Name___________________________________  Date __________________________  Period ___

Practice Quiz:  16.8, 16.9  = The Ka  - Kb Relationship; Acid-Base Properties of Salts
 
MULTIPLE CHOICE.  Choose the one alternative that best completes the statement or answers the question. 

1) 

The acid-dissociation constant, Ka, for gallic acid, C6H2(OH)3COOH, is 4.57 × 10 – 3. What is the base-dissociation constant, Kb, for the gallate ion? 

1) 

________
A) 

5.43 × 10 – 5 
B) 

7.81 × 10 – 6
C) 

2.19 × 10 – 12 
D) 

4.57 × 10 – 3
E) 

2.19 × 10 2 

2) 

The Ka  for HCN is 4.9 × 10 – 10. What is the value of Kb  for CN – ? 

2) 

________
A) 

4.0 × 10 – 6
B) 

2.0 × 10 9 
C) 

4.9 × 10 – 24 
D) 

2.0 × 10 – 5 
E) 

4.9 × 10 4 

3) 

Calculate the pOH of a 0.0827 M aqueous sodium cyanide solution at 25.0 °C. Kb  for CN –  is
 4.9 x 10 – 10. 

3) 

________
A) 

8.8 
B) 

9.3 
C) 

10 
D) 

1.1 
E) 

5.2 

4) 

Calculate the pH of 0.726 M anilinium hydrochloride (C6H5NH3Cl) solution in water, given that Kb  for aniline is 3.83 × 10 – 4. 

4) 

________
A) 

8.64 
B) 

12.2 
C) 

12.4 
D) 

1.77 
E) 

5.36 

5) 

The Ka  for formic acid (HCO2H) is 1.8 × 10 – 4. What is the pH of a 0.35 M aqueous solution of sodium formate (NaHCO2)? 

5) 

________
A) 

5.4 
B) 

4.2 
C) 

3.3 
D) 

8.6 
E) 

11 

6) 

Ka  for HX is 7.5 × 10 – 12. What is the pH of a 0.15 M aqueous solution of NaX? 

6) 

________
A) 

1.9 
B) 

6.0 
C) 

7.9 
D) 

12 
E) 

8.0 

7) 

Using the data in the table, which of the conjugate bases below is the strongest base?

 

7) 

________
A) 

CHO2 –  
B) 

ClO – 
C) 

OAc – 
D) 

F – 
E) 

OAc –  and CHO2 –   

8) 

Which of the following aqueous solutions has the highest [OH – ]? 

8) 

________
A) 

a solution with a pH of 3.0 
B) 

a 1 × 10 – 4  M solution of HNO3 
C) 

a 1 × 10 – 3  M solution of NH4Cl  
D) 

pure water 
E) 

a solution with a pOH of 12.0 

9) 

Of the following substances, an aqueous solution of _____ will form basic solutions. 

	NH4Cl     Cu(NO3)2     K2CO3     NaF

9) 

________
A) 

NaF only 
B) 

NaF, K2CO3 
C) 

K2CO3, NH4Cl   
D) 

NH4Cl  only  
E) 

NH4Cl , Cu(NO3)2      

10) 

Of the compounds below, a 0.1 M aqueous solution of ______ will have the highest pH. 

10) 

______
A) 

NaHS, Kb  of HS –  = 1.8 × 10 – 7  
B) 

NH4NO3, Kb  of NH3 = 1.8 × 10 – 5  
C) 

KCN, Ka  of HCN = 4.0 × 10 – 10  
D) 

NaOAc, Ka  of HOAc = 1.8 × 10 – 5  
E) 

NaClO, Ka  of HClO = 3.2 × 10 – 8   

11) 

An aqueous solution of ______ will produce a basic solution. 

11) 

______
A) 

KBr 
B) 

NH4ClO4 
C) 

NaCl 
D) 

Na2SO3 
E) 

NaHSO4 

SHORT ANSWER.  Write the word or phrase that best completes each statement or answers the question. 

