1
Chapter 17: Additional Aspects of Aqueous Equilibria Kahoot! 1.
Adding Br‐ to a saturated aqueous solution of ___ decreases its solubility in water. BaSO4, Li2CO3, PbS, AgBr 2. Which of the following mixtures could be used to prepare an effective buffer? HCl and KCl, HNO3 and KNO3, HCl and NH4Cl, NH3 and NH4Cl 3. Which 1 L solution has the greatest buffer capacity? 0.1 M NH3 and 0.1 M NH4Cl, 0.05 M NH3 and 0.05 M NH4Cl, 0.1 M NH3 and 0.01 M NH4Cl, 0.5 M NH3 and 0.5 M NH4Cl 4. Select the best acid or base to pair with its conjugate salt to prepare a 8.5 pH buffer. acetic acid; Ka = 1.8x10‐5, ammonia; Kb = 1.8x10‐5, hydroxylamine; Kb = 1.1x10‐8, citric acid; Ka = 7.5x10‐4 5. Select the correct representation of the Henderson‐Hasselbalch equation. Ka = [H+][A‐], Kw = [H+][OH‐] = 10‐14, pH = pKa + log [base]/[acid], pKa = pH + log[base]/[acid] 6. Which indicator is preferable when titrating a weak base with a strong acid? Methyl red (color change pH = 5), bromothymol blue (pH = 7), phenolphthalein (pH = 9), None of the above 7. Which indicator is preferable when titrating a weak acid with a strong base? Methyl red (color change pH = 5), bromothymol blue (pH = 7), phenolphthalein (pH = 9), None of the above 8. For BaCO3, Ksp = 5.0 x 10‐9. What is [Ba2+]? 7.1 x 10‐5 M, 1.0 x 10‐8 M, 2.5 x 10‐9 M, 5.0 x 10‐9 M 9. For BaF2, Ksp = 1.7 x 10‐6. What is [Ba2+]? 1.7 x 10‐6 M, 3.4 x 10‐6 M, 7.6 x 10‐3 M, 1.5 x 10‐2 M 10. Which of the following reagents will reduce the solubility of BaF2? NaCl, Ba(C2H3O2) 2, KOH, NH4Br 11. Which of the following reagents will increase the solubility of BaF2? HCl, HF, KOH, NH4Br Whiteboard Examples Example: Calculate the pH of a solution prepared by mixing equal volumes of 0.20 M CH3NH2 and 0.60 M CH3NH3Cl (Kb = 3.7 x 10‐4). What is the pH of 0.20 M CH3NH2 without addition of CH3NH3Cl? assuming we have 1 L of each 0.20 moles CH 3 NH 2
1 L
L
 0.10 M CH 3 NH 2
CH 3 NH 2 0 
2.0 L
0.60 moles CH 3 NH 3
1 L
L
CH 3 NH 3  
 0.30 M CH 3 NH 3

0
2.0 L
I C E Kb 
0.1
‐x
0.1 ‐ x
x  8.97  10
0.3
+x
0.3 + x
0
+x
+x
[CH 3 NH 3 ][OH  ] (0.3  x) x

 3.7  104
[CH 3 NH 2 ]
0.1  x
x 2  0.3004 x  3.7  105  0
‐
‐
‐
5
M OH

pH  14  log(8.97 105 )  9.95
‐‐‐‐ pH of 0.20 M CH3NH2: 2
Kb 
[CH 3 NH 3 ][OH  ]
x2

 3.7  104
[CH 3 NH 2 ]
0.1  x
x 2  3.7  104 x  3.7  105  0
3
x  8.24  10 M OH

pH  14  log(8.24  103 )  11.92
Buffer Example I: Calculate the pH and pOH of a 500.0mL solution containing 0.225M HPO42  and 0.225M PO43 at 25C where the Ka( HPO42  ) = 4.2x10-13.
 [ A ] 
 0.225M
13
pH  pK a  log 
   log(4.2  10 )  log 
 0.225M
 [ HA] 
pH  12.38

 
Buffer Example II: How would we prepare a pH = 4.44 buffer using CH3CO2H and CH3CO2Na? Ka = 1.8 x 10‐5  [CH 3CO2 ] 
 [ A ] 
5
pH  pK a  log 
log


  4.44  log(1.8  10 )  0.305

[
HA
]
[
CH
CO
H
]


