Chapter 16

Acids and Bases

 

16.2

In general the components of the conjugate acid-base pair are on opposite sides of the reaction arrow.  The base always has one fewer proton than the acid.

 

a.

The conjugate acid-base pairs are (1) HCN (acid) and CN- (conjugate base) and (2) CH3COO- (base) and CH3COOH (conjugate acid).

b.

(1) HCO (acid) and CO (conjugate base) and (2) HCO (base) and H2CO3 (conjugate acid).

c.

(1) H2PO (acid) and HPO (conjugate base) and (2) NH3 (base) and NH (conjugate acid).

d.

(1) HClO (acid) and ClO- (conjugate base) and (2) CH3NH2 (base) and CH3NH (conjugate acid).

e.

(1) H2O (acid) and OH- (conjugate base) and (2) CO (base) and HCO (conjugate acid).

16.3

Tables 16.1 and 16.2 of the text list important Brønsted acids and bases and their respective conjugates. 

a.

both (why?)

b.

base

c.

acid

d.

base

e.

acid

f.

base

g.

base

h.

base

i.

acid

j.

acid

16.4

The conjugate acid of any base is just the base with a proton added.

a.

H2S

b.

H2CO3

c.

HCO

d.

H3PO4

e.

H2PO

f.

HPO

g.

H2SO4

h.

HSO

i.

HSO

16.5

Recall that the conjugate base of a Brønsted acid is the species that remains when one proton has been removed from the acid.

a.

nitrite ion:  NO

b.

hydrogen sulfate ion (also called bisulfate ion):  HSO

c.

hydrogen sulfide ion (also called bisulfide ion):  HS-

d.

cyanide ion:  CN-

e.

formate ion:  HCOO-

16.6

a.

The Lewis structures of C2HO and C2O  are

 

                       

(The lone pairs on the oxygen atoms are not shown.)

b.

H+ and C2H2O4 can act only as acids, C2HO can act as both an acid and a base, and C2O can act only as a base.

16.7

The conjugate base of any acid is simply the acid minus one proton.

a.

CH2ClCOO-

b.

IO

c.

H2PO

d.

HPO

e.

PO

f.

HSO

g.

SO

h.

IO

i.

SO

j.

NH3

k.

HS-

l.

S2-

m.

OCl-

16.10

The strength of the H-X bond is the dominant factor in determining the strengths of binary acids.  As with the hydrogen halides (see Section 16.9 of the text), the H-X bond strength decreases going down the column in Group 6A.  The compound with the weakest H-X bond will be the strongest binary acid:      H2Se > H2S > H2O.

16.11

All the listed pairs are oxoacids that contain different central atoms whose elements are in the same group of the periodic table and have the same oxidation number.  In this situation the acid with the most electronegative central atom will be the strongest.

 

a.

H2SO4  >  H2SeO4

b.

H3PO4  >  H3AsO4

16.12

The CHCl2COOH is a stronger acid than CH2ClCOOH.  Having two electronegative chlorine atoms compared to one, will draw electron density away from the O–H group, making the O-H bond more polar.  The hydrogen atom in CHCl2COOH is more easily ionized compared to the hydrogen atom in CH2ClCOOH.

16.13

The conjugate bases are C6H5O- from phenol and CH3O- from methanol.  The C6H5O- is stabilized by resonance:

 

The CH3O- ion has no such resonance stabilization.  A more stable conjugate base means an increase in the strength of the acid.

16.18

Strategy:

The equilibrium concentrations of H+ and OH– must satisfy Equation 16.1, (25°C). Substitute the given concentrations and solve for [OH–]:

Solution:

a.

b.

c.

d.

16.19

Strategy:

The equilibrium concentrations of H+ and OH– must satisfy Equation 16.1, (25°C).  Substitute the given concentrations and solve for [H+]:

Setup:

 

Solution:

a.

b.

c.

d.

16.20

Strategy:

The equilibrium concentrations of H+ and OH– must satisfy Equation 16.1, but with the appropriate equilibrium constant for 50°C:(50°C). Substitute the given concentrations and solve for [OH–]:

 

Solution:

a.

b.

c.

d.

16.21

Strategy:

Use Equation 16.1 to calculate the [H3O+] concentrations.

 

Kw = [H3O+][OH–]

Setup:

Kw = 5.13 × 10–13 at 100°C.  Rearranging Equation 16.1 to solve for [H3O+] gives:

 

Solution:

a.

b.

c.

d.

16.24

[OH-]  =  0.62 M

 

16.25

[H+] = 1.4 ´ 10-3 M

 

16.26

a.

Ba(OH)2 is ionic and fully ionized in water.  The concentration of the hydroxide ion is 5.6 ´ 10-4 M (Why?  What is the concentration of Ba2+?) We find the hydrogen ion concentration.

 

 

The pH is then:   

pH  =  -log[H+] =  -log(1.8 ´ 10-11)  =  10.74

b.

Nitric acid is a strong acid, so the concentration of hydrogen ion is also 5.2 ´ 10-4 M.  The pH is:

 

                                                                pH  =  -log[H+] =  -log(5.2 ´ 10-4)  =  3.28

16.27

a.

HCl is a strong acid, so the concentration of hydrogen ion is also 0.0010 M.  (What is the concentration of chloride ion?)  We use the definition of pH.

 

pH  =  -log[H+] =  -log(0.0010)  =  3.00

b.

KOH is an ionic compound and completely dissociates into ions.  We first find the concentration of hydrogen ion.

 

 

The pH is then found from its defining equation

 

pH  =  -log[H+] =  -log[1.3 ´ 10-14]  =  13.89

16.28

Since pH = -log[H+], we write [H+] = 10-pH

a.

[H+] =  10-5.20  =  6.3 ´ 10-6 M

b.

pH  =  -log[H+] =  16.00

 

log[H+] =  -16.00

 

[H+] =  10-16.00  =  1.0 ´ 10-16 M

c.

We are given the concentration of OH- ions and asked to calculate [H+]. The relationship between [H+] and [OH-] in water or an aqueous solution is given by the ion-product of water, Kw (Equation 16.1 of the text).

 

Kw  =  1.0 ´ 10-14  =  [H+][OH-]

 

Rearranging the equation to solve for [H+], we write

 

 

Since the [OH-] < 1 ´ 10-7 M we expect the [H+] to be greater than 1 ´ 10-7 M.

16.29

Strategy:

Here we are given the pH of a solution and asked to calculate [H+].  Because pH is defined as
pH = -log[H+], we can solve for [H+] by taking the antilog of the pH; that is, [H+] = 10-pH.

Solution:

a.

