CHAPTER 16 ACIDS AND BASES

ACIDS AND BASES

PRACTICE EXAMPLES

1A (E)

(a) In the forward direction, HF is the acid (proton donor; forms F^), and H2O is the base (proton acceptor; forms H3O^). In the reverse direction, F^ is the base (forms HF), accepting a proton from H3O^, which is the acid (forms H2O).

(b) In the forward direction, HSO4 is the acid (proton donor; forms SO42), and NH3 is the base (proton acceptor; forms NH4). In the reverse direction, SO42 is the base (forms HSO4), accepting a proton from NH4, which is the acid (forms NH3).

(c) In the forward direction, HCl is the acid (proton donor; forms Cl^), and C2H3O2 is the base (proton acceptor; forms HC2H3O2). In the reverse direction, Cl^ is the base (forms HCl), accepting a proton from HC2H3O2, which is the acid (forms C2H3O2).

1B (E) We know that the formulas of most acids begin with H. Thus, we identify HNO₂ and HCO₃^ as acids.

     

     

+

2 2 2 3

2 +

3 2 3 3

HNO aq + H O(l) NO aq + H O aq ; HCO aq + H O(l) CO aq + H O aq



 





A negatively charged species will attract a positively charged proton and act as a base.

Thus PO₄^3 and HCO₃^ can act as bases. We also know that PO₄^3 must be a base because it cannot act as an acid—it has no protons to donate—and we know that all three species have acid-base properties.

     

     

3 2

4 2 4

3 2 2 3 2 2

PO aq + H O(l) HPO aq + OH aq ;

HCO aq + H O(l) H CO (aq) CO H O aq + OH aq

  

   





Notice that HCO₃^ is the amphiprotic species, acting as both an acid and a base.

2A (M) H O₃ ⁺ is readily computed from pH: H O₃ ⁺ = 10^pH H O₃ ⁺ = 10^2.85= 1.4 10 ^3M.

OH^ can be found in two ways: (1) from K_w = H O OH₃ ^{+ } , giving

14 12

w + 3 3

1.0 10

OH = = = 7.1 10

1.4 10 H O

K ^

 



   

    

M, or (2) from pH pOH+ = 14.00 , giving

pOH= 14.00pH= 14.00 2.85 = 11.15 , and then OH ^ = 10^pOH = 10^11.15 = 7.1 10 ^12 M.

2B (M) H O₃ ⁺ is computed from pH in each case: H O₃ ⁺ = 10^pH

H O M H O M

conc dil

3 + .

2.50 3

3 + .

3.10 4

= 10^ = 3.2 10 ^ = 10^ = 7.9 10 ^

All of the H O₃ ⁺ in the dilute solution comes from the concentrated solution.

3 +

+ 3 3 +

3 3

3.2 10 mol H O

amount H O = 1.00 L conc. soln = 3.2 10 mol H O 1 L conc. soln

 

  

Next we calculate the volume of the dilute solution.

3 +

3 4 +

3

1 L dilute soln

volume of dilute solution = 3.2 10 mol H O = 4.1 L dilute soln 7.9 10 mol H O



  

 Thus, the volume of water to be added is = 3.1 L.

Infinite dilution does not lead to infinitely small hydrogen ion concentrations. Since dilution is done with water, the pH of an infinitely dilute solution will approach that of pure water, namely pH = 7.

3A (E) pH is computed directly from the equation below.

 

+ +

3 3

H O , pH = log H O = log 0.0025 = 2.60

     

    .

We know that HI is a strong acid and, thus, is completely dissociated into H O₃ ⁺ and I^. The consequence is that I^ = H O₃ ⁺ = 0.0025 M. OH^ is most readily computed from pH: pOH = 14.00 pH = 14.00 2.60 = 11.40; OH   ^= 10^pOH = 10^11.40 = 4.0 10 ^12M 3B (M) The number of moles of HCl(g) is calculated from the ideal gas law. Then H O₃ ⁺ is

calculated, based on the fact that HCl(aq) is a strong acid (1 mol H O₃ ⁺ is produced from each mole of HCl).

+ 3 +

3

747 mmHg 1atm 0.535L

760 mmHg moles HCl(g) =

0.08206 L atm

(26.5 273.2) K mol K

moles HCl(g) 0.0214 mol HCl(g) = 0.0214 mol H O when dissolved in water [H O ] 0.0214 mol pH = -log(0.0214) = 1.670

 

 

 

 

 





4A (E) pH is most readily determined from pOH=log OH^ . Assume Mg OH

b g

2 is a strong base.