12) 

What is the pH of a sodium acetate solution prepared by adding 0.820 grams of sodium acetate to 100.0 ml of water at 25.0 oC? The Ka at 25.0 oC for acetic acid is 1.8 × 10-5. 

12) 

______________

13) 

What is the pH of a sodium formate solution prepared by adding 0.680 grams of sodium formate to 100.0 ml of water at 25.0 oC? The Ka at 25.0 oC for formic acid is 1.8 × 10-4. 

13) 

______________

 MULTIPLE CHOICE.  Choose the one alternative that best completes the statement or answers the question. 

14) 

The acid-dissociation constant of hydrocyanic acid (HCN) at 25.0 °C is 4.9 × 10 – 10. What is the pH of an aqueous solution of  0.080 M sodium cyanide (NaCN)?   

14) 

_____
A) 

2.89 
B) 

11.11 
C) 

1.3 × 10-3  
D) 

3.9 × 10-11  
E) 

7.8 × 10-12  

  16.8, 16.9 Extended Response Section

	15)  Although the acid-dissociation constant for phenol (C6H5OH) is listed in Appendix D, the base-dissociation 
constant for the phenolate ion (C6H5O – ) is not.  Explain why it is not necessary to list both Ka for phenol and      
Kb for the phenolate ion.

	16)  	a)  Given that Ka for acetic acid is 1.8 x 10 – 5  and that for hypochlorous acid is 3.0 x 10 – 8 , which is the 
                                   stronger acid?

	     	b)  Which is the stronger base, the acetate ion or the hypochlorite ion?

		c)  Calculate Kb values for C2H3O2 –   and ClO –  .

17)  Using data from Appendix D, calculate [OH – ] and pH for the following solutions:

a)  0.055 M Na2S

b)  0.085 M NaF

c)  A mixture that is 0.045 M in NaC2H3O2 and 0.055 M in Ba(C2H3O2)2.

18)  Predict whether aqueous solutions of the following substances are acidic, basic, or neutral:

a)  ammonium bromide

b)  iron (III) chloride

c)  sodium carbonate

d) potassium perchlorate

e)  sodium hydrogen oxalate

19)  An unknown salt is either NaF, NaCl, or NaOCl.  When 0.050 mol of the salt is dissolved in water to form 
       0.500 L of solution, the pH of the solution is 8.08.  What is the identity of the salt?

20)  Sorbic acid (HC6H7O2) is a weak monoprotic acid with Ka = 1.7 x 10 – 5 .  Its salt (potassium sorbate) is added to 
       cheese to inhibit the formation of mold.  What is the pH of a solution containing 11.25 g of potassium sorbate 

16.8, 16.9  Practice Test Answer Key
1) 

C 		  5)  D			  9)  B		13)  8.37

2) 

D 		  6)  D			10)  C		14)  B

3) 

E 		  7)  B			11)  D		

4) 

E 		  8)  D			12)  8.87

15)  For a conjugate acid/ base pair such as  C6H5OH / C6H5O –, Kb for the conjugate base is always Kw / Ka for the 
        conjugate acid.  Kb for the conjugate base can always be calculated from Ka for the conjugate acid, so a separate 
        list of Kb values is not necessary.

16)  	a)  Acetic acid is stronger because it has the larger Ka value.

	b)  Hypochlorite ion is the stronger base because the weaker acid, hypochlorous acid, has the stronger 
                    conjugate base.

	c)  Kb for C2H3O2 –  = Kw/ Ka  for HC2H3O2   = 1.0 x 10 – 14  / 1.8 x 10 – 5   =  5.6 x 10 – 10

                    Kb for ClO –    = Kw/ Ka  for HClO   = 1.0 x 10 – 14  / 3.0 x 10 – 8   =  3.3 x 10 – 7
                               (Note that Kb for ClO –    is greater than Kb for C2H3O2 – )

17)	a)  S2 – (aq)  +  H2O (l)    HS – (aq)  +  OH –  (aq)         Ka for HS –  = 1 x 10 – 19 (from Appendix D)