From Equation 16.2 of the text:

 

pH = -log[H+] =  2.42

 

log[H+] =  -2.42

 

To calculate [H+], we need to take the antilog of -2.42.

 

[H+] =  10-2.42  =  3.8 ´ 10-3 M

 

Check:  Because the pH is between 2 and 3, we can expect [H+] to be between 1 ´ 10-2 M and 1 ´ 10-3 M.  Therefore, the answer is reasonable.

b.

pH = -log[H+] =  11.24

 

log[H+] =  -11.21

 

To calculate [H+], we need to take the antilog of -11.21.

 

[H+] =  10-11.21  =  6.2 ´ 10-12 M

 

Check:  Because the pH is between 11 and 12, we can expect [H+] to be between 1 ´ 10-11 M and 1 ´ 10-12 M.  Therefore, the answer is reasonable.

c.

pH = -log[H+] =  6.96

 

log[H+] =  -6.96

 

To calculate [H+], we need to take the antilog of -6.96.

 

[H+] =  10-6.96  =  1.1 ´ 10-7 M

d.

pH = -log[H+] =  15.00

 

log[H+] =  -15.00

 

To calculate [H+], we need to take the antilog of -15.00.

 

[H+] =  10-15.00  =  1.0 ´ 10-15 M

16.30

a.

acidic

b.

neutral

c.

basic

16.31

 

pH

[H+]

Solution is:

< 7

> 1.0 ´ 10-7 M

acid

> 7

< 1.0 ´ 10-7 M

basic

= 7

= 1.0 ´ 10-7 M

neutral

 

 

 

16.32

 

KOH is a strong base and therefore ionizes completely.  The OH- concentration equals the KOH concentration, because there is a 1:1 mole ratio between KOH and OH-.

 

[OH-]  =  0.360 M

 

pOH  =  -log[OH-]  =  0.444

16.33

The pH can be found by using Equation 16.6 of the text.

 

pH  =  14.00 - pOH  =  14.00 - 9.40  =  4.60

 

The hydrogen ion concentration can be found as follows.

 

4.60  =  -log[H+]

 

Taking the antilog of both sides:

 

[H+] =  2.5 ´ 10-5 M

16.34

Molarity of the HCl solution is:

               

 

pH  =  -log(0.762)  =  0.118

16.35

We can calculate the OH- concentration from the pOH.

 

pOH  =  14.00 - pH  =  14.00 - 10.00  =  4.00

 

[OH-]  =  10-pOH  =  1.0 ´ 10-4 M

 

Since NaOH is a strong base, it ionizes completely.  The OH- concentration equals the initial concentration of NaOH.

 

[NaOH]  =  1.0 ´ 10-4 mol/L

 

So, we need to prepare 546 mL of 1.0 ´ 10-4 M NaOH.

 

This is a dimensional analysis problem.  We need to perform the following unit conversions.

 

mol/L  ®  mol NaOH  ®  grams NaOH

 

546 mL  =  0.546 L

 

16.38

a.

0.92.

b.

-0.38.    

c.

3.49.

16.39

a.

-0.009.    

b.

1.46.    

c.

5.82.

16.40

a.

[HBr] = [H3O+] = 10-0.12 = 0.76 M.    

b.

[HBr] = [H3O+] = 10-2.46 = 3.5 ´ 10-3 M.    

c.

[HBr] = [H3O+] = 10-6.27 = 5.4 ´ 10-7 M.

16.41

Strategy:

HNO3 is a strong acid.  Therefore, the concentration of HNO3 is equal to the concentration of hydronium ion.  We use Equation 16.3 to determine [H3O+]:

Solution:

a.

[HNO3] = [H3O+] = 10-4.21 = 6.2 ´ 10-5 M.

b.

[HNO3] = [H3O+] = 10-3.55 = 2.8 ´ 10-4 M.    

c.

[HNO3] = [H3O+] = 10-0.98 = 0.10 M.

16.42

a.

[OH-] = [KOH] = 0.066 M.

 

pOH = -log [OH-] = - log (0.066) = 1.18

 

pH = 14.00 - pOH = 14.00 - 1.18 = 12.82

b.

[OH-] = [NaOH] = 5.43 M.

 

pOH = -log [OH-] = - log (5.43) =  -0.735

 

pH = 14.00 - pOH = 14.00 - (-0.735) = 14.74

c.

[OH-] = 2[Ba(OH)2] = 2(0.74 M) = 1.48 M.

 

pOH = -log [OH-] = - log (1.48) = –0.170

 

pH = 14.00 - pOH = 14.00 - (–0.170) = 14.17

16.43

Strategy:

We use Equation 16.4 to determine pOH and Equation 16.6 to determine pH.  The hydroxide ion concentration of a monobasic strong base, such as LiOH and NaOH, is equal to the base concentration.  In the case of a dibasic base such as Ba(OH)2, the hydroxide concentration is twice the base concentration.

Solution:

a.

[OH-] = [LiOH] = 1.24 M.

 

pOH = -log [OH-] = - log (1.24) = -0.093

 

pH = 14.00 - pOH = 14.00 - (-0.093)= 14.09

b.

[OH-] = 2[Ba(OH)2] = 2(0.22 M) = 0.44 M.

 

pOH = -log [OH-] = - log (0.44) = 0.36

 

pH = 14.00 - pOH = 14.00 - 0.36 = 13.64

c.

[OH-] = [NaOH] = 0.085 M.

 

pOH = -log [OH-] = - log (0.085) =  1.07

 

pH = 14.00 - pOH = 14.00 - 1.07 = 12.93

16.44

We calculate the pOH using Equation 16.6 and the hydroxide ion concentration using Equation 16.5:

a.

pOH = 14.00 - pH = 14.00 - 9.78 = 4.22

 

[LiOH] = [OH-] = 10-pOH = 10-4.22 = 6.0 ´ 10-5 M

b.

[OH-] = 10-pOH = 10-4.22 = 6.0 ´ 10-5 M

 

[Ba(OH)2] = [OH-]==3.0 ´ 10-5 M

16.45

Strategy:

We use Equation 16.6 to determine pOH and Equation 16.5 to determine hydroxide ion concentration.  The concentration of a monobasic strong base such as LiOH is equal to the hydroxide concentration.  In the case of a dibasic base such as Ba(OH)2, the base concentration is half the hydroxide concentration.

Solution:

a.

pOH = 14.00 - pH = 14.00 - 11.04 = 2.96

 

[KOH] = [OH-] = 10-pOH = 10-2.96 = 1.1 ´ 10-3 M

b.

