Acids and Bases 2. HCl 3. H20 4. H30+ - West Virginia University

CHAPTER 2 Acids and Bases

1.

a. 1. +NH4

b. 1. -NH2

2. HCl 2. Br-

3. H20 3. N03-

4. H30+ 4. HO-

2.

if the lone pairs are not shown:

a. CH30H as an acid

CH30H + NH3

CH30 - + N+ H4

CH30H as a base b. NH3 as an acid

CH30H + HCl NH3 + HO-

+

CH30H + CI-

I

H

-NH2 + H20

NH3 as a base

NH3 + HBr ~

NH4 + Br-

if the lone pairs are shown:

a. CH30H as an acid CH30H as a base

CH3S?H + NH3

..

..

CH3QH +H~l:

CH3Q:- + NH4

+

CH30H + :Cl:-

I

H

b. NH3 as an acid NH3 as a base

NH3 + HO:-

NH3 + HBr:

:NH2 + H20: N+ H4 + :Br:

3.

a. The lower the pKa, the stronger the acid,so the compound with apKa = 5.2 is the stronger

acid.

b. The greater the dissociation constant, the stronger the acid, so the compound with an acid dissociation constant = 3.4 X 10-3 is the stronger acid.

4.

pKa = 4.82; therefore, K? = l.51 X 10-5

The greater the acid dissociation constant, the stronger the acid, so butyric acid (K; = 1.51 x 10-5)

is a weaker acid than vitamin C (K, = 6.76 x 10-5).

or

The lower the pKa, the stronger the acid, so butyric acid (pKa = 4.82) is a weaker acid than vitamin C (pKa = 4.17).

13

14 Chapter 2

b. HC03 - + HCl ~ c. C032- + 2 HCl ~

6.

A neutral solution has pH = 7. Solutions with pH < 7 are acidic; solutions with pH > 7 are basic.

a. basic

b. acidic

c. basic

7.

a. CH3COO- is the stronger base.

Because HCOOH is the stronger acid, it has the weaker conjugate base.

b. -NH2 is the stronger base. Because H20 is the stronger acid, it has the weaker conjugate base.

c. H20 is the stronger base. +

Because CH30H2 is the stronger acid, it has the weaker conjugate base.

8. The conjugate acids have the following relative strengths:

+ CH30H2

o II

> CH3COH

+ > CH3NH3 > CH30H

> CH3NH2

The bases, therefore, have the following relative strengths:

+

9.

CH3NH2 + CH30H ~

CH3NH3 + CH~O .)

pKa = 40 pKa = 15.5

The stronger acid of the two reactants will be the acid (that is, it is the one that will donate a proton); the weaker acid will accept the proton.

Chapter 2 15

10. Notice that in each case, the equilibrium goes away from the strong acid and toward the weak

acid.

0

II

a. CH3COH

+ H2O

....--- -1000.

0

II

CH3CO- + H30+

pKa =4.8

pKa = -1.7

0

II

CH3COH

+ H30+

pKa =-1.7

CH30H + HO-

pKa = 15.5

....--- -1000.

-~ -

+OH

II

CH3COH

pKa = -6.1

+ H2O

CH3O-

+ H2O

pKa = 15.7

The pKa values are so close that there will be essentially no difference in the equilibrium arrows.

CH30H

+ H3OpKa = -1.7

CH3NH2 + HOpKa =40

CH3NH2 + H30+ pKa =-1.7

b. HCl + H2O pKa =-7

....--- -1000.

.......-.?-.-

--..-.-."---.-..-."--

+ CH30H

H

+ H2O

pKa = -2.5

-

CH3NH + H2O

pKa = 15.7

+

CH3NH3 + H2O

pKa = 10.7

H30+ + cr

pKa = -1.7

NH3

+ H2O pKa = 15.7

....--- -1000.

+NH4 + HO-

pKa = 9.4

11. a. HBr is the stronger acid because bromine is larger than chlorine.

+

b. CH3CH2CH20H2

nitrogen.

is the stronger acid because oxygen is more electronegative than

c. The compound on the right (an alcohol) is a stronger acid than the compound on the left (an amine) because oxygen is more electronegative than nitrogen.

