AP Chemistry Unit 8- Homework Problems Ka and Kb Acid & Base ...

AP Chemistry
Unit 8- Homework Problems
K a and K b
Acid & Base Theory
1. What is the definition of an Arrhenius acid? H + releaser
2. What is the definition of an Arrhenius base? OH - releaser
3. What is the definition of a Bronsted acid? H + donator
4. What is the definition of a Bronsted base? H + acceptor
5. What is the conjugate base of:
a. HNO 2 NO 2
-1
b. H 2 SO 3
-1
HSO 3
+1
c. NH 4 NH 3
d. HCN CN -1
6. What is the conjugate acid of:
a. Cl -1 HCl
-2
b. HPO 4
-1
H 2 PO 4
c. OH -1 H 2 O
d. N 2 H 4
+1
N 2 H 5
7. In each equation below, identify each species as the acid, base, conjugate acid or conjugate base.
a. HNO 3 + PO -3 4 NO -1 3
-2
+ HPO 4
Acid Base CB CA
b. N -1 3 + H 2 SO 4 HSO -1 4 + HN 3
Base Acid CB CA
c. NH 3 + H 2 PO -1 3 NH +1 4
-2
+ HPO 3
Base Acid CA CB
d. H 2 C 2 O 4 + OH -1 -1
H 2 O + HC 2 O 4
Acid Base CA CB
8. For each species below, describe whether a solution of it would be acidic, basic, or neutral.
a. HCl Acid h. CaCl 2 Neutral
b. KOH Base i. Na 2 CO 3 Base
c. NaCl Neutral j. LiBr Neutral
d. NaC 2 H 3 O 2 Base k. KF Base
e. NH 4 Cl Acid l. NaNO 2 Base
f. Na 3 PO 4 Base m. K 2 SO 3 Base
g. KNO 3 Neutral n. KI Neutral
9. What is an amphiprotic substance? List 4 of them.
A substance that can act as either a Bronsted acid or Bronsted base

Auto-Ionization of water, pH, and strong acids
1. What is the equation for the auto-ionization of water? 2 H 2 O (l) H 3 O +1 (aq) + OH -1 (aq)
2. What is the equation for K w and what is its value? K w = [H 3 O + ][OH -1 ] = 1x10 -14
3. Because the pH scale is a logarithmic scale, every change in 1 is really a change in how much? 10
4. Using the pH square and the equations learned in class, fill in the blank spots in the chart below:
pH [H 3 O +1 ] [OH -1 ] pOH
3.5 3.16x10 -4 3.16x10 -11 10.5
8.04 9.2x10 -9 1.10x10 -6 5.96
3.67 2.14x10 -4 4.7x10 -11 10.33
5.8 1.58x10 -6 6.31x10 -9 8.2
5. According to the leveling effect:
a. What is the strongest acid species in water? H 3 O +1
b. What is the strongest base species in water? OH -1
6. What is the % dissociation of strong acids and bases in water? 100%
7. List 6 strong acids. HCl, HBr, HI, HNO 3 , HClO 4 , H 2 SO 4
8. List 6 strong bases. LiOH, NaOH, KOH, RbOH, CsOH, Ba(OH) 2 , Sr(OH) 2
Weak acids & bases, K a , K b , and pH
1. How are weak acids and bases different from strong ones? Weak acids dissociate