   

   

 

2 2

2 2

9.63 mg Mg OH 1000 mL 1g 1 mol Mg OH 2 mol OH OH =

100.0 mL soln 1 L 1000 mg 58.32g Mg OH 1 mol Mg OH OH = 0.00330 M; pOH = log 0.00330 = 2.481

pH = 14.000 pOH = 14.000 2.481 = 11.519

 



     

 

  

 

 

4B (E) KOH is a strong base, which means that each mole of KOH that dissolves produces one mole of dissolved OH aq^

b g

. First we calculate OH^ and the pOH. We then use

pH pOH+ = 14.00 to determine pH.

OH g KOH

g soln

mol KOH g KOH

mol OH mol KOH

g soln mL soln

mL

L M

 

   

= 3.00 100.00

1 56.11

1 1

1.0242 1

1000

1 = 0.548

 

pOH = log 0.548 = 0.261 pH = 14.000 pOH = 14.000 0.261 = 13.739 

5A (M) H O₃ ⁺ = 10^pH = 10^4.18 = 6.6 10 ^5 M.

Organize the solution using the balanced chemical equation.

Equation: HOCl aq + H O(l)

 

2   H O aq₃ ⁺

b g

+ OCl aq^

b g

Initial: 0.150 M ^— 0 M 0 M

Changes: 6.6 10 M ^{5 —} + 6.6 10 M ^5 + 6.6 10 M ^5 Equil: ≈ 0.150 M ^— 6.6 10 M ^5 6.6 10 M ^5



⁺

 

⁵



⁵



3 6.6 10 6.6 10 8

H O OCl

= 2.9 10

HOCl 0.150

Ka

 



      

   

  

5B (M) First, we use pH to determine OH^ . pOH= 14.00pH= 14.00 10.08 = 3.92 . OH^ = 10^pOH = 10^3.92 = 1.2 10 ^4 M. We determine the initial concentration of cocaine and then organize the solution around the balanced equation in the manner we have used before.

17 21 4 17 21 4

17 21 4

17 21 4

0.17 g C H O N 1000 mL 1mol C H O N

[C H O N] = = 0.0056 M

100 mL soln  1L 303.36 g C H O N

Equation: C H O N(aq) + H O(l) _{17 21 4 2}  C H O NH aq_{17 21 4} ⁺

b g

+ OH aq^

b g

Initial: 0.0056 M ^— 0 M 0 M

Changes: 1.2 10 M ^{4 —} +1.2 10 M ^4 +1.2 10 M ^4 Equil: ≈ 0.0055 M ^— 1.2 10 M ^4 1.2 10 M ^4

K_b    

 

  

C H O NH OH 

C H O N

17 21 4 +

17 21 4

4 4

1.2 10 1.2 10 6

0 0055 2 6 10

c hc h

. .

6A (M) Again we organize our solution around the balanced chemical equation.

Equation:

Initial: HC H FO (aq) + H O(l)^{2 2 2 2}

0.100 M ^—

 H O₃ ⁺(aq) +

 0M

C H FO_{2 2 2}^(aq) 0 M

Changes: x M ^— +x M +x M

Equil:

b

0.100 x

g

^{M — x M x M}

 

+

3 2 2 2 3

a

2 2 2

H O C H FO

= ; therefore, 2.6 10

0.100 HC H FO

K x x

x







    

     



 

 

We can use the 5% rule to ignore x in the denominator. Therefore, x = [H3O⁺] = 0.016 M, and pH = –log (0.016) = 1.8. Thus, the calculated pH is considerably lower than 2.89 (Example 16-6).

6B (M) We first determine the concentration of undissociated acid. We then use this value in a set-up that is based on the balanced chemical equation.

9 7 4 9 7 4

9 7 4

0.500 g HC H O 1 mol HC H O 1

2aspirin tablets× × × = 0.0171M

tablet 180.155g HC H O 0.325L Equation:

Initial: HC H O (aq) + H^{9 7 4} 2O(l) 0.0171 M ^—

  H O₃ ⁺(aq) +

0 M ^{9 7 4}

C H O (aq)^ 0 M Changes: x M ^— +x M +x M Equil:

b

0.0171 x

g

^{M — x M x M}

+

3 9 7 4 4

a

9 7 4

H O C H O

= = 3.3 10 =

HC H O 0.0171

K x x

x





 

 

 

 

  

    



2+ 3.3 10 4 5.64 10 = 06

x  ^   ^ (find the physically reasonable roots of the quadratic equation)

4 7 5

3.3 10 1.1 10 2.3 10

0.0022 M; pH log (0.0022) 2.66 x 2

  

     

    

7A (M) Again we organize our solution around the balanced chemical equation.

Equation:

Initial:

Changes

2 2 2 2

HC H FO (aq) + H O(l)

0.015 M ^—

x M ^—



H3O⁺ (aq) +

0 M +x M

C H FO_{2 2 2}^(aq) 0 M

+x M Equil:

b

0.015 x

g

^{M — x M x M}

  ^{  }

+ 2

3 2 2 2 3

a

2 2 2

H O C H FO

= = 2.6 10 =

HC H FO 0.015 0.015

x x x

K x



  

 

     



x= x²  0 015 2 6 10.  .  ^3 0 0062. M = [H O ]₃ ⁺ Our assumption is invalid:

0.0062 is not quite small enough compared to 0.015 for the 5% rule to hold. Thus we use another cycle of successive approximations.