[OH-] = 10-pOH = 10-2.96 = 1.1 ´ 10-3 M

 

[Ba(OH)2] = [OH-]==5.5 ´ 10-4 M

16.46

Strategy:

Construct an equilibrium table, and express the equilibrium concentration of each species in terms of x.  Solve for x using the approximation shortcut, and evaluate whether or not the approximation is valid.  If the shortcut is not appropriate, use the quadratic equation given in Appendix 1 to solve for x.  Use Equation 16.2 to determine pH.

Setup:

                                       

 

+

     

 

 

 

      

 

Initial (M):

1.8 ´ 10-5

 

 

0.00

0.00

Change (M):

-x

 

 

+x

+x

Equilibrium (M):

1.8 ´ 10-5 – x

 

 

x

x

 

According to Table 16.8, Ka for carbonic acid is 4.2 × 10–7.

Solution:

These equilibrium concentrations are then substituted into the equilibrium expression to give

 

 

At a first approximation, assume that x is small compared to 1.8 ´ 10-5 so that

1.8 ´ 10-5 – x ≈ 1.8 ´ 10-5.

 

 

 

                Note that this approximate value of x is 15% of 1.8 ´ 10-5, so it is probably not appropriate to have assumed that 1.8 ´ 10-5 – x ≈ 1.8 ´ 10-5.  Instead, attempt to solve for x without making any approximations.  Start by clearing the denominator of equation (1) and then expanding the result to get:

                                                                               

               

This equation is a quadratic equation of the form ax2 + bx + c = 0, the solutions of which are given by (see Appendix 1):

 

.

               

For this case, a = 1, b = 4.2 × 10–7, and c = –7.56×10–12.  Substituting these values into the quadratic formula gives:

 

 

 

 

The negative solution is not physically reasonable since there is no such thing as a negative concentration. The solution is x = 2.5 ×10–6 M.

 

According to the equilibrium table, x = [H+].  Therefore, [H+] = 2.5 ×10–6 M  and

 

pH = –log(2.5 ×10–6) = 5.6

16.47

Strategy:

HNO3 is a strong acid.  Given pH, use Equation 16.3 to solve for [H3O+].

 

[H3O+] = 10–pH

 

Use reaction stoichiometry to determine HNO3 concentration.

Solution:

[H3O+] = 10–4.65 = 2.24 × 10–5 M

 

The hydronium ion concentration is equal that of HNO3 according to the balanced equation:

 

HNO3(aq) + H2O(aq)    H3O+(aq) + NO(aq)

 

Therefore, [H3O+] = [HNO3] = 2.24 × 10–5 M

16.54

a.

strong acid

b.

weak acid

c.

strong acid (first stage of ionization)

d.

weak acid

e.

weak acid

f.

weak acid

g.

strong acid

h.

weak acid

i.

weak acid

16.55

a.

strong base

b.

weak base

c.

weak base

d.

weak base

e.

strong base

16.56

The maximum possible concentration of hydrogen ion in a 0.10 M solution of HA is 0.10 M.  This is the case if HA is a strong acid.  If HA is a weak acid, the hydrogen ion concentration is less than 0.10 M.  The pH corresponding to 0.10 M [H+] is 1.00.  (Why three digits?)  For a smaller [H+] the pH is larger than 1.00 (why?).

 

a.

false, the pH is greater than 1.00

b.

false, they are equal

c.

true

d.

false

 

16.57

Strategy:

Recall that a weak acid only partially ionizes in water.  We are given the initial quantity of a weak acid (CH3COOH) and asked to calculate the pH.  For this we will need to calculate [H+], so we follow the procedure outlined in Section 16.5 of the text.

Solution:

We set up a table for the dissociation.

 

 

C6H5COOH(aq) 

H+(aq)

+  C6H5COO-(aq)

Initial (M):

0.10

 

0.00

0.00

Change (M):

­-x

 

+x

+x

Equilibrium (M):

(0.10 - x)

 

x

x

 

 

 

Assuming that x is small compared to 0.10, we neglect it in the denominator:

 

 

x  =  2.55 ´ 10-3 M  =  [H+]

 

pH  =  -log(2.55 ´ 10-3)  =  2.59

 

This problem could also have been solved using the quadratic equation.  The difference in pH obtained using the approximation x » 0 is very small.  (Using the quadratic equation would have given pH = 2.60, a difference of less than 0.4%.)

16.58

We set up a table for the dissociation.                            

 

 

 

HF(aq)        

H+(aq)

+   F-(aq)

Initial (M):

0.15

 

0.00

0.00

Change (M):

­-x

 

+x

+x

Equilibrium (M):

(0.15 - x)

 

x

x

 

 

7.1 ´ 10-4 =

 

Assuming that x is small compared to 0.15, we neglect it in the denominator:

 

7.1 ´ 10-4  =

 

x  =  0.0103 M  =  [H+]

 

pH  =  -log(0.0103)  =  1.99

16.59

We set up a table for the dissociation.     

 

 

 

HCN(aq)         

H+(aq)

+   CN-(aq)

Initial (M):

0.095

 

0.00

0.00

Change (M):

­-x

 

+x

+x

Equilibrium (M):

(0.095 - x)

 

x

x

 

 

4.9 ´ 10-10 =

 

Assuming that x is small compared to 0.095, we neglect it in the denominator:

 

4.9 ´ 10-10  =

 

x  =  6.82 ´ 10-6 M  =  [H+]

 

pH  =  -log(6.82 ´ 10-6)  =  5.17

16.60

We set up a table for the dissociation.                            

 

 

C6H5OH(aq)

H+(aq)

+   C6H5O-(aq)

Initial (M):

0.34

 

0.00

0.00

Change (M):

­-x

 

+x

+x

Equilibrium (M):

(0.34 - x)

 

x

x

 

 

1.3 ´ 10-10 =

 

Assuming that x is small compared to 0.34, we neglect it in the denominator:

 

1.3 ´ 10-10  =

 

x  =  6.65 ´ 10-6 M  =  [H+]

 

 

pH  =  -log(6.65 ´ 10-6)  =  5.18

16.61

Strategy:

Formic acid is a weak acid, and some of it will ionize in solution according to the equilibrium

 

HF(aq) + H2O(l)    F–(aq) + H3O+(aq)

 

 

The percent ionization of a weak acid is given by Equation 16.7:

 

,

 

where [HA]0 is the initial concentration of the acid (that is, the concentration assuming there is no ionization). Using the Ka expression above, set up a concentration table and solve for [H+]. Then, use this value to compute the percent ionization.

Solution:

a.

 

 

HF(aq) +

H2O(l)

F–(aq)

+   H3O+(aq)

Initial (M):

0.016

 

 

0

0

Change (M):

–x

 

 

+ x

+ x

Equilibrium (M):

0.016 – x

 

 

x

x

                                       

 

 

As a first approximation, assume that x is small compared to 0.016 so that 0.016 – x » 0.016.