6. For a 0.055 M solution of HClO (K a = 2.9x10 -8 ) calculate:
a. pH
HA (aq) + H 2 O (l) H 3 O +1 (aq) + A -1 (aq)
I 0.055 0 0
C -x +x +x
E 0.055 x x
2.9x10 -8 = x 2 /0.055 x = 3.99x10 -5 pH = -log (3.99x10 -5 ) = 4.40
b. % dissociation
3.99x10 -5 x 100% = 0.0725%
0.055
7. For a 0.0034 M solution of C 5 H 5 N (K b = 1.7x10 -9 ) calculate:
a. pH
B (aq) + H 2 O (l) OH -1 (aq) + HB +1 (aq)
I 0.0034 0 0
C -x +x +x
E 0.0034 x x
1.7x10 -9 = x 2 /0.0034 x = 2.40x10 -6 pOH = -log (2.40x10 -6 ) = 5.62 pH = 8.38
b. % dissociation
2.40x10 -6 x 100% = 0.071%
0.0034
8. For a 0.15 M solution of H 2 C 6 H 6 O 6 (K a1 = 8x10 -5 K a2 = 1.6 x10 -12 ) calculate:
a. pH
H 2 A (aq) + H 2 O (l) H 3 O +1 (aq) + HA -1 (aq)
I 0.15 0 0
C -x +x +x
E 0.15 x x
8x10 -5 = x 2 /0.15 x = 3.46x10 -3 pH = -log (3.46x10 -3 ) = 2.46
b. Equilibrium concentrations of:
-1
i) HC 6 H 6 O 6 x = 3.46x10 -3
-2
ii) C 6 H 6 O 6 K a2 = 1.6x10 -12
9. For a 0.025 M solution of diprotic base, B (K b1 = 4x10 -6 , K b2 = 8x10 -10 ) calculate:
a. pH
B (aq) + H 2 O (l) OH -1 (aq) + HB +1 (aq)
I 0.025 0 0
C -x +x +x
E 0.025 x x
4x10 -6 = x 2 /0.025 x = 3.16x10 -4 pOH = -log (3.16x10 -4 ) = 3.5 pH = 10.5
b. Equilibrium concentrations of:
i) HB + x = 3.16x10 -4
ii) H 2 B +2 K b2 = 8x10 -10

Acid-Base Interactions
Strong-Strong
1. What is the pH when 50 mL of 0.1 M HCl is mixed with 50 mL of 0.1 M NaOH?
Acid = (50/100)(0.1 M) = 0.05 M
Base = (50/100)(0.1 M) = 0.05 M
Equal amounts of strong acid/base so should be neutral
H 3 O +1 (aq) + OH -1 (l) 2 H 2 O
I 0.05 0.05
C -0.05 -0.05
E 0 0
pH = 7
2. What is the pH when 250 mL of 0.2 M HNO 3 is mixed with 500 mL of 0.1 M KOH?
Acid = (250/750)(0.2 M) = 0.0667 M Base = (500/750)(0.1 M) = 0.0667 M
Equal amounts of strong acid/base so should be neutral
H 3 O +1 (aq) + OH -1 (l) 2 H 2 O
I 0.0667 0.0667
C -0.0667 -0.0667
E 0 0
pH = 7
3. What is the pH when 100 mL of 0.001 M HBr is mixed with 50 mL of 0.001 M NaOH?
Acid = (100/150)(0.001 M) = 6.67x10 -4 M
Base = (50/150)(0.001 M) = 3.33x10 -4 M
More acid than base so should be acidic
H 3 O +1 (aq) + OH -1 (l) 2 H 2 O
I 6.67x10 -4 3.33x10 -4
C -3.33x10 -4 -3.33x10 -4
E 3.33x10 -4 0
pH = -log (3.33x10 -4 ) = 3.47
4. What is the pH when 40 mL of 0.01 M HI is mixed with 120 mL of 0.008 M CsOH?
Acid = (40/160)(0.01 M) = 0.0025 M Base = (120/160)(0.008 M) = 0.006 M
More base than acid so should be basic
H 3 O +1 (aq) + OH -1 (l) 2 H 2 O
I 0.0025 0.006
C -0.0025 -0.0025
E 0 0.0035
pOH = -log (0.0035) = 2.46 pH = 11.54

