Solution to Exercises for Study Sheet 28
NOTE: [ ] = M
1. a) Since NaNO3 is a salt derived from strong acid (HNO3) and a strong base (NaOH) it has a
pH = 7.00
Strong Bases
Strong Acids
IAOH
HCl, HBr, HI, HClO3, HClO4,
st
IIA(OH)2 (except for: Be and Mg)
HNO3, H2SO4 (only 1 ionization step)
2O
Na2 SO4 ⎯H⎯⎯
→
b)
[ ]i
3.0
[ ]c
- 3.0
[ ]f
0
2 Na + (aq) +
SO4 2 − (aq)
0
0
+ 2(3.0)
+ 3.0
6.0
3.0
H −OH
−

SO4 2 − (aq) 

 HSO3 (aq) +
[ ]i
3.0
[ ]c
-x
[ ]e
3.0 - x
Ka Kb = Kw
Only ions that are derived from
either weak acids or weak bases
hydrolyze in water.
OH − (aq)
0
0
+x
+x
x
x
⇒ K b, SO 2− =
4
Kw
K a, HSO −
SO42-hydrolyzes in water because it is derived
from a weak acid (HSO4-).
Substitute equilibrium concentrations
into equilibrium expression.
[HSO3− ] [OH − ]
=
[SO4 2 − ]
4
Na+ doesn’t hydrolyze in water since it is
derived from a strong base (NaOH).
1 x 10 −14
= 8.3 x 10 −13
1.2 x 10 −2
=
4
K b, SO 2−
Na2SO4 is a water soluble salt that
completely ionizes in water.
8.3 x 10 −13 =
[x] [x]
[3.0 − x]
Assume that x is small
2.5 x 10
−12
= x
⇒
2
−
x =
[OH ]e = x = 1.58 x 10
−6
2.5 x 10
−12
= 1.58 x 10
−6
M
−
pOH = − log[OH ] = − log[1.58 x 10 −6 ] = 5.80
pH + pOH = 14 ⇒ pH + 5.80 = 14 ⇒ pH = 8.20
c)
2O
NH 4 Cl ⎯H⎯⎯
→ NH 4 + (aq) +
Cl − (aq)
[ ]i
0.5
0
0
[ ]c
- 0.5
+ 0.5
+ 0.5
[ ]f
0
0.5
0.5
H2O


NH 4 + (aq) 

 NH 3
[ ]i
0.5
[ ]c
-x
[ ]e
0.5 - x
(aq)
+
H 3O + (aq)
0
0
+x
+x
x
x
Note how the salt (Na2SO4)
derived from a strong base and
a weak acid has a pH > 7.
NH4Cl is a water soluble salt that
completely ionizes in water.
NH4+ hydrolyzes in water since it is derived
from a weak base (NH3).
Cl – doesn't hydrolyze in water because it is
derived from a strong acid (HCl).
Ka Kb = Kw
K a, NH +
4
⇒ K a, NH + =
4
K b, NH 3
1 x 10 −14
=
= 5.6 x 10 −10
−5
1.8 x 10
Substitute equilibrium concentrations
into equilibrium expression.
[NH 3 ] [H + ]
=
[NH 4 + ]
2.8 x 10 −10 = x 2
Kw
5.6 x 10 −10 =
[NOTE: H+= H3O+]
[x] [x]
[0.5 − x]
Assume that x is small
⇒
2.8 x 10 −10 = 1.67 x 10 −5
x =
[H + ]e = x = 1.67 x 10 −5 M
pH = − log[H + ] = − log[1.67 x 10 −5 ] = 4.78
2O
KC2 H 3O2 ⎯H⎯⎯
→ K + (aq) +
d)
C2 H 3O2 − (aq)
[ ]i
0.2
0
[ ]c
- 0.2
+ 0.2
+ 0.2
[ ]f
0
0.2
0.2
0.2
[ ]c
[ ]e
(aq)
0
-x
+x
+x
0.2 - x
x
x
K b, C H O −
3 2
⇒ K b, C H O − =
2
3 2
[HC2 H 3O2 ] [OH − ]
=
[C2 H 3O2 − ]
Kw
K a, HC2 H 3O2
K+ doesn’t hydrolyze in water since it is
derived from a strong base (KOH).
OH − (aq)
+
0
Ka Kb = Kw
2
KC2H3O2 is a water soluble salt that
completely ionizes in water.
0
H −OH

C2 H 3O2 − (aq) 