 

 

                Note that this approximate value of x is 10% of 0.016, so it is probably not appropriate to have assumed that 0.016 – x » 0.016. Instead, attempt to solve for x without making any approximations. Start by clearing the denominator of equation (1) and then expanding the result to get:

                                                                               

               

This equation is a quadratic equation of the form ax2 + bx + c = 0, the solutions of which are given by (see Appendix 1):

 

.

               

For this case, a = 1, b = 1.7×10–4, and c = –2.72×10–6. Substituting these values into the quadratic formula gives:

 

 

 

 

The negative solution is not physically reasonable since there is no such thing as a negative concentration. The solution is x = 1.6×10–3 M.

 

Finally,

 

.

Think About It:

Did the result obtained using the quadratic formula differ significantly from the approximate solution? What if the given Ka value had three or more significant digits instead of two?

b.

Proceed in the same manner as part (a). Note though that the initial concentration is even less than it was in part (a), making it even more necessary to avoid the approximate solution.

 

 

HF(aq) +

H2O(l)

F–(aq)

+   H3O+(aq)

Initial (M):

5.7×10–4

 

 

0

0

Change (M):

–x

 

 

+ x

+ x

Equilibrium (M):

5.7×10–4 – x

 

 

x

x

                                               

 

 

 

 

 

The negative solution is rejected, giving x = 6.3×103 M. Thus,

 

c.

Proceed in the same manner as part (a). Note though that the initial concentration is very large, and we may safely seek an approximate solution.

 

HF(aq) +

H2O(l)

F–(aq)

+   H3O+(aq)

Initial (M):

1.75

 

 

0

0

Change (M):

–x

 

 

+ x

+ x

Equilibrium (M):

1.75 – x

 

 

x

x

 

 

 

16.62

Strategy:

Phenol is a weak acid, and some of it will ionize in solution according to the equilibrium

 

C6H5OH(aq) + H2O(l)    C6H5O–(aq) + H3O+(aq)

 

 

The percent ionization of a weak acid is given by Equation 16.7:

 

,

 

where [HA]0 is the initial concentration of the acid (that is, the concentration assuming there is no ionization). Using the Ka expression above, set up a concentration table and solve for [H+]. Then, use this value to compute the percent ionization.

Solution:

a.

 

 

C6H5OH(aq) +

H2O(l)

C6H5O–(aq)

+   H3O+(aq)

 

Initial (M):

0.56

 

 

0

0

Change (M):

–x

 

 

+ x

+ x

Equilibrium (M):

0.56 – x

 

 

x

x

 

 

Since the initial concentration is very large compared to Ka, assume that 0.56 – x » 0.56. Then,

 

 

b.

Proceed as in part (a).

 

C6H5OH(aq) +

H2O(l)

C6H5O–(aq)

+   H3O+(aq)

 

Initial (M):

0.25

 

 

0

0

Change (M):

–x

 

 

+ x

+ x

Equilibrium (M):

0.25 – x

 

 

x

x

 

 

 

 

c.

Proceed as in part (a).

 

C6H5OH(aq) +

H2O(l)

C6H5O–(aq)

+   H3O+(aq)

 

Initial (M):

1.8×10–6

 

 

0

0

Change (M):

–x

 

 

+ x

+ x

Equilibrium (M):

1.8×10–6 – x

 

 

x

x

 

 

 

16.63

Strategy:

We are asked to compute Ka for the equilibrium

 

HA(aq)  +  H2O(l)      A–(aq)  +  H3O+(aq)

 

 

We are given the percent ionization of the acid, which is related to [H+] and [HA]0 by Equation 16.7:

  or    ,

 

where [HA]0 is the initial concentration of the acid (that is, the concentration assuming there is no ionization). Since the percent ionization (0.92%) is very small, it is safe to assume that

[HA] » [HA]0. Using equation (2) above for [H+], along with the fact that [H+] = [A–], substitute into equation (1) and evaluate Ka.

Solution:

 

 

 

 

16.64

Strategy:

The relevant equilibrium for the weak acid ionization is

                                       

HA(aq)  +  H2O(l)      A–(aq)  +  H3O+(aq)

 

               

where [H+] = [A–] (why?).

 

We are asked to find [HA]0 when the percent ionization is 2.5%, and [HA]0 is related to the percent ionization and [H+] by Equation 16.7:

 

     (2)

               

Since the percent ionization is small (2.5%), assume that [HA] » [HA]0. Solve equation (2) for [H+] and substitute this value into equation (1), then solve the resulting equation for [HA]0.

Solution:

 

 

Substituting [H+] for [A–],  for [H+], and [HA]0 for [HA] in equation (1), we get:

 

 

 

 

 

16.65

Strategy:

We use the pH to determine [H+]eq, and set up a table for the dissociation of the weak acid, filling in what we know.  The initial concentration of weak acid is 0.19 M.  The initial concentrations of H+ and the anion are both zero; and because they are both products of the ionization of the weak acid, their equilibrium concentrations are equal: [H+]eq = [A-]eq.

Solution:

[H+] = 10-4.52 = 3.02 ´ 10-5 M

 

 

HA(aq)

H+(aq)

+   A–(aq)

 

Initial (M):

0.19

 

0

0

Change (M):

­-3.02 ´ 10-5

 

+3.02 ´ 10-5

+3.02 ´ 10-5

Equilibrium (M):

(0.19 - 3.02 ´ 10-5)

 

3.02 ´ 10-5

3.02 ´ 10-5

 

 

The amount of HA ionized is very small compared to the initial concentration:

 

0.19 - 3.02 ´ 10-5 » 0.19

                               

Therefore, = 4.8 ´ 10-9

16.66

First we find the hydrogen ion concentration.

 

[H+] =  10-pH  =  10-6.20  =  6.3 ´ 10-7 M

 

If the concentration of [H+] is 6.3 ´ 10-7 M, that means that 6.3 ´ 10-7 M of the weak acid, HA, ionized because of the 1:1 mole ratio between HA and [H+] . 

 

Setting up a table:

               

               

 

HA(aq)

H+(aq)

+   A–(aq)

Initial (M):

0.010

 

0

0

Change (M):

­–6.3 × 10–7

 

+6.3 × 10–7

+6.3 × 10–7

Equilibrium (M):

≈ 0.010

 

6.3 × 10–7

6.3 × 10–7

 

Substituting into the acid ionization constant expression:

 

 

We have omitted the contribution to [H+] due to water.