5. What is the pH when 50 mL of 0.0025 M H 2 SO 4 is mixed with 75 mL of 0.0040 M LiOH?
Acid = 2*(50/125)(0.0025 M) = 0.0020 M
Base = (75/125)(0.0040 M) = 0.0024 M
More base than acid so should be basic
H 3 O +1 (aq) + OH -1 (l) 2 H 2 O
I 0.0020 0.0024
C -0.0020 -0.0020
E 0 0.0004
pOH = -log (0.0004) = 3.4 pH = 10.6
6. What is the pH when 20 mL of 0.00050 M HClO 4 is mixed with 20 mL of 0.00040 M Ba(OH) 2 ?
Acid = (20/40)(0.00050 M) = 0.00025 M Base = 2* (20/40)(0.0004 M) = 0.0004 M
More base than acid so should be basic
H 3 O +1 (aq) + OH -1 (l) 2 H 2 O
I 0.00025 0.0004
C -0.00025 -0.00025
E 0 0.00015
pOH = -log (0.00015) = 3.8 pH = 10.2
Strong-Weak
7. What is the pH when 25 mL of 0.25 M HCl is mixed with 25 mL of 0.25 M NH 3 ? (K b = 1.8x10 -5 )?
Acid = (25/50)(0.25 M) = 0.125 M Base = (25/50)(0.25 M) = 0.125 M
Equal amounts of strong acid/weak base so should be acidic
Shortcut hint- Equivalence point of a strong acid weak base gives a normal K a problem
(you can skip the actual rxn and go straight to K a of conj. acid of weak base)
ACTUAL RXN- GOES ALMOST ALL THE WAY (water is a product; math is bad)
H 3 O +1 (aq) + NH 3 (aq) H 2 O (l) + NH +1 4 (aq)
I 0.125 0.125 0
C -0.125 -0.125 +0.125
E 0 0 0.125
NOW FLIP RXN AND DO K a (K a = 1x10 -14 /1.8x10 -5 ) = 5.56x10 -10
NH +1 4 (aq)+ H 2 O (l) H 3 O +1 (aq) + NH 3 (aq)
I 0.125 0 0
C -x +x +x
E 0.125 x x
5.56x10 -10 = x 2 /0.125 x = 8.33x10 -6 pH = -log (8.33x10 -6 ) = 5.08

8. What is the pH when 20 mL of 0.020 M HNO 3 is mixed with 20 mL of 0.030 M C 6 H 5 NH 2 (K b = 4x10 -10 )?
Acid = (20/40)(0.02 M) = 0.01 M Base = (20/40)(0.030 M) = 0.015 M
Not at equivalence point. Not at midpoint.
ACTUAL RXN- GOES ALMOST ALL THE WAY (water is a product; math is bad)
H 3 O +1 (aq) + C 6 H 5 NH 2 (aq) H 2 O (l) + C 6 H 5 NH +1 3 (aq)
I 0.01 0.015 0
C -0.01 -0.01 +0.01
E 0 0.005 0.01
NOW FLIP RXN AND DO K a (K a = 1x10 -14 /4x10 -10 ) = 2.5x10 -5
C 6 H 5 NH +1 3 (aq) + H 2 O (l) H 3 O +1 (aq) + C 6 H 5 NH 2 (aq)
I 0.01 0 0.005
C -x +x +x
E 0.01 x 0.005
2.5x10 -5 = x(0.005/0.01) x = 5x10 -5 pH = -log (5x10 -5 ) = 4.30
Shortcut hint- Simple buffer problem. Write K a equation and plug in amounts after strong acid is
neutralized (you can skip the actual rxn)
Strong acid: 0.01 M neutralized is turned into a weak acid
Weak base: 0.015 M -0.01 M neutralized by acid so 0.005 left
Now plug straight into 2 nd ICE table above and solve
9. What is the pH when 50 mL of 0.040 M NaOH is mixed with 50 mL of 0.040 M HN 3 (K a = 1.9x10 -5 )?
Base = (50/100)(0.04 M) = 0.02 M Acid = (50/100)(0.04 M) = 0.02 M
Equal amounts of strong base/weak acid so should be basic
ACTUAL RXN- GOES ALMOST ALL THE WAY (water is a product; math is bad)
OH -1 (aq) + HN 3 (aq) H 2 O (l) + N -1 3 (aq)
I 0.02 0.02 0
C -0.02 -0.02 +0.02
E 0 0 0.02
NOW FLIP RXN AND DO K b (K b = 1x10 -14 /1.9x10 -5 ) = 5.26x10 -10
N -1 3 (aq)+ H 2 O (l) OH -1 (aq) + HN 3 (aq)
I 0.02 0 0
C -x +x +x
E 0.02 x x
5.26x10 -10 = x 2 /0.02 x = 3.24x10 -6
pOH = -log (3.242x10 -6 ) = 5.49 pH = 8.51
Shortcut hint- Equivalence point of a strong base weak acid gives a normal K b problem
(you can skip the actual rxn and go straight to K b of conj. base of weak acid)