                                   Kw                                       1 x 10 – 14 
                    Kb =                             →   Kb =                        =  1 x 105 , note that Kb > > 1, so equilibrium lies far to the right
                              Ka  for  HS –                         1 x 10 – 19                   ..essentially S 2–  hydrolyzes almost 100% , so we can 
                                                                                                           safely ignore the small amount that does NOT 
                                                                                                           hydrolyze; therefore, starting [S2 – ] = equilibrium[HS – ],     
                                                                                                           and equilibrium[HS – ] = [OH– ]  =  0.055 M      

          [OH – ] [HS – ]                                  (0.055 M – X)2                              (0.055 M)2              
 Kb =                                →   1 x 10 5   =                               ≈    1 x 10 5    =                           =   3.0 x 10– 8 M = equilibrium [S2 – ]
               [S2 – ]                                                      X                                                  X 

3.0 x 10– 8 M is the amount of S2 – that did NOT hydrolyze….this amount should be subtracted from the assumed [OH – ] of 0.055 M, but it’s such a miniscule amount it will not change the [OH – ] enough to make any difference in the pH calculation
                                                                                                         
        pOH =  –  log  [OH– ]    →    pOH = –  log  (0.055)  =  1.25964  =  1.26 

        14 = pH  +  pOH   →  14 – 1.26 = pH   →  pH = 12.74

(Note, Kb HS –   = 1.05 x 10 – 7 ,  [OH – ] produced by further hydrolysis of HS –  = 7.6 x 10 – 5; the second hydrolysis step does not make a significant contribution to the total [OH – ] or pH.)

F.Y.I.
Ka HS –   = 1 x 10 – 19   (Appendix D)  		Ka  H2S  = 9.5 x 10 – 8  (Appendix D)  
Kb   S2 –   = 1 x 10 5				Kb  HS –  = 1.1 x 10 – 7

16.8, 16.9 Practice Test Answer Key Continued:

17) Continued

b)  F –   (aq)  +  H2O (l)    HF (aq)  +  OH –  (aq)         Ka for HF = 6.8 x 10 – 4 (from Appendix D)

                                   Kw                                       1 x 10 – 14 
                    Kb =                             →   Kb =                        =  1.5 x 10 – 11 , note that Kb < < 1, so equilibrium lies far to the
                 Ka  for  HF                        6.8 x 10 – 4                              left,…..essentially F –   does NOT hydrolyze so
                                                                                                         we can safely ignore the small amount that does

        [OH – ] [HF]                                               X2                                                      X2
           Kb =                                →   1.5 x 10 – 11   =                           ≈    1.5 x 10 – 11    =                          =   1.1 x 10 – 6 M = [OH – ]
            [F – ]                                               0.085 M – X                                          0.085 M

        pOH =  –  log  [OH– ]    →    pOH = –  log  (1.1 x 10 – 6 )  =  5.9586 = 5.95 (obscure rounding rule again!)

        14 = pH  +  pOH   →  14 – 5.95 = pH   →  pH = 8.05

c)  C2H3O2 –   (aq)  H2O (l)     HC2H3O2 (aq)  +  OH –  (aq)            Ka for HC2H3O2  = 1.8 x 10– 5 

                              Kw                                             1 x 10 – 14 
        Kb =                                         →   Kb =                        = 5.6 x 10 – 10 , note that Kb < < 1, so equilibrium lies  far to the left,….
     Ka  for  HC2H3O2                         1.8 x 10– 5                               essentially C2H3O2 –   does NOT hydrolyze, so we can 
                                                                                          safely ignore the small amount that does
 
   [OH – ] [HC2H3O2]                                                      X2                                                                 X2                                                            
     Kb =                                         →   5.6 x 10 – 10   =                                                    ≈  5.6 x 10 – 10    =                                             =  9.3 x 10 – 6 
      [C2H3O2 –    ]                                           [(0.045 M + 2(0.055 M)] – X                                     0.155 M