16.67

A pH of 3.26 corresponds to a [H+] of 5.5 ´ 10-4 M.  Let the original concentration of formic acid be x.  If the concentration of [H+] is 5.5 ´ 10-4 M, that means that 5.5 ´ 10-4 M of HCOOH ionized because of the 1:1 mole ratio between HCOOH and H+.

 

 

 

HCOOH(aq)   

H+(aq)

+   HCOO-(aq)

 

Initial (M):

x

 

0

0

Change (M):

­-5.5 ´ 10-4

 

+5.5 ´ 10-4

+5.5 ´ 10-4

Equilibrium (M):

x - (5.5 ´ 10-4)

 

5.5 ´ 10-4

5.5 ´ 10-4

 

 

Substitute Ka and the equilibrium concentrations into the ionization constant expression to solve for x.

 

 

 

x  =  [HCOOH]  =  2.3 ´ 10-3 M

16.68

A pH of 5.26 corresponds to a [H+] of 5.5 ´ 10-6 M.  Let the original concentration of weak acid (HA) be x.  If the concentration of [H+] is 5.5 ´ 10-6 M, that means that 5.5 ´ 10-6 M of HA ionized because of the 1:1 mole ratio between HA and H+.


 

 

 

HA(aq)      

H+(aq)

+   A-(aq)

 

Initial (M):

x

 

0

0

Change (M):

­-5.5 ´ 10-6

 

+5.5 ´ 10-6

+5.5 ´ 10-6

Equilibrium (M):

x - (5.5 ´ 10-6)

 

5.5 ´ 10-6

5.5 ´ 10-6

 

 

Substitute Ka and the equilibrium concentrations into the ionization constant expression to solve for x.

 

 

 

x  =  [HA]  =  6.4 ´ 10-6 M

16.69

At 37°C, Kw = 2.5 ´ 10-14.  [H+][OH-] = 2.5 ´ 10-14.  Because in neutral water the concentrations of hydronium and hydroxide are equal, we can solve for pH as follows:

 

[H+] = [OH-] = x

 

x2 = 2.5 ´ 10-14

 

 

pH = -log(1.58 ´ 10-7) = 6.80

16.72

a.

We construct the usual table.

 

 

 

Initial (M):

0.10

 

 

0.00

0.00

Change (M):

–x

 

 

+ x

+ x

Equilibrium (M):

(0.10 - x)

 

 

x

x

 

 

                                                                 

 

 

Assuming (0.10 - x) » 0.10, we have:

 

x  =  1.3 ´ 10-3 M  =  [OH-]

 

pOH  =  -log(1.3 ´ 10-3)  =  2.89

 

pH  =  14.00 - 2.89  =  11.11

b.

By following the identical procedure, we can show:  pH = 8.96.

16.73

Strategy:

Weak bases only partially ionize in water.

 

B(aq) + H2O(l)    BH+(aq) + OH-(aq)

 

Note that the concentration of the weak base given refers to the initial concentration before ionization has started.  The pH of the solution, on the other hand, refers to the situation at equilibrium.  To calculate Kb, we need to know the concentrations of all three species, [B], [BH+], and [OH-] at equilibrium.  We ignore the ionization of water as a source of OH- ions.

Solution:

We proceed as follows.

 

Step 1:   The major species in solution are B, OH-, and the conjugate acid BH+.

 

Step 2:   First, we need to calculate the hydroxide ion concentration from the pH value.  Calculate the pOH from the pH.  Then, calculate the OH- concentration from the pOH.

 

pOH  =   14.00 - pH  = 14.00 - 10.66  =  3.34

 

pOH  =  -log[OH-]

 

-pOH  =  log[OH-]

 

Taking the antilog of both sides of the equation,

 

10-pOH  =  [OH-]

 

[OH-]  =  10-3.34  =  4.57 ´ 10-4 M

 

Step 3:   If the concentration of OH- is 4.57 ´ 10-4 M at equilibrium, that must mean that        4.57 ´ 10-4 M of the base ionized.  We summarize the changes.

 

                                       

B(aq)       +

H2O(l)

BH+(aq)   

+   OH-(aq)

 

Initial (M):

0.30

 

 

0

0

Change (M):

-4.57 ´ 10-4

 

 

+4.57 ´ 10-4

+4.57 ´ 10-4

Equilibrium (M):

0.30 - (4.57 ´ 10-4)

 

 

4.57 ´ 10-4

4.57 ´ 10-4

 

Step 4:   Substitute the equilibrium concentrations into the ionization constant expression to solve for Kb.

 

 

Kb= 6.97 ´ 10-7

16.74

A pH of 11.22 corresponds to a [H+] of 6.03 ´ 10-12 M and a [OH-] of 1.66 ´ 10-3 M.

 

Setting up a table:                                        

                                       

Initial (M):

x

 

 

0.00

0.00

Change (M):

-1.66 ´ 10-3

 

 

+1.66 ´ 10-3

+1.66 ´ 10-3

Equilibrium (M):

x - (1.66 ´ 10-3)

 

 

1.66 ´ 10-3

1.66 ´ 10-3

 

 

 

 

 

 

Assuming 1.66 ´ 10-3 is small relative to x, then

 

x  =  0.15 M  =  [NH3]

16.75

We set up an equilibrium table to solve for [OH-], and use pOH to get pH.  We use B and BH+ to represent the weak base and its protonated form, respectively.

 

                                       

Initial (M):

0.61

 

 

0.00

0.00

Change (M):

-x

 

 

+x

+x

Equilibrium (M):

0.61 - x

 

 

x

x

 

 

                                                                 

 

                               

 

 

 

 

Assuming x is small relative to 0.61, then

 

 

x  =  0.0096 M  =  [OH-]

 

pOH = –log(0.0096) = 2.02

 

pH = 14.00 - 2.02 = 11.98

16.76

We use the pH given in the problem to solve for pOH and then for [OH-].  We then set up an equilibrium table and fill in what we know:

 

pOH = 14.00 - 10.88 = 3.12

 

[OH-] = 10-3.12 = 7.59 ´ 10-4 M

 

 

                                       

Initial (M):

0.19

 

 

0.00

0.00

Change (M):

-7.59 ´ 10-4

 

 

+7.59 ´ 10-4

+7.59 ´ 10-4

Equilibrium (M):

0.19 - 7.59 ´ 10-4

 

 

7.59 ´ 10-4

7.59 ´ 10-4

 

 

 

Kb 3.0 ´ 10-6

16.77

a.

HA has the largest Ka value.  Therefore, A- has the smallest Kb value. 

b.

HB has the smallest Ka value.  Therefore, B- is the strongest base.

16.80

a.

C- has the weakest conjugate acid (HC). 

b.

In order of decreasing base strength: C- > A- > B-.

16.81

Strategy:

We calculate the Kb value for a conjugate base using Equation 16.7.