10. What is the pH when 120 mL of 0.0075 M KOH is mixed with 200 mL of 0.0050 M HOCl (K a = 3.5x10 -8 )?
Base = (120/320)(0.0075 M) = 0.00281 M Acid = (200/320)(0.005 M) = 0.00312 M
Not at equivalence point. Not at midpoint.
ACTUAL RXN- GOES ALMOST ALL THE WAY (water is a product; math is bad)
OH -1 (aq) + HOCl (aq) H 2 O (l) + OCl -1 (aq)
I 0.00281 0.00312 0
C -0.00281 -0.00281 +0.00281
E 0 0.00031 0.00281
NOW FLIP RXN AND DO K b (K b = 1x10 -14 /3.5x10 -8 ) = 2.89x10 -7
OCl -1 (aq) + H 2 O (l) OH -1 (aq) + HOCl (aq)
I 0.00281 0 0.00031
C -x +x +x
E 0.00281 x 0.00031
2.89x10 -7 = x(0.00031/0.00281) x = 2.59x10 -6
pOH = -log (2.59x10 -6 ) = 5.59 pH = 8.41
Shortcut hint- Simple buffer problem. Write K a equation and plug in amounts after strong base is
neutralized (you can skip the actual rxn)
Strong base = 0.00281 turns into weak base
Weak acid = 0.00312 neutralized and left with 0.00312-0.00281 = 0.00031
So:
HOCl (aq) + H 2 O (l) H 3 O + (aq) + OCl -1 (aq)
I 0.00031 0 0.00281
C -x +x +x
E 0.00031 x 0.00281
3.5x10 -8 = x(0.00281/0.00031) x = 3.86x10 -9 pH = 8.41
Midpoint
11. What is the pH when 25 mL of 0.050 M HCl is mixed with 50 mL of 0.050 C 5 H 5 N (K b = 1.5x10 -9 )?
acid = (25/75)(0.050 M) = 0.0166 M base = (50/75)(0.050 M) = 0.0333 M
At midpoint! (Half as much strong acid as base)
pH = pKa so K a = (1x10 -14 /1.5x10 -9 ) = 6.67x10 -6 so pH = - log (6.67x10 -6 ) = 5.18
12. What is the pH when 40 mL of 0.80 M NaOH is mixed with 80 mL of 0.8 M HNO 2 (K a = 4.5x10 -4 )?
base = (40/120)(0.80 M) = 0.266 M acid = (80/120)(0.80 M) = 0.5333 M
At midpoint! (Half as much strong base as acid)
pH = pKa so pH = - log (4.5x10 -4 ) = 3.35

13. What is the pH when 30 mL of 0.006 M HNO 3 is mixed with 30 mL of 0.012 M (CH 3 ) 3 N (K b = 7.4x10 -5 )?
acid = (30/60)(0.006 M) = 0.003 M base = (30/60)(0.012 M) = 0.006 M
At midpoint! (Half as much strong acid as base)
pH = pKa so K a = (1x10 -14 /7.4x10 -5 ) = 1.35x10 -10 so pH = - log (1.35x10 -10 ) = 9.87
14. What is the pH when 100 mL of 0.0078 M KOH is mixed with 100 mL of 0.0156 M C 6 H 5 CO 2 H (K a = 6.3x10 -5 )
base = (100/200)(0.0078 M) = 0.0039 M
acid = (100/200)(0.0156 M) = 0.0078 M
At midpoint! (Half as much strong base as acid)
pH = pKa so pH = - log (6.3x10 -5 ) = 4.2
K sp and pH
1. What is the pH of a saturated solution of Mn(OH) 2 K sp = 1.9x10 -13 ?
K sp = [Mn +2 ][OH -1 ] 2 1.9x10 -13 = 4x 3
x= 3.67x10 -5 so [OH -1 ] = 7.32 x10 -5 so pOH = 4.14 so pH = 9.86
2. What is the pH of a saturated solution of Ca(OH) 2 K sp = 5.5x10 -5 ?
K sp = [Ca +2 ][OH -1 ] 2 5.5x10 -5 = 4x 3
x= 0.0240 so [OH -1 ] = 0.0480 so pOH = 1.32 so pH = 12.68
3. At what pH will a 0.0075 M solution of Mg(OH) 2 begin to form a precipitate (K sp = 5.6x10 -12 )?
K sp = [Mg +2 ][OH -1 ] 2 5.6x10 -12 = [0.0075][x] 2
x= 2.73x10 -5 so [OH -1 ] = 2.73x10 -5 so pOH = 4.56 so pH = 9.44
4. At what pH will a 0.00042 M solution of Pb(OH) 2 begin to form a precipitate (K sp = 1.4x10 -15 )?
K sp = [Pb +2 ][OH -1 ] 2 1.4x10 -15 = [0.00042][x] 2
x= 1.83x10 -6 so [OH -1 ] = 1.83x10 -6 so pOH = 5.74 so pH = 8.26