  

  

3 3 +

a 3

3 3 +

a 3

= = 2.6 10 = (0.015 0.0062) 2.6 10 0.0048 M = [H O ] 0.015 0.0062

= = 2.6 10 = (0.015 0.0048) 2.6 10 0.0051 M = [H O ] 0.015 0.0048

K x x x

K x x x

 

 

    



    



  

3 3 +

a = = 2.6 10 = (0.015 0.0051) 2.6 10 0.0051 M = [H O ]3

0.015 0.0051

K x x  ^ x    ^ 



Two successive identical results indicate that we have the solution.

pH=log H O₃ ⁺ =log

b

0.0051 = 2.29

g

. The quadratic equation gives the same result (0.0051 M) as this method of successive approximations.

7B (M) First we find [C5H11N]. We then use this value as the starting base concentration in a set-up based on the balanced chemical equation.

C H N mg C H N

mL soln

mmol C H N

mg C H N M

5 11 5 11 5 11

5 11

=114 315

1

85.15 = 0.00425

 Equation:

Initial:

Changes:

Equil:

5 11 2

C H N(aq) + H O(l) 0.00425 M ^—

x M ^—



^0.00425x



^{M —}

 C H NH (aq) +_{5 11} ⁺ 0 M

+ Mx Mx

OH^(aq)

0 M + Mx Mx

 

+

5 11 3

b

5 11

C H NH OH

= = 1.6 10 =

C H N 0.00425 0.00425

x x x x

K x



      

     

 We assumed that x 0.00425 x 0 0016 0 00425 0 0026.  .  . M The assumption is not valid. Let’s assume x  0 0026. x 0 0016 0 00425 0 0026. ( .  . ) 0 0016. Let’s try again, with x 0 0016.

x 0 0016 0 00425 0 0016. ( .  . ) 0 0021. Yet another try, with x 0 0021. x 0 0016 0 00425 0 0021. ( .  . ) 0 0019. The last time, with x 0 0019.

0.0016 (0.00425 0.0019) 0.0019 M = [OH ]

x   ^

pOH= log[H O ] 3 ^  log(0.0019) 2.72 pH= 14.00pOH= 14.00 2.72 = 11.28 We could have solved the problem with the quadratic formula roots equation rather than by successive approximations. The same answer is obtained. In fact, if we substitute

x = 0.0019 into the K_b expression, we obtain 0.0019 / 0.00425 0.0019 = 1.5 10

b g b

^{2 }

g

^{ 3}

compared to K_b = 1.6 10 ^3. The error is due to rounding, not to an incorrect value.

Using x = 0.0020 gives a value of 1.8 10 ^3, while using x = 0.0018 gives 1.3 10 ^3.

8A (M) We organize the solution around the balanced chemical equation; a M is [HF]initial. Equation:

Initial:

Changes:

Equil:

HF(aq) + H O(l) 2

M

a ^—

M

x ^—



^{a x}



^{M —}

  H O (aq)₃ ⁺

0 M + Mx Mx

+

F (aq)^ 0 M + Mx x M

  ^{  }

+ 2

3 4

a

H O F

= = = 6.6 10

HF

x x x

K a x a



    

     

 x= a6 6 10.  ^4 For 0.20 M HF, a = 0.20 M x= 0.20 6.6 10  ^4 0.011 M

0.011 M

% dissoc= 100%=5.5%

0.20 M 

For 0.020 M HF, a = 0.020 M x= 0 020 6 6 10.  .  ^4 0 0036. M

We need another cycle of approximation: x = ( .0 020 0 0036 . )6 6 10.  ^4 0 0033. M Yet another cycle with x 0.0033 M : x = ( .0 020 0 0033 . )6 6 10.  ^4 0 0033. M

0.0033M

% dissoc = 100% = 17%

0.020M 

As expected, the weak acid is more dissociated.

8B (E) Since both H O₃ ⁺ and C H O_{3 5 3}^ come from the same source in equimolar amounts, their concentrations are equal. H O₃ ⁺ = C H O_{3 5 3}^ = 0.067 0.0284 M= 0.0019M

K_a H O C H O HC H O

= = 0.0019 0.0019

0.0284 0.0019 = 1.4 10

3 +

3 5 3

3 5 3

4



LNM OQP

b gb

^

g

^{ }

9A (M) For an aqueous solution of a diprotic acid, the concentration of the divalent anion is very close to the second ionization constant:

2

6

2 a

OOCCH COO K = 2.0 10 M

  

   

  . We

organize around the chemical equation.