Solution:

Kb(CN-) = 2.0 ´ 10-5

 

Kb(F-) = 1.4 ´ 10-11

 

Kb(CH3COO-) = 5.6 ´ 10-10

 

Kb(HCO) = 2.4 ´ 10-8

16.84

If  we can assume that the equilibrium concentration of hydrogen ion results only from the first stage of ionization.  In the second stage this always leads to an expression of the type:

 

 

where c represents the equilibrium hydrogen ion concentration found in the first stage.  If  we can assume (c ± y) » c, and consequently

Think About It:

Is this conclusion also true for the second stage ionization of a triprotic acid like H3PO4?

 

16.85

The pH of a 0.040 M HCl solution (strong acid) is:  pH = -log(0.040) = 1.40.  Follow the procedure for calculating the pH of a diprotic acid to calculate the pH of the sulfuric acid solution.

Strategy:

Determining the pH of a diprotic acid in aqueous solution is more involved than for a monoprotic acid.  The first stage of ionization for H2SO4 goes to completion.  We follow the procedure for determining the pH of a strong acid for this stage.  The conjugate base produced in the first ionization (HSO4-) is a weak acid.  We follow the procedure for determining the pH of a weak acid for this stage.

Solution:

We proceed according to the following steps.

 

Step 1:     H2SO4 is a strong acid.  The first ionization stage goes to completion.  The ionization of H2SO4 is

 

H2SO4(aq)  ®  H+(aq) + HSO (aq)

 

The concentrations of all the species (H2SO4, H+, and HSO4-) before and after ionization can be represented as follows.

 

 

   

Initial (M):

0.040

 

0

0

Change (M):

­-0.040

 

+0.040

+0.040

Equilibrium (M):

0

 

0.040

0.040

 

 

Step 2:   Now, consider the second stage of ionization.  HSO4- is a weak acid.  Set up a table showing the concentrations for the second ionization stage.  Let x be the change in concentration.  Note that the initial concentration of H+ is 0.040 M from the first ionization.

 

 

 

Initial (M):

0.040

 

0.040

0

Change (M):

­-x

 

+x

+x

Equilibrium (M):

0.040 - x

 

0.040 + x

x

 

 

Write the ionization constant expression for Ka.  Then, solve for x.  You can find the Ka value in Table 16.8 of the text.

 

 

Since Ka is quite large, we cannot make the assumptions that

 

0.040 - x  »  0.040     and     0.040 + x  »  0.040

 

Therefore, we must solve a quadratic equation.

 

x2 + 0.053x - (5.2 ´ 10-4)  =  0

 

 

 

x  =  8.5 ´ 10-3 M      or      x  =  -0.062 M

 

The second solution is physically impossible because you cannot have a negative concentration.  The first solution is the correct answer.

 

Step 3:   Having solved for x, we can calculate the H+ concentration at equilibrium.  We can then calculate the pH from the H+ concentration.

 

[H+]  =  0.040 M + x  =  [0.040 + (8.5 ´ 10-3)]M  =  0.049 M

 

pH  =  -log(0.049)  =  1.31

Think About It:

Without doing any calculations, could you have known that the pH of the sulfuric acid would be lower (more acidic) than that of the hydrochloric acid?

16.86

There is no H2SO4 in the solution because HSO has no tendency to accept a proton to produce H2SO4.  (Why?)  We are only concerned with the ionization:

 

 

HSO(aq) 

+   SO(aq)

Initial (M):

0.20

 

0.00

0.00

Change (M):

­-x

 

+x

+x

Equilibrium (M):

(0.20 - x)

 

+x

+x

 

 

 

 

Solving the quadratic equation:

 

x  =  [H+]  =  [SO]  =  0.045 M

 

[HSO]  =  (0.20 - 0.045) M  =  0.16 M

16.87

For the first stage of ionization:

                                         

 

   

+   HCO (aq)

 

Initial (M):

0.025

 

0.00

0.00

Change (M):

­-x

 

+x

+x

Equilibrium (M):

(0.025 - x)

 

+x

+x

 

 

 

x  =  1.0 ´ 10-4 M

 

For the second ionization,

 

 

    

 

Initial (M):

1.0 ´ 10-4

 

1.0 ´ 10-4

0.00

Change (M):

­-y

 

+y

+y

Equilibrium (M):

(1.0 ´ 10-4) - y

 

(1.0 ´ 10-4) + y

y

                                         

 

 

y  =  4.8 ´ 10-11 M

 

Since HCO3- is a very weak acid, there is little ionization at this stage.  Therefore we have:

 

[H+]  =  [HCO3-]  =  1.0 ´ 10-4 M  and  [CO32-]  =  y  =  4.8 ´ 10-11 M

 

16.88

Although phosphoric acid is polyprotic, because its first and second ionization constants differ by significantly more than a factor of 1000 (Ka1/Ka2 =120,000), we can determine pH from the first ionization alone.  (The contributions of the second and third ionizations to the hydronium ion concentration are negligible.)  We set up an equilibrium table:

 

 

 

   

+   H2PO (aq)

Initial (M):

0.25

 

0.00

0.00

Change (M):

­-x

 

+x

+x

Equilibrium (M):

(0.25 - x)

 

+x

+x

 

 

We begin by assuming that x is small enough to neglect in the denominator.  This gives

 

x = = 0.00188

 

We then iterate the solution until we get a constant value for x.  For the first iteration, we substitute the first value of x into the denominator and solve for x again:

 

 

x = 0.043

 

Substituting into the denominator again gives the second iteration:

 

 

x = 0.039

 

Third iteration:

 

 

x = 0.040

 

Fourth iteration:

 

 

x = 0.040

 

The value of x converges at 0.040 so [H+] = 0.040 M and pH = -log(0.040) = 1.40.

 

Alternatively, we could have solved for [H+] using the quadratic equation.

16.89

Oxalic acid is diprotic.  Because its first and second ionization constants differ only by a factor of about 1000, we consider both the first and second ionizations to determine [H+].  We set up an equilibrium table:

 

 

  

+  

Initial (M):

0.25

 

0.00

0.00

Change (M):

­-x

 

+x

+x

Equilibrium (M):

(0.25 - x)

 

+x

+x

 

 

Using the quadratic equation to solve for x we find

 

x = 0.099

 

For the second ionization:

 

 

+  

Initial (M):

0.099

 

0.099

0.00

Change (M):

­-y

 

+y

+y

Equilibrium (M):

(0.099 - y)

 

0.099+y

y

 

 

 

We neglect y with respect to 0.099 and find

 

y = 6.1 ´ 10-5

 

The concentration of hydronium ion after the second ionization is therefore

 

0.099 + 6.1 ´ 10-5 » 0.099 M

 

and

 

pH = -log(0.099) = 1.00

 

Note that the second ionization did not contribute enough to the hydronium concentration to change the value of pH.