Titration Curves
1. Draw a titration curve of a strong base being titrated by a strong acid. Be sure to label a pH of 7 in the
appropriate spot.
pH
7
mL acid added
2. Draw a titration curve of a strong acid being titrated by a strong base. Be sure to label a pH of 7 in the
appropriate spot.
pH
7
mL acid added
3. Draw a titration curve of a weak base being titrated by a strong acid. Be sure to label a pH of 7 in the
appropriate spot.
pH
7
mL acid added

4. Draw a titration curve of a weak acid being titrated by a strong base. Be sure to label a pH of 7 in the
appropriate spot.
pH
7
mL acid added
7. What is the significance of the midpoint of a titration? It is pH = pK a

8. If 40 mL of weak acid HA was titrated as the graph below shows, calculate:
a. Show the pH at the equivalence point Eq. Pt @30 mL pH = 8.5
b. Show the pH at the midpoint Mid Pt @15 mL pH = 5
c. Calculate the original M of HA M a V a = M b V b
M a (40 mL) = (0.10 M)(30 mL) M a = 0.075 M
d. Calculate K a of HA pK a = pH at midpoint so pK a = 5 or K a = 1x10 -5
Titration of 40 mL of ??? M weak acid HA
with 0.10 M NaOH
Eq. Pt @30 mL pH = 8.5
Mid Pt @15 mL pH = 5

9. If 25 mL of weak base B was titrated as the graph below shows, calculate:
a. Show the pH at the equivalence point Eq. Pt @60 mL pH = 4
b. Show the pH at the midpoint Mid Pt @ 30 mL pH = 8
c. Calculate the original M of B M a V a = M b V b (0.50 M)(60 mL) = (? M)(25 mL) M b = 1.2 M
d. Calculate K b of B pK a = pH at midpoint so pK b = pOH at midpoint or K b = 1x10 -6
Titration of 25 mL of ??? M weak base B
with 0.50 M HCl
Mid Pt @ 30 mL pH = 8
Eq. Pt @60 mL pH = 4
HCl

Buffers
1. What is the definition of a buffer? A solution that resists changes in pH.
2. What kind of an acid and base are present in a buffer? Weak acid and conj base or weak base
and conj acid
3. The K a of carbonic acid is 4.3x10 -7 .
a. What is the pH of a solution that is 0.25 M H 2 CO 3 and 0.25 M NaHCO 3 ?
-log (4.3x10 -7 ) = 6.4
b. What would the ratio of acid to base be if you wanted to buffer at exactly a pH of 6?
4.3x10 -7 = 1x10 -6 (B/A) B:A = 0.43:1 or A:B = 2.32:1
4. The K a of hydrozoic acid is 1.9x10 -5 .
a. What is the pH of a solution that is 0.25 M HN 3 and 0.75 M N -1 3 ?
K a = 1.9x10 -5 = [H 3 O +1 ](0.25/0.75) = 5.7x10 -5 pH = -log(5.7x10 -5 ) = 4.24
b. What would the ratio of acid to base be if you want to buffer exactly at a pH of 5?
1.9x10 -5 = 1x10 -5 (B/A) B:A = 1.9:1 or A:B = 0.53:1
5. The K b of aniline is 4x10 -10 .
a. What is the pH of a solution that is 0.050 M C 6 H 5 NH 2 and 0.050 M C 6 H 5 NH +1 3 ?
pOH = -log (4x10 -10 ) = 9.4 so pH = 4.6
or
pH = -log (2.5x10 -5 ) = 4.6
b. What would the ratio of acid to base be if you want to buffer exactly at a pH of 5?
2.5x10 -5 = 1x10 -5 (B/A) B:A = 2.5:1 or A:B = 0.4:1
6. The K b of ethyleamine is 4.4x10 -4 .
a. What is the pH of a solution that is 0.25 M C 5 H 5 N and 1.25 M C 5 H 5 NH +1 ?
K a = 1x10 -14 /4.4x10 -4 = 2.27x10 -11
2.27x10 -11 = [H 3 O +1 ] (0.25/1.25) = 1.14x10 -10 pH = -log(1.14x10 -10 ) = 9.95
b. What would the ratio of acid to base be if you want to buffer exactly at a pH of 10?
2.5x10 -11 = 1x10 -10 (B/A) B:A = 0.25:1 or A:B = 4:1
7. Which of the following buffer systems would you choose if you want to buffer at a pH of 2?
-1
a. HC 3 H 5 O 2 / C 3 H 5 O 2 K a = 1.35x10 -5
-1
b. HNO 2 / NO 2 K a = 7.2x10 -4
c. HIO / IO -1 K a = 2.3x10 -11
-1
d. H 3 PO 4 / H 2 PO 4 K a = 7.6x10 -3
You would pick system d. The K a value is closest to the pH you want to buffer at.
Describe how to make the buffer so it is exactly at a pH of 2.
7.6x10 -3 = 1x10 -2 (B/A) so: B:A = 0.76:1 or A:B = 1.316
8. Lactic acid has a K a of 8.3x10 -4 . A buffer is made so that it is 0.50 M in HC 3 H 5 O 3 and 0.50
M in C 3 H 5 O -1 3 .
a. What is the pH of the buffer? pH = -log(8.3x10 -4 ) = 3.1
b. What will the pH of the buffer be if 0.15 M of HCl are added?
0.15 M HCl will react away 0.15 M of the weak base and add 0.15 M of the weak acid
8.3x10 -4 = [H 3 O +1 ](0.35/0.65) = 1.54x10 -3 pH = -log(1.54x10 -3 ) = 2.8