Equation:

Initial:

Change:

Equil:

 

2 2 2

CH COOH (aq) + H O(l) 1.0 M ^—

x M ^— (1.0 –x) M ^—



+

H O (aq) 3

0 M + Mx

M x

+

 

2 2

HCH COO ^(aq) 0 M

+ Mx M x

3 2 2 2 3

2 2

[H O ][HCH (COO) ] ( )( )

1.4 10 [CH (COOH) ] 1.0 1.0

a

x x x

K x

 

     



3 2 +

3 2

1.4 10 3.7 10 M [H O ] [HOOCCH COO ]

x  ^   ^   ^ x 1.0M is a valid assumption.

9B (M) We knowKa₂ 



doubly charged anion



for a polyprotic acid. Thus,

2

5 2

a = 5.3 10 C O2 4

K  ^   ^ . From the pH, H O^{3 +}= 10^pH = 10^0.67 = 0.21 M. We also recognize that HC O_{2 4}^ = H O₃ ⁺ , since the second ionization occurs to only a very small extent. We note as well that HC O_{2 4}^ is produced by the ionization of H C O_{2 2 4}. Each mole of HC O_{2 4}^ present results from the ionization of 1 mole of H C O_{2 2 4}. Now we have sufficient information to determine the K . _a1

1

+

3 2 4 2

a

2 2 4

H O HC O 0.21 0.21

= = = 5.3 10

H C O 1.05 0.21

K





 

 

 

 

  

    



10A (M) H SO_{2 4} is a strong acid in its first ionization, and somewhat weak in its second, with

2

2

a = 1.1 10 = 0.011

K  ^ . Because of the strong first ionization step, this problem involves determining concentrations in a solution that initially is 0.20 M H O₃ ⁺ and 0.20 M HSO₄^. We base the set-up on the balanced chemical equation.

Equation: _{HSO aq H O}

4^

b g

+ 2 ^{(l)  H O aq3 +}

b g

⁺ SO (aq) 4^2-

Initial: 0.20 M ^— 0.20 M 0 M Changes: x M ^— + Mx + Mx Equil:



^0.20x



^{M —}



^{0.20 +x}



^{M x M}

 

2

+ 2

3 4

a

4

H O SO 0.20 + 0.20

= = = 0.011 , assuming that 0.20M.

0.20 0.20

HSO

= 0.011M Try one cycle of approximation:

x x x

K x

x x





 

  

    

  

 



0.011 0.20 + 0.011

0.20 0.011 =0.21

0.19 = 0.19 0.011

0.21 = 0.010

 

b g



b g

^{x x x M}

The next cycle of approximation produces the same answer 0.010M = SO ₄^2 ,

+

3 4

H O = 0.010 + 0.20 M = 0.21 M, HSO ^= 0.20 0.010 M = 0.19 M

  

   

10B (M) We know that H SO_{2 4} is a strong acid in its first ionization, and a somewhat weak acid in its second, with K_a2 = 1.1 10 = 0.011 ^2 . Because of the strong first ionization step, the problem essentially reduces to determining concentrations in a solution that initially is 0.020 M H O₃ ⁺ and 0.020 M HSO₄^. We base the set-up on the balanced chemical equation. The result is solved using the quadratic equation.

Equation: _{HSO aq H O}

4^

b g

+ 2 ^{(l)  H O aq3 +}

b g

^{+ SO}4 ^aq 2

b g

Initial: 0.020 M ^— 0.020 M 0 M Changes: x M ^— + Mx + Mx Equil:



^0.020x



^{M —}



^{0.020 +x}



^{M Mx}

 

2

+ 2

3 4 2 4

a

4

H O SO 0.020 +

= = = 0.011 0.020 + = 2.2 10 0.011

0.020 HSO

K x x x x x

x







 

 

     



 

 

+

4 3

HSO ^ = 0.020 0.0060 = 0.014 M H O = 0.020 + 0.0060 = 0.026 M

    

 

(The method of successive approximations converges to = 0.006 Mx in 8 cycles.) 11A (E)

(a) CH NH NO_{3 3}⁺ ₃^ is the salt of the cation of a weak base. The cation, CH NH , will _{3 3}⁺ hydrolyze to form an acidic solution



CH NH + H O^{3 3+ 2} CH NH + H O^{3 2 3 +}



, while NO3

–, by virtue of being the conjugate base of a strong acid will not hydrolyze to a detectable extent. The aqueous solutions of this compound will thus be acidic.

(b) NaI is the salt composed of the cation of a strong base and the anion of a strong acid, neither of which hydrolyzes in water. Solutions of this compound will be pH neutral.

(c) NaNO₂ is the salt composed of the cation of a strong base that will not hydrolyze in water and the anion of a weak acid that will hydrolyze to form an alkaline solution



^{NO + H O2 2 HNO + OH2 }



. Thus aqueous solutions of this compound will be basic (alkaline).

