Hints for Assignment 8, CHEM 1P81


1 and 2 Hints. Conjugate acid-base pairs differ by a single proton (H+). If you are given a base and need its conjugate acid, add one H+. If you are given an acid and need its conjugate base, remove one H+.

Example: What is the conjugate base for CH4? Answer: remove one H+ to get CH3- as the conjugate base for CH4.

Example: What is the conjugate acid of the peroxide ion, O22-? Answer: add one H+ to get HO2- as the conjugate acid for O22-.
 
 

3. Hints. Refer to the Hints above for the definitions of conjugate acid-base pairs.
Example: In the following acid-base reaction, identify the Bronsted acid and base on the left, and their conjugate partners on the right:

CH3CH2OH + NH2- ---> CH3CH2O- + NH3

Acid on left
Base on left
Acid on right
Base on right
CH3CH2OH
NH2-
NH3
CH3CH2O-

The acid on the left (CH3CH2OH) is paired with its conjugate base partner (CH3CH2O- ) on the right. Similarly, the base on the left (NH2-) gains a proton to become its conjugate acid partner (NH3)on the right.

4. Example. For the reaction above in Question 3, predict whether the equilibrium position will lie predominantly to the left or to the right.
Answer. Bronsted acid-base reactions have an equilibrium position that lies on the side of the equation having the weaker conjugate acid and weaker conjugate base. According to Table 17.4 in Kotz and Treichel, CH3CH2OH (= C2H5OH) is a stronger acid than NH3, and NH2- is a stronger base than CH3CH2O-
CH3CH2OH +
NH2- --->
CH3CH2O- + 
NH3
stronger acid
stronger base
weaker base
weaker acid

Thus this equilibrium is expected to lie predominantly to the right.
 
 

5. Example. In a Bronsted acid-base reaction, the stronger acid reacts with the stronger base to produce a weaker acid and a weaker base. Thus, whether a reaction "goes" or "does not go" can be used to determine the relative strength of the Bronsted acids and bases in the reaction. From the data below, determine the order of the strengths of the three Bronsted acids.
1. H2O2 + C6H5O- ---> C6H5OH + HO2- does not go

2. H3C6H5O7 + C6H5O- ---> H2C6H5O7- + C6H5OH goes

3. H3C6H5O7 + HO2- ---> H2O2 + H2C6H5O7- goes

Answer. Since reaction 1 does not go, we know that C6H5OH is a stronger acid than H2O2, and HO2- is a stronger base than C6H5O-. Similarly, reactions 2 and 3 tell us that H3C6H5O7 is a stronger acid than either C6H5OH or H2O2. Thus the order of acid strengths is

H3C6H5O7 > C6H5OH > H2O2

6. Hint. See Example 17.4 in Kotz and Treichel 4th edition, and "Problem Solving Tips and Ideas" on the same page. This question is an exercise in using your calculator to find logarithms and antilogarithms, and remembering the correct definitions of pH and pOH.
pH = - log [H+]

pOH = - log [OH-]

[H+] [OH-] = Kw = 1 x 10-14

acid solutions have pH < 7; basic solutions have pH >7

7. Example. A 0.300 M solution of a weak acid HA is 2.00 % ionized. What are the H+, A-, and HA concentrations at equilibrium? What is Ka?

Setup. We willl neglect the amount of hydrogen ion produced by water in its autoionization, since it is very small. If the HA solution is 2.00% ionized, that means 2.00% of 0.300 M HA (= 0.006 M) has dissociated into its ions. From the equation, this means that 0.006 M H3O+ and 0.006 M A- are produced
Equation 
HA +
H2O --->
H3O+
A-
Initial 
0.300 M
0
0
Change 
-0.00600 M
0.00600 M
0.00600 M
Final 
0.294 M
0.00600 M
0.00600 M

The definition of Ka is

Ka = [H3O+] [A-] / [HA]
Substitute the equilibrium values from the table to get Ka = 1.22 x 10-4
8. The Ka for an acid HA is 1.5 x 10-6. What is the pH of a 0.250 M solution of the acid? What is the % ionization of the acid at this concentration?

Setup and Answer

 
Equation 
HA +
H2O --->
H3O+
A-
Initial 
0.250 M
0
0
Change 
-x M
x M
x M
Final 
0.250 - x M
x M
x M
Ka = [H3O+] [A-] / [HA] = 1.5 x 10-6
= [x] [x] / [0.250 - x]
Solve to get x = H3O+ = 6.1 x 10-4

Thus the pH = - log[H+] = 3.21

The % ionization is 100 x (amount ionized / starting amount)

= 100 x 6.1 x 10-4 / 0.250 = 0.245% ionized.

9. Example. A 0.500 M solution of a weak base B has a pH of 10.10. The equation for ionization is

B (aq) + H2O ---> BH+ + OH-

What are the BH+, OH-, and B concentrations at equilibrium? Calculate Kb for the base.

Answer. First note that Kb = [BH+] [OH-] / [B]. We are given the pH but we need the hydroxide concentration. We get it in the same way as in Question 6 above: pH = 10.10, pOH = 3.90, [OH-] = 1.26 x 10-4. This is the equilibrium concentration of hydroxide, so 1.26 x 10-4 M of base had to react to produce this amount of hydroxide.
Equation 
B +
H2O --->
BH+
OH-
Initial 
0.500 M
0
0
Change 
-1.26 x 10-4
1.26 x 10-4
1.26 x 10-4
Equilibrium 
0.500 M
1.26 x 10-4
1.26 x 10-4

Now that we have the equilibrium concentrations, we can calculate Kb.

Kb = [BH+] [OH-] / [B]

= (1.26 x 10-4)( 1.26 x 10-4) / (0.500)

= 3.17 x 10-8

10. Example. A weak base B has a pKb equal to 4.50. Calculate the hydroxide concentration, the pH, and the % ionization for a 0.150 M solution.
Answer. First write the equation for ionization of a weak base; then write the expression for Kb so you will know what values you are looking for.
B (aq) + H2O ---> BH+ + OH-
Kb = [BH+] [OH-] / [B]

pKb = 4.50 = - log Kb

Therefore log Kb = -4.50 and Kb = 3.16 x 10-5

Equation 
B +
H2O --->
BH+
OH-
Initial 
0.150 M
0
0
Change 
-x
x
x
Equilibrium 
0.150 - x
x
x
Kb = [BH+] [OH-] / [B] = 3.16 x 10-5
= [x] [x] / [0.150 - x]
Solve to get x = [OH-] = 2.2 x 10-3

Thus pOH = 2.66 and pH = 14 - pOH = 11.34

% ionization is calculated as in Question 8; it is 100 x (amount of base ionized / initial amount of base)

= 100 x 2.2 x 10-3 / 0.150 = 1.45%
 
 

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Last revised: March 15 2001 by M. F. Richardson
© Brock University, 2001