9. Arsenous acid has a K a of 6.6x10 -10 . A solution is made so that it is 50 mL of 1.2 M in
HAsO 2 .
a. What is the pH of the solution?
6.6x10 -10 = x 2 /1.2 x = [H 3 O +1 ] = 2.8x10 -5 pH = 4.55
b. How many grams of NaAsO 2 must be added so the solution is a buffer at a pH of 10?
Assume no change in volume.
6.6x10 -10 = 1x10 -10 ([base]/1.2 M acid) so [base] = 7.92 M
[base] = 7.92 M so 7.92 moles/L * 0.05 L = 0.396 moles NaAsO 2
0.396 moles NaAsO 2 * 129.9 g/mole = 51.4 g NaAsO 2
10. Methylamine has a K b of 4.2x10 -4 . A buffer is made so that it is 0.75 M in CH 3 NH 2 and
0.75 M in CH 3 NH +1 3 .
a. What is the pH of the buffer?
pOH = -log(4.2x10 -4 ) = 3.4 so pH = 10.6
or
pH = -log(2.4x10 -11 ) = 10.6
b. What will the pH of the buffer be if 0.25 M of HNO 3 are added?
0. 25 M HNO 3 will react away 0.25 M of the weak base and add 0.25 M of the weak acid
2.4x10 -11 = [H 3 O +1 ](0.50/1.0) = 4.8x10 -11 pH = -log(4.8x10 -11 ) = 10.3
11. Quinoline has a K b of 6.3x10 -10 . A solution is made so that it is 250 mL of 1.75 M C 9 H 7 N.
a. What is the pH of the solution?
6.3x10 -10 = x 2 /1.75 x = 3.32x10 -5 pOH = 4.48 pH = 9.52
b. How many grams of C 9 H 7 NHCl must be added so that the solution is a buffer exactly
at a pH of 5? Assume no change in volume.
K a = 1x10 -14 /6.3x10 -10 = 1.59x10 -5
1.59x10 -5 = 1x10 -5 (1.75 M/[acid]) so [acid] = 1.1 M
[acid] = 1.1 M so 1.1 moles/L * 0.25 L = 0.275 moles acid
0.275 moles acid * 165.5 g/mole = 45.5 grams C 9 H 7 NHCl

Lewis Acids & Bases
1. What is the definition of a Lewis Acid? Electron pair acceptor
2. What is the definition of a Lewis Base? Electron pair donor
3. What is an amphoteric substance? A substance that can be a Bronsted base or a Lewis acid
4. Give 4 examples of amphoteric substances. Al(OH) 3 Zn(OH) 2 Cr(OH) 3
Sn(OH) 4
5. Aluminum hydroxide, Al(OH) 3 , is a famous amphoteric substance.
a. Write an equation showing Al(OH) 3 acting as a Bronsted base
Al(OH) 3 + 3 H 3 O +1 6 H 2 O + Al +3
b. Write an equation showing Al(OH) 3 acting as a Lewis acid
Al(OH) 3 + OH -1 [Al(OH) 4 ] -1
6. In the following equations, which species is acting as the Lewis acid and the Lewis base?
a. NH 3 + BF 3 BF 3 NH 3
LB LA
b. SnCl 4 + 2 Cl -1 [SnCl 6 ] -2
LA LB
c. H + + OH -1 H 2 O
LA LB