12. a. Because HF is the weakest acid, F- is the strongest base. b. Because HI is the strongest acid, I- is the weakest base.

16 Chapter 2

13. a. oxygen

As you saw in Problem 11, the size of an atom is more important than its electronegativity in determining stability. So even though oxygen is more electronegative than sulfur, H2S is a stronger acid than H20, and CH3SH is a stronger acid than CH30H. Because the sulfur atom is larger, the electrons associated with the negatively charged sulfur are spread out over a greater volume, thereby causing it to be a more stable base. The more stable the base, the stronger is its conjugate acid.

14. a. HO-; if the atoms are the same, the negatively charged one is a stronger base than the neutral one.

b. NH3; H30+ is a stronger acid than +NH4 because oxygen is more electronegative than nitrogen; the stronger the acid, the weaker its conjugate base.

c. CH30-;

o II

CH3COH is a stronger acid than CH30H.

d. CH30-; CH3SH is a stronger acid than CH30H because sulfur is larger than oxygen.

15. a. CH3COO-

+

b. CH3CH2NH3

16. a. 1. neutral 2. neutral 3. charged 4. charged 5. charged 6. charged

c. H2O d. Br-

e. +NH4

f. HC N

b. 1. charged 2. charged 3. charged 4. charged 5. neutral 6. neutral

g. N02h. N03-

c. 1. neutral 2. neutral 3. neutral 4. neutral 5. neutral 6. neutral

17. a. Because the pH of the solution is greater than the pKa value of the carboxylic acid group, the group will be in its basic form (without its proton). Because the pH of the solution is less than the pKa value of the ammonium group, the group will be in its acidic form (with its proton).

b. No, because that would require a weaker acid (the +NH3 group) to lose a proton more readily than a stronger acid (the COOH group).

18. a. The basic form of the buffer (CH3COO-) removes added H+ .

b. The acidic form of the buffer (CH3COOH) removes added HO-.

CH3COOH + HO- ~

CH3COO- + H20

Due to the rapid equilibrium, the added H+ or HO- readily reacts with the species (CH3COO- or CH3COOH) in the solution and thereby the effect on the solution's pH is minimized.

19.

a. ZnCI2

..

+ CH3QH

.-.-.-.-.--..-.-..-.

-

ZnCI2

~

1+ CH3QH

b.

FeBr3 + ~.

::8r:-

~

-

FeBr3

I

:Br:

c. A~ICI3 + :C..I: ~.....---- AI ICl-3

:C]:

Chapter 2 17

20. a, b, c, and hare Brensted acids (protonating-donating acids). Therefore, they react with HOby donating a proton to it.

d, e, f, and g are Lewis acids. They react with HO- by accepting a pair of electrons from it.

I

a. CH30- + H2O

e. CH30H

b. NH3 + H2O

f. HO-FeBr3

c. CH3NH2 + H2O

g. HO-AIC13

d. HO-BF3

h. CH3COO- + H2O

21. Ifthe pH of the solution is less than the pKa of the compound, the compound will be in its acidic

form (with its proton). If the pH of the solution is greater than the pKa of the compound, the

compound will be in its basic form (without its proton).

o

II

+

a. at pH = 3 CH3COH

b. at pH = 3

CH3CH2NH3

c. at pH = 3 CF3CH20H

?II _

at pH = 6 CH3CO

at pH = 6

+

CH3CH2NH3

at pH = 6 CF3CH20H

?II

at pH = 10 CH3CO-

+

at pH= 10 CH3CH2NH3

at pH = 10 CF3CH20H

?II

at pH = 14 CH3CO-

atpH= 14 CH3CH2NH2

atpH= 14 CF3CH20-

0

II

22. a. CH3COH + CH3O-

---.-.--..

?II

CH3CO- + CH30H

b. CH3CH20H + -NH2 ---.-.--.. CH3CH20- + NH3

0

II

c. CH3COH + CH3NH2

0

---.-.--..

II

CH3CO-

+

+

CH3NH3

d. CH3CH20H + HCl

---.-.--..'

+

CH3CH20H2

+

cr

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