11B (E) Even without referring to the K values for acids and bases, we can predict that the reaction that produces H O₃ ⁺ occurs to the greater extent. This, of course, is because the pH is less than 7, thus acid hydrolysis must predominate.

We write the two reactions of H PO_{2 4}^ with water, along with the values of their equilibrium constants.

 

⁺

 

²

 

₂ ⁸

2 4 2 3 4 a

H PO ^ aq + H O(l)H O aq + HPO ^ aq K = 6.3 10 ^

     

1

14 12

2 4 2 3 4 b w 3

a

1.0 10

H PO aq + H O(l) OH aq + H PO aq = = = 1.4 10 7.1 10

K K K

   



 

 



As predicted, the acid ionization occurs to the greater extent.

2 2

4

0.031 0.00096 + 0.00088

+ 0.031 0.00022 = 0 = = 0.0060 M = SO

x x x   2  ^

12A (M) From the value of pK we determine the value of K_{b b} and then K_a for the cation.

cocaine _b ^p _a ^w

b

: = 10 = 10 = 3.9 10 = =1.0 10

3.9 10 = 2.6 10

8.41 9 14

9

K K K 6

K

 K   

 

 

 

codeine _b ^p _a ^w

b

: = 10 = 10 = 1.1 10 = =1.0 10

1.1 10 = 9.1 10

7.95 8 14

8

K K K 7

K

 K   



  

 

(This method may be a bit easier:

pK_a = 14.00pK_b = 14.00 8.41 = 5.59, K_a = 10^5.59 = 2.6 10 ^6) The acid with the larger K_a will produce the higher H⁺ , and that solution will have the lower pH. Thus, the solution of codeine hydrochloride will have the higher pH (i.e., codeine hydrochloride is the weaker acid).

12B (E) Both of the ions of NH CN aq₄

b g

react with water in hydrolysis reactions.

     

¹⁴

+ + w 10

4 2 3 3 a 5

b

1.0 10

NH aq + H O(l) NH aq + H O aq = = = 5.6 10

1.8 10 K K

K

 



 

 



 

2

   

b ^{w 14}10 ⁵

a

1.0 10

CN aq + H O(l) HCN aq + OH aq = = = 1.6 10

6.2 10 K K

K

   



 

 



Since the value of the equilibrium constant for the hydrolysis reaction of cyanide ion is larger than that for the hydrolysis of ammonium ion, the cyanide ion hydrolysis reaction will proceed to a greater extent and thus the solution of NH CN aq₄

b g

will be basic (alkaline).

13A (M) NaF dissociates completely into sodium ions and fluoride ions. The released fluoride ion hydrolyzes in aqueous solution to form hydroxide ion. The determination of the equilibrium pH is organized around the balanced equation.

Equation:

Initial:

Changes:

Equil:

-

F (aq) + H O(l) 2

0.10 M ^— x M ^—



^0.10x



^{M —}



HF(aq) 0 M

+ Mx Mx

+

OH^(aq) 0 M

+ Mx Mx

14 2

11

4 -

1.0 10 [HF][OH ] ( )( )

1.5 10

6.6 10 [F ] (0.10 ) 0.10

w b

a

K x x x

K K x

 





       

 

11 6 6

= 0.10 1.5 10 1.2 10 M=[OH ]; pOH= log(1.2 10 ) 5.92 pH = 14.00 pOH = 14.00 5.92 = 8.08 (As expected, pH > 7)

x   ^   ^{ }   ^ 

 

13B (M) The cyanide ion hydrolyzes in solution, as indicated in Practice Example 16-12B. As a consequence of the hydrolysis, OH^ = HCN . OH^ can be found from the pH of the solution, and then values are substituted into the K_b expression for CN^, which is then solved for CN^ .

pOH = 14.00 pH = 14.00 10.38 = 3.62 

 

-pOH 3.62 4

OH = 10 = 10^ = 2.4 10 M = HCN^

  

 

 

K_b HCN OH

CN CN CN M

= = 1.6 10 = 2.4 10

= 2.4 10

1.6 10 = 3.6 10

5

4 2 4 2

5

3

















   

 

c h c h

14A (M) First we draw the Lewis structures of the four acids. Lone pairs have been omitted since we are interested only in the arrangements of atoms.

HClO₄ should be stronger than HNO₃. Although Cl and N have similar electronegativities, there are more terminal oxygen atoms attached to the chlorine in perchloric acid than to the nitrogen in nitric acid. By virtue of having more terminal oxygens, perchloric acid, when ionized, affords a more stable conjugate base. The more stable the anion, the more easily it is formed and hence the stronger is the conjugate acid from which it is derived. CH FCOOH₂ will be a stronger acid than CH BrCOOH₂ because F is a more electronegative atom than Br.