            [OH –  ] = 9.3 x 10 – 6 M

        pOH =  –  log  [OH– ]    →    pOH = –  log  (9.3 x 10 – 6)  =  5.0315 = 5.03 
       
       14 = pH  +  pOH   →  14 – 5.03 = pH   →  pH = 8.97

18) 	a)  NH4Br =  acidic;  NH4+ is a weak acid;  Br –  is a negligible bas

b) FeCl3 =  acidic;  F3+ is a highly charged metal cation and a Lewis acid;  Cl –  is a negligible base   

c)  Na2CO3 = basic;  CO32 –  is the conjugate base of the weak acid, HCO3 – ; Na+ is negligible 

d) KClO4 = neutral;  both ClO4 –  and K+ are negligible

e)  NaHC2O4 = acidic; HC2O4 –  is amphoteric, but Ka for the acid dissociation (6.4 x 10 – 5 ) is much greater 
                                       than Kb for the base hydrolysis (1.0 x 10 – 14 / 5.9 x 10 – 2  = 1.7 x 10 – 13 ); Na+ is 
                                       negligible
F.Y.I.
Ka HC2O4 –   = 6.4 x 10 – 5   (Appendix D)  		Ka  H2C2O4 = 5.9 x 10 – 2  (Appendix D)  
Kb C2H42 –  = 1.7 x 10 – 13				Kb  HC2O4 –   = 1.6 x 10 – 10 
16.8, 16.9 Practice Test Answer Key Continued:

19)  Estimate pH using relative base strength and then calculate to confirm prediction.

	A pH of 8.08 is weakly basic, first look for a salt that will result in a basic solution:

	NaF  =   basic;  Na+ is negligible, F –  is the conjugate base of the weak acid HF

	NaCl = neutral; Na+ is negligible, Cl –   is a negligible base

	NaOCl = basic; Na+ is negligible, OCl –  is the conjugate base of the weak acid HOCl

	Ka of HF = 6.8 x 10 – 4 
                                                               HF is a stronger acid than HOCl,  therefore F –   is a weaker base than OCl – 
	Ka of HOCl = 3.0 x 10 – 8 

	F –  (aq)  +  H2O (l)    HF (aq)  +  OH –  (aq)           KbHF = 1 x 10 – 14  / 6.8 x 10 – 4   = 1.5 x 10 – 11 

                         [OH – ] [HF]                                                         X2                                                                                                              X2
	Kb =                                 →       1.5 x 10 – 11  =                                          ≈    1.5 x 10 – 11   = 
		[F –  ]                                                         0.050 moL  –  X                                         0.050 mol/ 0.500 L
                                                                                              0.500 L

                                                          X2
              →      1.5 x 10 – 11   =                              =   1.2 x 10 – 6  M = [OH – ]
                                                    0.10 M

	pOH =  –  log [OH – ]  →     pOH =  –  log (1.2 x 10 – 6  M)  →  pOH = 5.9208  =  5.92

	14 = pH  +  pOH   →  14 = pH  +  5.92   →  pH = 8.08…..The unknown is NaF

20)     11.25 g KC6H7O2             1 mole KC6H7O2           1 mol C6H7O2 – 
                                                                                                                           = [C6H7O2 – ] = 0.0428 M
                  1.75 L                150.23   g KC6H7O2           1 mole KC6H7O2   

C6H7O2 –  (aq) +   H2O (l)    HC6H7O2 (aq)  +  OH –  (aq)            Ka = 1.7 x 10  – 5   ,  Kb = 5.9 x 10 – 10

            [OH – ] [HC6H7O2  ]                                                          X2                                                                                          X2
Kb =                                            →       5.9 x 10 – 10  =                                         ≈   5.9 x 10 – 10  =                        = 5.0 x 10 – 6  M = [OH – ]
 [C6H7O2 –]                                                         0.0428 M  –  X                                         0.0428 M

	pOH =  –  log [OH – ]  →     pOH =  –  log (5.0 x 10 – 6  M)  →  pOH = 5.30103  =  5.30

	14 = pH  +  pOH   →  14 = pH  +  5.30   →  pH = 8.70