16.90

1.

The two steps in the ionization of a weak diprotic acid are:

 

H2A(aq) + H2O(l)    H3O+(aq) + HA-(aq)

 

HA-(aq) + H2O(l)    H3O+(aq) + A2-(aq)

 

The diagram that represents a weak diprotic acid is (c).  In this diagram, we only see the first step of the ionization, because HA- is a much weaker acid than H2A.

2.

Both (b) and (d) are chemically implausible situations.  Because HA- is a much weaker acid than H2A, you would not see a higher concentration of A2- compared to HA-.

16.94

a.

The K+ cation does not hydrolyze.  The Br- anion is the conjugate base of the strong acid HBr.  Therefore, Br- will not hydrolyze either, and the solution is neutral, pH » 7.

b.

Al3+ is a small metal cation with a high charge, which hydrolyzes to produce H+ ions.  The NO 

anion does not hydrolyze.  It is the conjugate base of the strong acid, HNO3.  The solution will be

acidic, pH < 7.

c.

The Ba2+ cation does not hydrolyze.  The Cl- anion is the conjugate base of the strong acid HCl.  Therefore, Cl- will not hydrolyze either, and the solution is neutral, pH » 7.

d.

Bi3+ is a small metal cation with a high charge, which hydrolyzes to produce H+ ions.  The NO 

anion does not hydrolyze.  It is the conjugate base of the strong acid, HNO3.  The solution will be

acidic, pH < 7.

16.95

Strategy:

In deciding whether a salt will undergo hydrolysis, ask yourself the following questions:  Is the cation a highly charged metal ion or an ammonium ion?  Is the anion the conjugate base of a weak acid?  If yes to either question, then hydrolysis will occur.  In cases where both the cation and the anion react with water, the pH of the solution will depend on the relative magnitudes of Ka for the cation and Kb for the anion.

Solution:

We first break up the salt into its cation and anion components and then examine the possible reaction of each ion with water.

a.

The Na+ cation does not hydrolyze.  The Br- anion is the conjugate base of the strong acid HBr.  Therefore, Br- will not hydrolyze either, and the solution is neutral.

b.

The K+ cation does not hydrolyze.  The SO anion is the conjugate base of the weak acid HSO and will hydrolyze to give HSO and OH-.  The solution will be basic.

c.

Both the NH and NO ions will hydrolyze.  NH is the conjugate acid of the weak base NH3, and NO is the conjugate base of the weak acid HNO2.  Using ionization constants from Tables 16.6 and 16.7, and Equation 16.7, we find that the Ka of NH4+                           (1.0 ´ 10-14/1.8 ´ 10-5 = 5.6 ´ 10-10) is greater than the Kb of NO                                       (1.0 ´ 10-14/4.5 ´ 10-4 = 2.2 ´ 10-11).  Therefore, the solution will be acidic.

d.

Cr3+ is a small metal cation with a high charge, which hydrolyzes to produce H+ ions.  The NO anion does not hydrolyze.  It is the conjugate base of the strong acid, HNO3.  The solution will be acidic.

16.96

There are two possibilities:  (i)  MX is the salt of a strong acid and a strong base so that neither the cation nor the anion react with water to alter the pH and (ii)  MX is the salt of a weak acid and a weak base with Ka for the acid equal to Kb for the base.  The hydrolysis of one would be exactly offset by the hydrolysis of the other.

16.97

There is an inverse relationship between acid strength and conjugate base strength.  As acid strength decreases, the proton accepting power of the conjugate base increases.  In general the weaker the acid, the stronger the conjugate base.  All three of the potassium salts ionize completely to form the conjugate base of the respective acid.  The greater the pH, the stronger the conjugate base, and therefore, the weaker the acid.

 

The order of increasing acid strength is HZ < HY < HX.

16.98

The acid and base reactions are:

 

                                            acid:           HPO(aq)    H+(aq) + PO(aq)

 

                                            base:          HPO(aq) + H2O(l)    H2PO(aq) + OH-(aq)

 

Ka for HPO is 4.2 ´ 10-13.  Note that HPO is the conjugate base of H2PO, so Kb is 1.6 ´ 10-7.  Comparing the two K's, we conclude that the monohydrogen phosphate ion is a much stronger proton acceptor (base) than a proton donor (acid).  The solution will be basic.

16.99

HCO    H+ + CO                                    Ka  =  4.8 ´ 10-11

 

HCO + H2O    H2CO3 + OH-                    

 

HCO has a greater tendency to hydrolyze than to ionize (Kb > Ka).  The solution will be basic (pH > 7).

16.100

The salt, sodium acetate, completely dissociates upon dissolution, producing 0.36 M [Na+] and 0.36 M [CH3COO-] ions.  The CH3COO- ions will undergo hydrolysis because they are a weak base.

 

                                       

     

 

Initial (M):

0.36

 

 

0.00

0.00

Change (M):

-x

 

 

+x

+x

Equilibrium (M):

(0.36 - x)

 

 

+x

+x

 

 

 

Assuming (0.36 - x) » 0.36, then

 

x  =  [OH-]  =  1.4 ´ 10-5

 

pOH  =  -log(1.4 ´ 10-5)  =  4.85

 

pH  =  14.00 - 4.85  =  9.15

16.101

The salt ammonium chloride completely ionizes upon dissolution, producing 0.42 M [NH] and 0.42 M [Cl-] ions.  NH will undergo hydrolysis because it is a weak acid (NH is the conjugate acid of the weak base, NH3).

 

Step 1:   Express the equilibrium concentrations of all species in terms of initial concentrations and a single unknown x, that represents the change in concentration.  Let (-x) be the depletion in concentration (mol/L) of NH.  From the stoichiometry of the reaction, it follows that the increase in concentration for both H3O+ and NH3 must be x.  Complete a table that lists the initial concentrations, the change in concentrations, and the equilibrium concentrations.

 

                                       

 

 

 

 

 

 

Initial (M):

0.42

 

 

0.00

0.00

Change (M):

-x

 

 

+x

+x

Equilibrium (M):

(0.42 - x)

 

 

+x

+x

 

 

 

 

 

 

 

 

 

Step 2:   You can calculate the Ka value for NH from the Kb value of NH3.  The relationship is

 

Ka ´ Kb  =  Kw

                                                or

 

 

Step 3:   Write the ionization constant expression in terms of the equilibrium concentrations.  Knowing the value of the equilibrium constant (Ka), solve for x.