The F atom withdraws additional electron density from the O — H bond, making the bond easier to break, which leads to increased acidity.

14B (M) First we draw the Lewis structures of the first two acids. Lone pairs are not depicted since we are interested in the arrangements of atoms.

H PO_{3 4} and H SO_{2 3} both have one terminal oxygen atom, but S is more electronegative than P. This suggests that



1



2

2 3 a

H SO = 1.3 10K  ^ should be a stronger acid than



1



3

3 4 a

H PO = 7.1 10K  ^ , and it is. The only difference between CCl CH COOH_{3 2} and CCl FCH COOH_{2 2} is the replacement of Cl by F. Since F is more electronegative than Cl, CCl CH COOH should be a weaker acid than CCl FCH COOH.

O N O

O

H O Cl O

O H

O

O C C

O

F H

H

H

O C C

O

Br H

H

H

O P O

O

H H

O H

O S O

O

H H

15A (M) We draw Lewis structures to help us decide.

(a) Clearly, BF₃ is an electron pair acceptor, a Lewis acid, and NH₃ is an electron pair donor, a Lewis base.

(b) H O₂ certainly has electron pairs (lone pairs) to donate and thus it can be a Lewis base. It is unlikely that the cation Cr³⁺ has any accessible valence electrons that can be donated to another atom, thus, it is the Lewis acid. The product of the reaction, [Cr(H2O)6]³⁺, is described as a water adduct of Cr³⁺.

Cr³⁺(aq) + H H 6 O

OH₂

OH₂ OH₂ OH₂

H₂O H₂O

Cr

3+

15B (M) The Lewis structures of the six species follow.

Both the hydroxide ion and the chloride ion have lone pairs of electrons that can be donated to electron-poor centers. These two are the electron pair donors, or the Lewis bases. Al OH

b g

3 and SnCl₄ have additional spaces in their structures to accept pairs of electrons, which is what occurs when they form the complex anions [Al(OH)4]^- and [SnCl6]^{2 -}. Thus, Al OH

b g

3 and SnCl₄ are the Lewis acids in these reactions.

B F

F

F

N H

H

 H F B

F

F N H

H

H O

H H

Sn Cl Cl Cl

Cl O H

+ Al O

H O H

O H

Al O

H O H

O H O

H -

2 Cl Sn

Cl Cl Cl

Cl Cl

Cl 2-

2-

INTEGRATIVE EXAMPLE

A. (D) To confirm the pH of rainwater, we have to calculate the concentration of CO2(aq) in water and then use simple acid-base equilibrium to calculate pH.

Concentration of CO2 in water at 1 atm pressure and 298 K is 1.45 g/L, or

2 2

2 2

1.45 g CO 1 mol CO

0.0329 M CO L 44.0 g CO 

Furthermore, if the atmosphere is 0.037% by volume CO2, then the mole fraction of CO2 is 0.00037, and the partial pressure of CO2 also becomes 0.00037 atm (because

2 2

CO CO atm

P  P )

From Henry’s law, we know that concentration of a gas in a liquid is proportional to its partial pressure.

CO2

1 1

C(mol / L) H P , which can rearrange to solve for H:

0.0329 M

H 0.0329 mol L atm

1 atm

 

 

   

Therefore, concentration of CO2(aq) in water under atmospheric pressures at 298 K is:

2

1 1 5

CCO (mol / L) 0.0329 mol L atm  ^  ^ 0.00037 atm = 1.217 10 M ^ Using the equation for reaction of CO2 with water:

CO2 (aq) + 2 H2O (l) ^ H3O⁺ (aq) + HCO₃^ (aq)

1.217×10^-5 0 0

-x +x +x

1.217×10^-5–x x X

 

 

2 7

a1 5

6

6 3

K x 4.4 10

1.217 10 x

Using the quadratic formula, x = 2.104 10 M pH log H O log 2.104 10 5.68 5.7







 

  

 



 

       

We can, of course, continue to refine the value of [H3O⁺] further by considering the dissociation of HCO3–, but the change is too small to matter.

B. (D)

(a) For the acids given, we determine values of m and n in the formula EOm(OH)n. HOCl or Cl(OH) has m = 0 and n = 1. We expect Ka  10^-7 or pKa  7, which is in good

agreement with the accepted pKa = 7.52. HOClO or ClO(OH) has m = 1 and n = 1. We expect Ka  10^-2 or pKa  2, in good agreement with the pKa = 1.92. HOClO2 or

ClO2(OH) has m = 2 and n = 1. We expect Ka to be large and in good agreement with the accepted value of pKa = -3, Ka = 10^-pKa = 10³. HOClO3 or ClO3(OH) has m = 3 and n = 1. We expect Ka to be very large and in good agreement with the accepted Ka = -8, pKa = -8, Ka = 10^-pKa = 10⁸ which turns out to be the case.