3
2


[CH 3CO2 ]
0.305
 10
 0.496
[CH 3CO2 H ]
Therefore, in order to make a 4.44 buffer solution we need 0.496 moles of CH3CO2Na for every mole of CH3CO2H Comprehensive Example:Calculate the pH of each of the following solutions: a. 0.100 M HC3 H 5O2 , K a  1.3  105  H   C3 H 5O2 
HC3 H 5O2  H   C3 H 5O2  K a    
 HC3 H 5O2 
x2
1.3  105 
 x  1.1103 M H 
0.100  x
b. 0.100 M NaC3 H 5O2 pH  2.96
NaC3 H 5O2  Na   C3 H 5O2

K w OH   HC3 H 5O2 
C3 H 5O  H 2 O  OH  HC3 H 5O2  K b 

Ka
C3 H 5O2 
x2
x2
7.7  1010 

 x  8.8  106 M OH  pH  14  pOH  8.94
0.100  x 0.100
c. a mixture containing a. & b.  H   C3 H 5O2 
HC3 H 5O2  H   C3 H 5O2  K a    
 HC3 H 5O2 

2

1.3  105 
 0.100  x  x
0.100  x
 x  1.3  105 M H 
pH  4.89
  A 



using Henderson-Hasselbach, pH  pKa  log 

 HA 

 0.100 
pH   log1.3  105  log 
  4.89
 0.100 
d. a mixture containing c. and 0.020 mol of NaOH when strong base is added to an acid containing solution it will 3
neutralize the acid and so all of the OH‐ will react completely with our propanoic acid HC3 H 5O2  OH   H 2 O  C3 H 5O2 HC3 H 5O2 OH  C3 H 5O2 B C A 0.100
‐0.020
0.080
0.020
0.100
‐0.020
+0.02
0
0.120
0.120


pH   log1.3  105  log 
  5.06  0.080 
e. a mixture containing c. and 0.020 mol of HCl H   C3 H 5O2  HC3 H 5O2 HC3 H 5O2 C3 H 5O2 H B C A 0.100
‐0.020
0.080
0.020
0.100
‐0.020
+0.02
0
0.120
0.080


pH   log1.3  105  log 
  4.71  0.120 
Titration Example I: What is [NH3] if 22.35mL of 0.1145 M HCl were needed to titrate a 100.0mL sample? NH 3  HCl  NH 4  Cl 
[ NH 3 ]  0.02235 L 
0.1145 moles HCl 1 mole NH 3
1


 0.02259 M 1 mole HCl 0.1000 L
L
Titration Example II – Strong with Strong: A 15.0 mL sample of 0.200 M NaOH is titrated with 0. 250 M of HCl. Calculate the pH of the mixture after 10.0, and 20.0 mL of acid have been added. If you are not given a dissociation constant this should remind you that the acid/base is strong. For 10.0 mL of HCl & 15.0 mL of NaOH: 0.250 mol H aq 
1L
10.0 mL 

 0.00250 mol H aq 
1000 mL
1L

15.0 mL 
0.200 mol OH  aq 
1L

 0.00300 mol OH aq 
1000 mL
1L
H aq 
 OH aq   H 2 Ol 
B
0.00250
0.00300
C
-0.00250
-0.00250
A
0
0.00050
moles
0.0005
OH   
 0.0200 M
0.025L
pH  14  log(0.02)  12.30
For 20.0 mL of HCl
0.250 mol H aq 
1L
20.0 mL 

 0.00500 mol H aq 
1000 mL
1L
H aq 
B
C
A
0.00500
-0.00300
0.00200
-
 OH aq   H 2 Ol 
0.00300
-0.00300
0
-
4
0.002 moles
 0.05714 M
0.035L
pH   log(0.05714)  1.24
 H   
Titration Example III – Strong with Weak: A 25.0 mL sample of 0.100 M acetic acid (HC2H3O2) is titrated with 0.125 M of NaOH. Calculate the pH of the mixture after 0.0,10.0, 20.0, and 30.0 mL of base have been added. (Ka = 1.8 x 10‐5) For 0.0 mL it is the same as a weak acid:  H   C3 H 3O2 
HC3 H 3O2  H   C3 H 3O2  K a    
 HC3 H 3O2 
x2
0.100  x
assume x << 0.100
1.8  105 
x2
1.8  10 
 x  1.34  103 M H 
0.100
1.34  103
verify assumption:
 100%  1.34%  5%
0.100
pH   log 1.1 103   2.87
5
For 10.0 mL: Initially we have:
HC2 H 3O2 aq   NaOH  aq   NaC2 H 3O2 aq   H 2 O l 
Na  is a spectator ion so we are really looking at:
HC2 H 3O2 aq   OH aq   C2 H 3O2 aq   H 2 Ol 
This is in fact a neutralization reaction. For 10.0 mL of NaOH:
25.0 mL 
0.100 mol HC2 H 3O2 aq 
1L