Combination Problems
1.
a) A 0.10 M solution of it has a starting pH of 3. If it was a strong acid it should have a pH of 1.
Also, the equivalence point is at about a pH of 9. It it was a strong acid, the equivalence point
should be at 7 through this titration with a strong base NaOH.
b) Find the pH at the midpoint of the titration. The equivalence point is at 25 so the midpoint is
at 12.5 mL. The pH at this point is 5 so pK a = 5 so K a = 1x10 -5
c) See above line in red
d) i) M a V a = M b V b (0.10)(25 mL) = (0.20 M)(x mL V b = 12.5 mL
ii) Less NaOH will be needed so the concentration will be higher resulting in a more
concentrated base and so a higher equivalence point

2.
a) NH 3 + H 3 O +1 NH +1 4 + H 2 O
b) Starting pH: 1.8x10 -5 = x 2 /0.10 x = 0.00134 pOH = 2.87 pH = 11.13
Equivalence point: (30 mL) (0.1 M) = (x mL)(0.2 M) x mL = 15 mL
NH 3 at equivalence point: (0.10 M)(30/45) = 0.0667 M
HCl at equivalence point: (0.20 M)(15/45) = 0.0667 M
pH at equivalence point: 1x10 -14 /1.8x10 -5 = x 2 /0.0667 x = 6.1x10 -6 pH = 5.2
c) Since the equivalence point is at 4.71, methyl red will be the closest to that as it changes color
at a pH of 5.5
d) Since K b for NH 3 is 1.8x10 -5 that means that K a for NH +1 4 is 1x10 -14 /1.8x10 -5 = 5.6x10 -10 .
Since K b is larger than K a , the solution will be basic

3. For the reaction:
NH 3 (aq) + H 2 O (l) NH 4 +1 (aq) + OH -1 (aq)
In 0.0180 M NH 3 at 25 o C, the [OH -1 ] is 5.6x10 -4 M.
a) Write the equilibrium-constant expression for the reaction above.
K b = [NH 4 +1 ][OH -1 ]/[NH 3 ]
b) Determine the pH of 0.0180 M NH 3
pOH = -log (5.6x10 -4 ) = 3.25 pH = 10.75
c) Determine the value of K b for NH 3
K b = (5.6x10 -4 ) 2 /(0.0180) = 1.74x10 -5
d) Determine the % ionization of NH 3
% Ionization = (5.6x10 -4 )/(0.0180) *100% = 3.11%
e) In an experiment, a 20.0 mL sample of 0.0180 M NH 3 was placed in a flask and
titrated to the equivalence point and beyond using 0.0120 M HCl
i) Determine the volume of 0.0120 M HCl that was added to reach the equivalence pt
M a V a = M b V b (0.0120)(V a ) = (0.0180)(20 mL) V a = 30 mL
ii) Determine the pH of the solution in the flask after a total of 15.0 mL of 0.0120 M
HCl was added
This is the midpoint of the titration so the midpoint pH = pK a
K b = 1.74x10 -5 so K a = 1x10 -14 /1.74x10 -5 = 5.747x10 -10
pH = -log (5.747x10 -10 ) = 9.24
iii) Determine the pH of the solution in the flask after a total of 40.0 mL of 0.0120 M
was added.
HCl: 0.0120 M (40/60) = 0.008 M
NH 3 : 0.0180 M (20/60) = 0.006 M
H 3 O +1 + NH 3 NH +1 4 + H 2 O
0.008 0.006
0.002 0
pH = log (0.002) = 2.70