(b) The formula H3AsO4 can be rewritten as AsO(OH)3, which has m = 1 and n = 3. The expected value is Ka = 10^-2.

(c) The value of pKa = 1.1 corresponds to Ka = 10^-pKa = 10^-1.1 = 0.08, which indicates that m = 1. The following Lewis structure is consistent with this value of m.

O P

OH H OH

EXERCISES

Brønsted-Lowry Theory of Acids and Bases 1. (E)

(a) HNO₂ is an acid, a proton donor. Its conjugate base is NO₂^. (b) OCl^ is a base, a proton acceptor. Its conjugate acid is HOCl.

(c) NH₂^ is a base, a proton acceptor. Its conjugate acid is NH₃. (d) NH₄⁺ is an acid, a proton donor. It’s conjugate base is NH₃.

(e) CH NH_{3 3}⁺ is an acid, a proton donor. It’s conjugate base is CH NH_{3 2}.

2. (E) We write the conjugate base as the first product of the equilibrium for which the acid is the first reactant.

(a) HIO aq + H O(l)3

 

2 IO3^

 

aq + H O aq3 ⁺

 

(b) C H COOH aq + H O(l)6 5

 

2 C H COO aq + H O aq6 5 ^

 

3 ⁺

 

(c) HPO4^2

 

aq + H O(l)2 PO4^3

 

aq + H O aq3 ⁺

 

(d) ^{C H NH }

 

aq + H O(l)C H NH aq + H O aq

 

⁺

 

3. (E) The acids (proton donors) and bases (proton acceptors) are labeled below their formulas. Remember that a proton, in Brønsted-Lowry acid-base theory, is H⁺. (a) HOBr(aq) + H O(l) ² H O (aq) + OBr (aq) ^{3 +}

acid base acid base

 



(b)

+ 2

4 2 3 4

HSO (aq) + H O(l) H O (aq) + SO (aq) acid base acid base

  

(c) HS (aq) + H O(l) ² H S(aq) + OH (aq) ² base acid acid base

  

(d) C H NH (aq) + OH (aq) ^{6 5 3} C H NH (aq) + H O(l) ^{6 5 2 2} acid base base acid

  

4. (E) For each amphiprotic substance, we write both its acid and base hydrolysis reaction.

Even for the substances that are not usually considered amphiprotic, both reactions are written, but one of them is labeled as unlikely. In some instances we have written an oxygen as  to keep track of it through the reaction.

 

2 +

2 3 2 2

H + H O^{ }+ H O unlikely H + H O^ H + OH^

    

+ +

4 2 3 3 4

NH + H O^ NH + H O NH has no pairs that can be donatede^

+ +

2 2 3 2 2 3

H + H OH + H O^ H + H OH  + OH^

2 +

2 3 2 2

HS + H O^ S + H O^ HS + H O^ H S + OH^

2 2 2 2

NO cannot act as an acid, (no protons)^ NO + H O^ HNO + OH^

2 +

3 2 3 3 3 2 2 3

HCO + H O^ CO ^+ H O HCO + H O^ H CO + OH^

 

+ +

2 3 2 2 unlikely

HBr + H OH O + Br^ HBr + H OH Br + OH^

5. (E) Answer (b), NH₃, is correct. HC H O_{2 3 2} will react most completely with the strongest base. NO₃^ and Cl^ are very weak bases. H O₂ is a weak base, but it is amphiprotic, acting as an acid (donating protons), as in the presence of NH₃. Thus, NH₃ must be the strongest base and the most effective in deprotonating HC H O_{2 3 2}.

6. (M) Lewis structures are given below each equation.

(a) 2NH l3

 

NH + NH4⁺ 2^

+ 2 H N H

H

H N H H H

H N H ^



(b) 2HF l

 

 H F + F2 ^{+ }

F

H H F F

H F

H

+ +

(c) 2CH OH l3

 

CH OH + CH O3 2⁺ 3 ^

(d) 2HC H O l2 3 2

 

H C H O + C H O2 2 3 2⁺ 2 3 2^

(e) 2 H SO ( ) _{2 4} l  2 H SO_{3 4}^HSO₄^

7. (E) The principle we will follow here is that, in terms of their concentrations, the weaker acid and the weaker base will predominate at equilibrium. The reason for this is that a strong acid will do a good job of donating its protons and, having done so, its conjugate base will be left behind. The preferred direction is:

strong acid + strong base  weak (conjugate) base + weak (conjugate) acid

(a) The reaction will favor the forward direction because OH^ (a strong base)  NH₃ (a weak base) and NH (relatively strong weak acid) H O (very weak acid)₄^  ₂ .

(b) The reaction will favor the reverse direction because HNO₃HSO₄^ (a weak acid in the second ionization) (acting as acids), and SO₄^2 NO₃^ (acting as bases).

