [H3O+][C2H3O2] / [HC2H3O2] = _________________ If we assume that ______________, then 0.30 x becomes ___________. The problem becomes easier, since we dont have to use the quadratic formula to solve it. Ka = __________= _________, so x = 2.3 103 x = [H3O+], so pH = log(2.3 103) = _____
2015 Pearson Education, Inc. Acids and Bases Example (concluded) 4) Ka = [H3O+][C2H3O2] / [HC2H3O2] = (x)(x) / (0.30 x) If we assume that x << 0.30, then 0.30 x becomes 0.30. The problem becomes easier, since we dont have to use the quadratic formula to solve it. Ka = 1.8 105 = x2 / 0.30, so x = 2.3 103 x = [H3O+], so pH = log(2.3 103) = 2.64 2015 Pearson Education, Inc.
Acids and Bases Your Turn: Try to solve it by yourself. After two minutes, turn to a neighbor and tell them what you did. Question: What is the pH of a 0.0200 M aqueous solution of HBr? 2015 Pearson Education, Inc. a. b. c. d.
1.00 1.70 2.30 12.30 Acids and Bases What is the pH of a 0.0200 M aqueous solution of HBr? a. b. c. d. 1.00 1.70
2.30 12.30 Acids and Bases 2015 Pearson Education, Inc. Again: What is the pH of a 0.0200 M aqueous solution of HF? The Ka of HF is 6.8 104. a. b. c. d. 1.70
2.43 3.17 12.30 2015 Pearson Education, Inc. Acids and Bases What is the pH of a 0.0200 M aqueous solution of HF? The Ka of HF is 6.8 104. a. b. c. d.
1.70 2.43 3.17 12.30 2015 Pearson Education, Inc. Acids and Bases Strong vs. Weak Acids Another Comparison Strong Acid: [H+]eq _____ [HA]init Weak Acid: [H+]eq _____ [HA]init This creates a difference in _______________ and in rates of chemical reactions.
Acids and Bases 2015 Pearson Education, Inc. Strong vs. Weak Acids Another Comparison Strong Acid: [H+]eq = [HA]init Weak Acid: [H+]eq < [HA]init This creates a difference in conductivity and in rates of chemical reactions. Acids and Bases 2015 Pearson Education, Inc.
Polyprotic Acids Polyprotic acids have _________________proton. It is always easier to ___________ the first proton than any successive proton. If the factor in the Ka values for the first and second dissociation has a difference of 3 or greater, the pH generally depends only on the first dissociation. Acids and Bases 2015 Pearson Education, Inc. Polyprotic Acids Polyprotic acids have more than one acidic proton. It is always easier to remove the first proton than any successive proton. If the factor in the Ka values for the first and second
dissociation has a difference of 3 or greater, the pH generally depends only on the first dissociation. Acids and Bases 2015 Pearson Education, Inc. Polyprotic Acids Acids and Bases 2015 Pearson Education, Inc. Weak Bases Ammonia, NH3, is a ______________. Like weak acids, weak bases have an
equilibrium constant called the _______________________________ Equilibrium calculations work the same as for acids, using the base dissociation constant instead. Acids and Bases 2015 Pearson Education, Inc. Weak Bases Ammonia, NH3, is a weak base. Like weak acids, weak bases have an equilibrium constant called the base dissociation constant. Equilibrium calculations work the same as for acids, using the base dissociation constant instead.
Acids and Bases 2015 Pearson Education, Inc. Base Dissociation Constants Acids and Bases 2015 Pearson Education, Inc. Example What is the pH of 0.15 M NH3? 1) NH3 + H2O H NH4+ + OH 2) Kb = [NH4+][OH] / [NH3] = ___________ 3) REACTION
Initial Change Equilibriu m 2015 Pearson Education, Inc. NH3 (M) NH4+ (M) OH (M) 0.15 0 0
x +x +x 0.15 x x x Acids and Bases Example What is the pH of 0.15 M NH3?
1) NH3 + H2O H NH4+ + OH 2) Kb = [NH4+][OH] / [NH3] = 1.8 105 3) REACTION Initial Change Equilibriu m 2015 Pearson Education, Inc. NH3 (M) NH4+ (M) OH (M) 0.15 0
0 x +x +x 0.15 x x x Acids and Bases
Example (completed) 4) 1.8 10 5 = ______________ If we assume that ____________________. Then: _________________ and: x = __________ Note: x is the molarity of OH , so log(x) will be the pOH ___________and [14.00 pOH] is pH ____________ Acids and Bases 2015 Pearson Education, Inc. Example (completed) 4) 1.8 10 5 = x2 / (0.15 x) If we assume that x << 0.15, 0.15 x = 0.15. Then: 1.8 105 = x2 / 0.15
and: x = 1.6 103 Note: x is the molarity of OH , so log(x) will be the pOH (pOH = 2.80) and [14.00 pOH] is pH (pH = 11.20). Acids and Bases 2015 Pearson Education, Inc. What is the pH of a 0.0400 M aqueous solution of KOH? a. b. c. d. 12.60
10.30 4.00 1.40 Acids and Bases 2015 Pearson Education, Inc. What is the pH of a 0.0400 M aqueous solution of KOH? a. b. c. d. 12.60 10.30
4.00 1.40 Acids and Bases 2015 Pearson Education, Inc. Types of Weak Bases Two main categories 1) Neutral substances with an Atom that has a nonbonding pair of electrons that can accept H+ (like ammonia and the amines) 2) Anions of weak acids Acids and
Bases 2015 Pearson Education, Inc. Relationship between Ka and Kb For a conjugate acidbase pair, Ka and Kb are related in this way: Ka Kb = Kw Therefore, if you know one of them, you can calculate the other. 2015 Pearson Education, Inc. 2015 Pearson Education, Inc. Acids and Bases
Memorize thisbut only one side Acids and Bases 2015 Pearson Education, Inc.