4. For the reaction:
C 6 H 5 NH 2 (aq) + H 2 O (l) C 6 H 5 NH 3 +1 (aq) + OH -1 (aq)
a) Write the equilibrium constant expression, K b , for the reaction above.
K b = [C 6 H 5 NH 3 +1 ][OH -1 ]/[C 6 H 5 NH 2 ]
b) A sample of C 6 H 5 NH 2 is dissolved in water to produce 25.0 mL of a 0.010 M solution.
The pH of the solution is 8.82. Calculate K b .
b) pH = 8.82 pOH = 5.18 [OH -1 ] = 10 -5.18 = 6.61 x10 -6
C 6 H 5 NH 2 + H 2 O
+
C 6 H 5 NH 3 + OH -
[] o 0.10 0 0
-6.61 x10 -6 +6.61 x10 -6 +6.61 x10 -6
[] eq 0.1 6.61 x10 -6 6.61 x10 -6
K b = (6.61 x10 -6 ) 2 /0.1 = 4.37 x10 -10
c) The solution prepared in part (b) is titrated with 0.10 M HCl. Calclulate the pH of the
solution when 5.0 mL of the acid has been added.
c) Base = (0.1) (25/30) = 0.0833
Acid = (0.1) (5/30) = 0.0167
C 6 H 5 NH 2 + H 3 O +
+
C 6 H 5 NH 3
[] o 0.0833 0.0167 0
-0.0167 -0.0167 +0.0167
[] eq 0.0666 0 0.0167
+ H 2 O
C 6 H 5 NH 2 + H 2 O
+
C 6 H 5 NH 3 + OH -
[] o 0.0666 0.0167 0
-x +x +x
[] eq 0.0666 0.0167 x
4.37 x10 -10 = [C 6 H 5 NH 3 + ][OH -1 ]/[ C 6 H 5 NH 2 ]
4.37 x10 -10 = [0.0167][x]/[ 0.0666]
X = 1.74 x10 -9 pOH = -log 1.74 x10 -9 = 8.76 pH = 5.24
d) Calculate the pH at the equivalence point of the titration in (c)
d) C 6 H 5 NH 2 + H 3 O +
+
C 6 H 5 NH 3 + H 2 O
[] o 0.05 0.05 0
-0.05 -0.05 +0.05
[] eq 0 0 0.05
+
C 6 H 5 NH 3 + H 2 O C 6 H 5 NH 2 + H 3 O +
[] o 0.05 0 0
-x +x +x
[] eq 0.05 x x
K a = 1x10 -14 /4.37x10 -10 = 2.29 x10 -5
X = 0.00107 pH = -log 0.00107 = 2.97
x 2 /0.05 = 2.29 x10 -5 = K a

e) The pK a values for several indicators are given below. Which is most suitable for this
titration? Justify.
Erythrosine pK a = 3
Litmus pK a = 7
Thymolphthalein pK a = 10
e) As the pH changes at about 3, erythrosine is the best as it changes at the same pH value.
5. Hypochlorous acid, HOCl, is a weak acid. The K a for HOCl is:
K a = [H 3 O +1 ][OCl -1 ] = 3.2x10 -8
[HOCl]
a) Write a chemical equation showing how HOCl behaves as an acid in water.
a) HOCl (aq) + H 2 O (l) H 3 O + (aq) + OCl -1 (aq)
b) Calculate the pH of a 0.175 M solution of HOCl
3.2 x10 -8 = x 2 /0.175 x = 7.48 x10 -5 pH = -log x = log 7.48 x10 -5 = 4.13
c) Write the net ionic equation for the reaction between HOCl and NaOH
c) HOCl (aq) + OH -1 (aq) H 2 O (l) + OCl -1 (aq)
d) In an experiment, 20.00 mL of 0.175 M HOCl is titrated with 6.55 mL of 0.435 M NaOH
i) Calculate the number of moles of NaOH added
di) 0.0655 L x 0.435 moles/L = 0.0285 moles OH - added
ii) Calculate the [H 3 O +1 ] in the flask after the NaOH has been added
dii)
acid = 0.175 (20/26.55) = 0.132 M
base = 0.435 (6.55/26.55) = 0.107 M
HOCl (aq) + OH -1 (aq) H 2 O (l) + OCl -1 (aq)
[] o 0.132 0.107 0
-0.107 -0.107 +0.107
[] eq 0.025 0 0.107
HOCl (aq) + H 2 O (l) H 3 O + (aq) + OCl -1 (aq)
[] o 0.025 0 0.107
-x +x +x
[] eq 0.025 x 0.107
3.2x10 -8 = x(0.107/0.025) x = 7.48x10 -9 pH = 8.13
ii) Calculate the [OH -1 ] in the flask after the NaOH has been added
1x10 -14 /7.48x10 -9 = 1.34x10 -6 M OH -1