(c) The reaction will favor the reverse direction because HC H O_{2 3 2} CH OH₃ (not usually thought of as an acid) (acting as acids), and CH O₃ ^ C H O_{2 3 2}^ (acting as bases).

2 H C O H

H C O H H

 H

H H

H

H C O H

 H 

2 H C C

H

H C C H H

 H

O O

O ^

H O

H H

H C C H H

O

O ^

S O

O

O H O

H S

O O

O H O

H + O S +

O

O H O

S O

O

O H O

H

H

 +

8. (M) The principle we follow here is that, in terms of their concentrations, the weaker acid and the weaker base predominate at equilibrium. This is because a strong acid will do a good job of donating its protons and, having done so, its conjugate base will remain in solution. The preferred direction is:

strong acid + strong base  weak (conjugate) base + weak (conjugate) acid

(a) The reaction will favor the forward direction because HC H O_{2 3 2} (a moderate acid)

HCO₃^ (a rather weak acid) (acting as acids) and CO₃^2 C H O_{2 3 2}^ (acting as bases).

(b) The reaction will favor the reverse direction because HClO₄ (a strong acid)

 HNO₂ (acting as acids), and NO₂^ ClO₄^ (acting as bases).

(c) The reaction will favor the forward direction, because H CO_{2 3} HCO₃^ (acting as acids) (because K₁K₂) and CO₃^2 HCO₃^ (acting as bases).

Strong Acids, Strong Bases, and pH

9. (M) All of the solutes are strong acids or strong bases.

(a) _{+ 3} ⁺

3 3

3

1 mol H O

H O = 0.00165 M HNO 0.00165 M

1 mol HNO

   

 

14 w 12

+ 3

1.0 10

OH = = = 6.1 10 M

0.00165M H O

K ^

  

  

   

(b) 1 mol OH

OH = 0.0087 M KOH 0.0087 M

1 mol KOH

 

   

 

14

+ w 12

3

1.0 10

H O = = = 1.1 10 M

0.0087M OH

K ^ _



   

   

(c) ^OH = 0.00213 M Sr OH

 

^{2 2 mol OH}

 

₂ 0.00426 M 1mol Sr OH

 

   

 

14

+ w 12

3

1.0 10

H O = = = 2.3 10 M

0.00426 M OH

K ^ _



   

   

(d) _{+ 4 3} ⁺ ₄

3

1mol H O

H O = 5.8 10 M HI 5.8 10 M

1mol HI

 

     

 

14 11

w + 4 3

1.0 10

OH = = 1.7 10 M

5.8 10 M H O

K ^

 



    

    

10. (M) Again, all of the solutes are strong acids or strong bases.

(a)

 

+

+ 3

3

1 mol H O

H O = 0.0045 M HCl = 0.0045 M 1 mol HCl

pH = log 0.0045 = 2.35

  

 



(b)

 

+

+ 4 3 4

3 3

3 4

1 mol H O

H O = 6.14 10 M HNO = 6.14 10 M

1 mol HNO pH = log 6.14 10 = 3.21

 



    

 

 

(c)

 

1 mol OH

OH = 0.00683 M NaOH = 0.00683M 1 mol NaOH

pOH = log 0.00683 = 2.166 and pH = 14.000 pOH = 14.000 2.166 = 11.83

 

  

 

  

(d)

   

 

3 3

2

2 3

2 mol OH

OH = 4.8 10 M Ba OH = 9.6 10 M

1 mol Ba OH

pOH = log 9.6 10 = 2.02 pH = 14.00 2.02 = 11.98

   



    

 

  

11. (E)

   

   

2 2

2 2

2 2

2 2

3.9 g Ba OH 8H O 1000 mL 1 mol Ba OH 8H O 2 mol OH OH =

100 mL soln 1 L 315.5 g Ba OH 8H O 1 mol Ba OH 8H O

= 0.25 M



  

  

 

 

 

 

14

+ w 14

3

1.0 10 14

H O = = = 4.0 10 M

0.25 M OH

OH^K pH = log 4.0 10 = 13.40









 

    

   

12. (M) The dissolved Ca OH

b g

2 is completely dissociated into ions.

pOH= 14.00pH= 14.00 12.35 = 1.65

pOH 1.65 2

OH^ = 10^ = 10^ = 2.2 10 M OH = 0.022 M OH^{  }

  

 

   

   

 

 

2 2 2

2 2

2

2

1 mol Ca OH 74.09 g Ca OH 1000 mg Ca OH 0.022 mol OH

solubility =

1 L soln 2 mol OH 1 mol Ca OH 1 g Ca OH

8.1 10 mg Ca OH =

1 L soln



   



In 100 mL the solubility is 1 L 8.1 10 mg Ca OH²

 

²

 

₂

100 mL = 81 mg Ca OH

1000 mL 1 L soln

  