 HC2 H 3O2
[ ]i
Note how the salt (NH4Cl) derived from a
weak base and a strong acid has a pH < 7.
=
C2H3O2- hydrolyzes in water because it is
derived from a weak acid (HC2H3O2).
1 x 10 −14
= 5.6 x 10 −10
1.8 x 10 −5
Substitute equilibrium concentrations
into equilibrium expression.
5.6 x 10 −10 =
[x] [x]
[0.2 − x]
Assume that x is small
1.1 x 10
−10
= x
2
⇒
x =
1.1 x 10
−10
= 1.06 x 10
−5
[OH − ]e = x = 1.06 x 10 −5 M
pOH = − log[OH − ] = − log[1.06 x 10 −5 ] = 4.98
pH + pOH = 14 ⇒ pH + 4.98 = 14 ⇒ pH = 9.02
2.
b(CH 3 )3 N
s(CH
3 )3
NH + Cl −
• K b, (CH 3 )3 N = [OH − ]
Weak base buffer formula; b and s can either be in moles or M.
0.25 mol
• 7.4 x 10 −5 = [OH − ] = 4.63 x 10 −5 M
0.40 mol
pOH = − log[OH − ] = − log[4.63 x 10 −5 ] = 4.33
pH + pOH = 14
⇒
Note how the salt (KC2H3O2) derived from a
strong base and a weak acid has a pH > 7.
pH + 4.33 = 14
⇒
pH = 9.67
3.
aHNO2
• K a, HNO2 = [H + ]
sNaNO2
Weak acid buffer formula; a and s can either be in moles or M.
0.20 M
• 4.5 x 10 −4 = 10 −3.80
s
0.20
• 4.5 x 10 −4 = 0.568 M
10 −3.80
s = [NaNO2 ] =
acid
+
4. a)
[H + ] = 10 − pH
(CH 3 )3 NH Cl
−
base
+
NaOH
moli
0.40
0.025
molc
- 0.025
- 0.025
molf
0.375
0
⎯⎯
→
For pH calculations, we don’t need to
keep track of substances that have a
pH = 7 because [H+] = [OH-].
+
(CH 3 )3 N
0.25
NaCl
+
H 2O
⎛ 0.500 mol ⎞
? mol NaOH = 50.0 mL ⎜
= 0.025 mol
⎝ 1000 mL ⎟⎠
+ 0.025
0.275
These are the constituents of a buffer (i.e., a
weak base and it’s common ion salt).
b(CH 3 )3 N
s(CH
3 )3
NH + Cl −
• K b, (CH 3 )3 N = [OH − ] 0.275 mol
• 7.4 x 10 −5 = [OH − ] = 5.42 x 10 −5 M
0.375 mol
pOH = − log[OH − ] = − log[5.42 x 10 −5 ] = 4.27
pH + pOH = 14
⇒
pH + 4.27 = 14
base
acid
(CH 3 )3 N +
b)
HCl ⎯⎯
→
moli
0.25
0.025
molc
- 0.025
- 0.025
molf
0.225
0
⇒
pH = 9.73
NOTE: The addition of a small
amount of strong base slightly
raised the pH of the buffer (i.e.,
from 9.67 to 9.73).
(CH 3 )3 NH + Cl −
0.40
⎛ 0.500 mol ⎞
? mol HCl = 50.0 mL ⎜
= 0.025 mol
⎝ 1000 mL ⎟⎠
+ 0.025
0.425
These are the constituents of a buffer (i.e., a
weak base and it’s common ion salt).
b(CH 3 )3 N
s(CH
3 )3
NH + Cl −
• K b, (CH 3 )3 N = [OH − ]
0.225 mol
• 7.4 x 10 −5 = [OH − ] = 3.92 x 10 −5 M
0.425 mol
pOH = − log[OH − ] = − log[3.92 x 10 −5 ] = 4.41
pH + pOH = 14
⇒
pH + 4.41 = 14
⇒
pH = 9.59
NOTE: The addition of a small
amount of strong acid slightly
lowered the pH of the buffer
(i.e., from 9.67 to 9.59).
For pH calculations, we don’t need to keep track of