 0.00250 mol HC2 H 3O2 aq 
1000 mL
1L

0.125 mol OH  aq 
1L

 0.00125 mol OH aq 
1000 mL
1L
Now we set up a “BCA” table or before, change, after table 10.0 mL 
HC2 H 3O2 aq   OH aq   C2 H 3O2 aq   H 2O l  B 0.00250
0.00125
0
‐
C ‐0.00125
‐0.00125
‐
A 0.00125
0
+0.00125
‐
So, we have neutralized the acid with the given amount of base and are now ready to apply the Henderson‐Hasselbalch equation:  0.0125 
  A  
 0.0350 L 


5
   log 1.8  10   log 
pH  pK a  log 
  4.74   HA 
 0.0125 


 0.0350 L 
Time for 20.0 mL of NaOH: 0.125 mol OH aq 
1L

 0.00250 mol OH aq  20.0 mL 
1000 mL
1L
5
HC2 H 3O2 aq   OH aq   C2 H 3O2 aq   H 2 Ol  B 0.00250
0.00250
0
‐
C ‐0.00250
‐0.00250
‐
A 0 0
+0.00250
‐
All of the acid has reacted with all the base and so we are now at the equivalence point. However, since we had a weak acid and a strong base we should expect our pH to be higher than 7. To find the pH we need an “ICE” table. C2 H 3O2 aq    0.00250 moles / 0.045L  0.0556 M 

C2 H 3O2 aq   H 2 O l   HC2 H 3O2 aq   OH aq  I 0.0556
‐
0
0
C ‐x ‐
+x
+x
E 0.0556‐x
‐
+x
+x
 HC2 H 3O2 aq   OH aq   K
14
x2

  w  10
10
5.6
10



Kb  
0.0556  x
K a 1.8  105
C2 H 3O2 aq  


x2
assume x  0.0556 5.6 10 10 
x  OH aq    5.6  10 6 M 0.0556
5.6  106
 100%  0.01%  5%
ck assumption :
0.0556
pH  14  log(5.6  106 )  8.75
Finally, 30.0 mL 
0.125 mol OH  aq 
1L
30.0 mL 

 0.00375 mol OH aq  1000 mL
1L
Now, we have more strong base than acid so we can set up the BCA table and then determine hydroxide concentration and pH directly. HC2 H 3O2 aq   OH aq   C2 H 3O2 aq   H 2 Ol  B C A 0.00250
0.00375
‐0.00250
‐0.00250
0 0.00125
moles
0.00125
OH   
 0.0227 M
0.055 L
pH  14  log(0.0227)  12.36
0
+0.00375
‐
‐
‐
Solubility Example: Determine the equilibrium concentrations (and solubilities) of BaF2(s), Ksp = 1.7x10‐6. 1.7  106 BaF2( s )  Ba(2aq )  2 F(aq )
K sp  [ S ][2S ]2
s 2s 1.7  106  4S 3  S  0.0075M
[ Ba 2  ]  0.0075M
and [ F  ]  2  0.0075  0.015M 0.0075 moles 137.327 g
g

 1.03
L
mol
L
0.015 moles 18.998 g
g

 0.285
F :
L
mol
L
Common‐Ion Solubility Example: Calculate the solubility of calcite (CaCO3) in 0.00100 M of Na2CO3 and in just plain water (Ksp = 4.5x10‐9 at 25C). Ba 2  :
6
CaCO3( s )  Ca(2aq )  CO(2aq )
Initial
Change
Eq
K sp  [ s ][0.001  s ]
Ca(2aq )
CO(2aq )
0
+s
s
0.0010
+s
0.0010+s
assume 0.001 >> s
K sp  4.5  109  0.001s  s  4.5  106 M
4.5  106
 100%  0.45%  5%
0.001
Therefore the solubility of CaCO3 is 4.5x10‐6M with [Ca2+] = 4.5x10‐6M & [CO32‐] = 0.0010 M How does this compare with the solubility of just CaCO3? Ca(2aq )
CO(2aq )
ck :
Initial Change
Eq 0
+s
s
0.0010
+s
+s
K sp  [ s ][ s ]
4.5  109  s 2  s  6.71 105 M
Ca 2    CO32    6.71 10 5 M
6.71 10 5 M  4.5 10 6 M thereby demonstrating how common-ion represses solubility