 

 

 

x  =  [H+]  =  1.5 ´ 10-5 M

 

pH  =  -log(1.5 ´ 10-5)  =  4.82

 

Since NH4Cl is the salt of a weak base (aqueous ammonia) and a strong acid (HCl), we expect the solution to be slightly acidic, which is confirmed by the calculation.

16.102

Aqueous NaF contains the F- ion, which is the conjugate base of a weak acid (HF)  We use Equation 16.7 to get Kb for the F- ion: 

 

Kb(F-) = 1.4 ´ 10-11

 

We construct an equilibrium table to determine the concentration of hydroxide:

 

                                       

F-(aq)      +

H2O(l)

HF(aq)

+   OH–(aq)

 

Initial (M):

0.082

 

 

0.00

0.00

Change (M):

-x

 

 

+x

+x

Equilibrium (M):

0.082 - x

 

 

x

x

 

 

 

 

 

 

Neglecting x in the denominator gives

 

x =  = 1.07 ´ 10-6

 

Therefore, [OH-] =  1.07 ´ 10-6 M, pOH = -log(1.07 ´ 10-6) = 5.97

 

and

 

pH = 14.00 - 5.97 = 8.03

16.103

Strategy:

We are asked to determine the pH of a salt solution.  We must first determine the identities of the ions in solution and decide which, if any, of the ions will hydrolyze.  In this case, the ions in solution are C2H5NH and I-.  The iodide ion, I-, is the anion of the strong acid, HI.  Therefore it will not hydrolyze.  The C2H5NH ion, however, is the conjugate acid of the weak base C2H5NH2 (ethylamine). It hydrolyzes to produce H+ and the weak base C2H5NH2:

 

C2H5NH(aq)  +  H2O(l)       C2H5NH2(aq)  +  H3O+(aq)

 

In order to determine the pH, we must first determine the concentration of hydronium ion produced by the hydrolysis.  For this, we need the ionization constant, Ka, of the C2H5NH ion.

Solution:

Using the Kb value for ethylamine and Equation 16.7, we calculate Ka for C2H5NH:

 

Ka =

 

We then construct an equilibrium table and fill in what we know to find [H3O+]:

 

 

 

 

 

 

 

 

 

 

 

 

 

Initial (M):

0.91

 

 

0.00

0.00

Change (M):

-x

 

 

+x

+x

Equilibrium (M):

(0.91 - x)

 

 

x

x

 

 

Assuming that x is very small compared to 0.91, we can write

 

 

and

 

x = [H3O+] = 4.05 ´ 10-6 M

 

pH = -log(4.05 ´ 10-6) = 5.39

16.106

Metal ions with high oxidation numbers are unstable.  Consequently, these metals tend to form covalent bonds (rather than ionic bonds) with oxygen.  Covalent metal oxides are acidic while ionic metal oxides are basic.  The latter oxides contain the O2- ion which reacts with water as follows:

 

O2- + H2O  ®  2OH-

16.107

The most basic oxides occur with metal ions having the lowest positive charges (or lowest oxidation numbers).

a.

Al2O3 < BaO  < K2O

b.

CrO3 < Cr2O3 < CrO

16.108

a.

2HCl(aq) + Zn(OH)2(s)  ®  ZnCl2(aq) + 2H2O(l)

b.

2OH-(aq) + Zn(OH)2(s)  ®  Zn(OH)(aq)

16.109

Al(OH)3 is an amphoteric hydroxide.  The reaction is:

 

Al(OH)3(s) + OH-(aq)  ®  Al(OH)(aq)

 

This is a Lewis acid-base reaction.  Can you identify the acid and base?

16.110

According to Section 16.11, metal oxides change from basic to amphoteric to acidic moving from left to right across a period.  Metal oxides become more basic moving from top to bottom within a group.

16.111

a.

The basic metallic oxides react with water to form metal hydroxides:

 

Li2O(s) + H2O(l)    2LiOH(aq)

b.

As in part (a) the basic metal oxide reacts to form a metal hydroxide:

 

CaO(s) + H2O(l)    Ca(OH)2(aq)

c.

The reaction between an acidic oxides and water is:

 

SO3(g) + H2O(l)    H2SO4(aq)

16.114

a.

Lewis acid; see the reaction with water shown in Section 16.12 of the text.

b.

Lewis base; water combines with H+ to form H3O+.

c.

Lewis base.

d.

Lewis acid; SO2 reacts with water to form H2SO3.  Compare to CO2 above.  Actually, SO2 can also act as a Lewis base under some circumstances.

e.

Lewis base; see the reaction with H+ to form ammonium ion.

f.

Lewis base; see the reaction with H+ to form water.

g.

Lewis acid; does H+ have any electron pairs to donate?

h.

Lewis acid; compare to the example of NH3 reacting with BF3.

16.115

AlCl3 is a Lewis acid with an incomplete octet of electrons and Cl- is the Lewis base donating a pair of electrons.

16.116

a.

Both molecules have the same acceptor atom (boron) and both have exactly the same structure (trigonal planar).  Fluorine is more electronegative than chlorine so we would predict based on electronegativity arguments that boron trifluoride would have a greater affinity for unshared electron pairs than boron trichloride.

b.

Since it has the larger positive charge, iron(III) should be a stronger Lewis acid than iron(II).

16.117

By definition Brønsted acids are proton donors, therefore such compounds must contain at least one hydrogen atom.  In Problem 16.114, Lewis acids that do not contain hydrogen, and therefore are not Brønsted acids, are CO2, SO2, and BCl3.  Can you name others?

16.118

a.

Fe is the Lewis acid; CO is the Lewis base. 

b.

BCl3 is the Lewis acid; NH3 is the Lewis base. 

c.

Hg2+ is the Lewis acid; I- is the Lewis base.

16.119

a.

AlBr3 is the Lewis acid; Br- is the Lewis base. 

b.

Cr is the Lewis acid; CO is the Lewis base. 

c.

Cu2+ is the Lewis acid; CN- is the Lewis base.

16.120

Loss of the first proton from a polyprotic acid is always easier than the subsequent removal of additional protons.  The ease with which a proton is lost (i.e., the strength of the acid) depends on the stability of the anion that remains.  An anion with a single negative charge is more easily stabilized by resonance than one with two negative charges.

16.121

A strong acid, such as HCl, will be completely ionized, choice (b).

 

A weak acid will only ionize to a lesser extent compared to a strong acid, choice (c).

 

A very weak acid will remain almost exclusively as the acid molecule in solution.  Choice (d) is the best choice.

16.122

The direction should favor formation of F-(aq) and H2O(l).  Hydroxide ion is a stronger base than fluoride ion, and hydrofluoric acid is a stronger acid than water.

16.123