substances that have a pH = 7 because [H+] = [OH-].
5.
NaF
+
⎯⎯
→
HCl
+
HF
moli
0.071
0.020
molc
- 0.020
- 0.020
+ 0.020
molf
0.051
0
0.020
NaCl
⎛ 1 mol NaF ⎞
? mol NaF = 3.0 g NaF ⎜
= 0.071 mol
⎝ 42.0 g NaF ⎟⎠
0
⎛ 2.0 mol HCl ⎞
? mol HCl = 10.0 mL ⎜
= 0.020 mol
⎝ 1000 mL ⎟⎠
These are the constituents of a buffer (i.e., a
weak acid and it’s common ion salt).
a HF
• K a, HF = [H + ]
sNaF
0.020 mol
• 6.5 x 10 −4 = [H + ]
0.051 mol
[H + ] = 2.55 x 10 −4 M
pH = − log [H + ] = − log[2.55 x 10 −4 ] = 3.59
base
NH 3
6. a)
acid
+
HCl
⎯⎯
→
moli
0.040
0.010
0.050
molc
- 0.010
- 0.010
+ 0.010
molf
0.030
0
0.060
b NH 3
sNH 4 Cl
⎛ 0.20 mol NH 3 ⎞
? mol NH 3 = 200 mL ⎜
= 0.040 mol
⎝ 1000 mL ⎟⎠
NH 4 Cl
⎛ 0.25 mol NH 3 ⎞
? mol NH 4 Cl = 200 mL ⎜
= 0.050 mol
⎝ 1000 mL ⎟⎠
⎛ 1 mol HCl ⎞
? mol HCl = 10. mL ⎜
= 0.010 mol
⎝ 1000 mL ⎟⎠
These are the constituents of a buffer (i.e., a
weak base and it’s common ion salt).
• K b, NH 3 = [OH − ]
0.030 mol
• 1.8 x 10 −5 = [OH − ]
0.060 mol
[OH − ] = 9.0 x 10 −6 M
pOH = − log [OH − ] = − log[9.0 x 10 −6 ] = 5.05
pH + pOH = 14
⇒
pH + 5.05 = 14
⇒
For pH calculations, we don’t need to keep
track of substances (i.e., NaCl and H2O)
that have a pH = 7 because [H+] = [OH-].
NH3 + H2O
acid
b)
NH 4 Cl
base
+
NaOH
⎯⎯
→
NH 4OH
moli
0.0375
0.0125
0.030
molc
- 0.0125
- 0.0125
+ 0.0125
molf
0.0250
0
0.0425
These are the constituents of a buffer (i.e., a
weak base and it’s common ion salt).
pH = 8.95
+
NaCl
⎛ 0.20 mol NH 3 ⎞
? mol NH 3 = 150 mL ⎜
= 0.0300 mol
⎝ 1000 mL ⎟⎠
⎛ 0.25 mol NH 4 Cl ⎞
? mol NH 4 Cl = 150 mL ⎜
⎟⎠ = 0.0375 mol
1000 mL
⎝
⎛ 1 mol NaOH ⎞
? mol NaOH = 0.50 g NaOH ⎜
= 0.0125 mol
⎝ 40.0 g NaOH ⎟⎠
b NH 3
sNH 4 Cl
• K b, NH 3 = [OH − ]
0.0425 mol
• 1.8 x 10 −5 = [OH − ]
0.0250 mol
[OH − ] = 3.06 x 10 −5 M
pOH = − log [OH − ] = − log[3.06 x 10 −5 ] = 4.51
pH + pOH = 14
⇒
pH + 4.51 = 14
⇒
pH = 9.49
For pH calculations, we don’t need to keep track of
substances that have a pH = 7 because [H+] = [OH-].
acid
7.
HCl
base
+
NaOH
moli
0.00320
0.00625
molc
- 0.00320
- 0.00320
molf
0
⎯⎯
→
NaCl
+
H 2O
⎛ 0.100 mol HCl ⎞
? mol HCl = 32.0 mL ⎜
= 0.00320 mol
⎝ 1000 mL ⎟⎠
0.00305
Since NaOH is a strong base,
mol NaOH = mol OH-.
⎛ 0.250 mol NaOH ⎞
? mol NaOH = 25.0 mL ⎜
⎟⎠ = 0.00625 mol
1000 mL
⎝
⎡ 0.00305 mol OH − ⎤
pOH = − log [OH ] = − log ⎢
⎥ = 1.27
−3
⎣ 57.0 x 10 L ⎦
−
pH + pOH = 14
⇒
pH + 1.27 = 14
⇒
pH = 12.70
Total Volume
32.0 mL + 25.0 mL = 57.0 mL
57.0 x 10-3 L