Ch 12.7 Acids and Bases
Supplemental Instruction
Iowa State University
Leader:
Course:
Instructor:
Date:
Grant DeRocher
Chem 167
Houk
04/21/13
Know Kw= 1.0 x 10-14
12.71, 12.73, 12.74, 12.75, 12.79, 12.80,
1. Write the formula of the conjugate base of each of the following acids: (a)HNO3 (b)H2O
(c) HSO4- (d) H2CO3 (e) H3O+
(a) NO3(b) OH(c) SO42(d) HCO3(e) H2O
2. For each of the following reactions, indicate the Bronsted-Lowry acids and bases. What
are the conjugate acid-base pairs?
(a) CN-(aq) + H2O(l)
HCN(aq) + OH-(aq)
(b) HCO3-(aq) + H3O+(aq)
H2CO3-(aq) + H2O(l)
(c) CH3COOH(aq) + HS-(aq)
CH3COO-(aq) + H2S(aq)
(a) CN-(aq) BL BASE+ H2O(l)BL ACID HCN(aq) conj acid+ OH-(aq) conj base
(b) HCO3-(aq) + H3O+(aq)
H2CO3-(aq) + H2O(l)
(c) CH3COOH(aq) + HS (aq)
CH3COO-(aq) + H2S(aq)
Other two work the same as the first one. Use bronsted Lawry definition of acids and bases
3. What are the products of the following acid-base reactions? Indicate the acid and its
conjugate base and the base and its conjugate acid.
(a) HClO4 + H2O
(b) NH4+ + H2O
(c) HCO3- + OH(a) ClO4- + H3O+ (b) NH3 + H3O+ (c) CO32- + H2O
4. Write chemical equations and equilibrium expressions for the reactions of each of the
following weak acids with water:
(a) CH3COOH
(b) C2H5COOH
(c) HF
(d) HClO
(e) H2CO3
Ka= [products]/[reactants] make sure to write the equation first to find Ka
H2O becomes H3O+(aq)
5. Hydrofluoric acid is a weak acid used in the building industry to etch patterns into glass
for elegant windows. Because it dissolves glass, it is the only inorganic acid that must be
stored in plastic containers. A 0.1 M solution of HF has a pH of 2.1. Calculate [H3O+] in
this solution.
The definition of pH is: pH=-log[H3O+] rearranging [H3O+] = 10-pH pH=2.1
[H3O+] = 10-2.1 = 8x10-3 (one sig fig)
6. Calculate the pH of a 0.10 M solution of propanoic acid and determine its present
ionization. The ionization constant Ka for propanoic acid is 1.3x10-5.
C2H5COOH(aq) + H2O
C2H5COO-(aq) + H3O+
Initial
0.10 M
0.00 M
0.00 M
Change
-X
+X
+X
Equilibrium
0.10 M
X
X
The equilibrium constant expression is: Ka= [C2H5COO-][H3O+]/[C2H5COOH]=1.3x10-5
[X][X]/[0.10-X]=1.3x10-5
Assume that .10-X is approximately .10. Ka=X2/.10=1.3x10-5.
X2=1.3x10-6 so X=1.1x10-3 M = [H3O+] = [C2H5COO-]
The pH= -log[H3O+]= -log[1.1x10-3]= 2.9
Percent ionization is defined as % ionization = [Anion]eq/[acid]init x 100
[C2H5COO-]/[ C2H5COOH] x 100 = [1.1x10-3/[ .10] x 100= 1.1%
7. Acrylic acid is used in the polymer industry in the production of acrylates. Its Ka is
5.6x10-5 . What is the pH of a 0.11 M solution of acrylic acid, CH2CHCOOH?
CH2CHCOOH(aq) + H2O
Initial
0.11 M
Change
-X
Equilibrium
0.11 M
CH2CHCOO-(aq) +
H3O+
0.00 M
0.00 M
+X
+X
X
X
The equilibrium constant expression is: Ka= [CH2CHCOO-][H3O+]/[ CH2CHCOOH]=5.6x10-5
[X][X]/[0.10-X]=5.6x10-5
Assume that .10-X is approximately .10. Ka=X2/.10=5.6x10-5.
X2=6.2x10-6 so X=2.5x10-3 M = [H3O+] = [CH2CHCOOH]
The pH= -log[H3O+]= -log[2.5x10-3]= 2.6
8. Morphine, an opiate derived from opium poppy, has the molecular formula C7H19NO3. It
is a weakly basic amine, with a Kb of 1.6x10-6. What is the pH of a 0.0045 M solution of
morphine?
C2H19NO3
Initial
Change
Equilibrium
+
C7H20NO3+
0.00 M
+X
X
H2O
0.0045 M
-X
0.0045 M
+
OH0.00 M
+X
X
The equilibrium constant expression is: Ka= [C7H20NO3+][OH-]/[ C2H19NO3]=1.6x10-6
[X][X]/[0.10-X]=1.6x10-6
Assume that .10-X is approximately .10. Ka=X2/.10=1.6x10-6.
X2=7.2x10-9 so X=8.5x10-5 M = [OH-] = [C7H20NO3+]
We need to calculate the [H3O+] to find the pH.
[H3O+][OH-] =Kw = 1.0x10-14 ,
[H3O+]=Kw/[OH-]= 1.0x10-14 / 8.5x10-5= 1.2x10-10 M
The pH= -log[H3O+ ]= -log[1.2x10-10]= 9.9
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Session 31 Answers. Chapter 12.6 4/14/2013