15 Titration Curves for Complex Acid/Base Systems
(1) Two acids or two bases of different strength.
(2) An acid or a base that has two or more acids or bases of different strengths
(3) An amphiprotic substance
15A Mixtures of Strong and Weak Acids or Strong and weak
Bases
Ex. 15-1,2 Calculate the pH of a mixture (25 mL) that is 0.1200 M in HCl and
0.0800 M in HA (Ka = 1.00 × 10-4) during its titration with 0.1000 M KOH
at (a) 0.00 mL (b) 5.00 mL and (c) 29.00 mL.
(a) 0.00 mL KOH added
[H3O+] = cHCl + [A-] = 0.1200 + [A-] ≈ 0.1200 M, pH = 0.92
(assume [A-] << 0.1200M)
Check this assumption:
Ka
[A - ]
1.00 × 10 −4
[A - ]
−4
=
=
= 8.33 × 10 , [HA ] =
[HA ] [ H 3 O + ]
0.1200
8.33 × 10 −4
cHA = [HA] + [A-] = 0.0800 M
[A - ]
+ [A - ] ≈ (1.20 × 10 3 )[A - ] = 0.0800 M , [A-] = 6.7× 10-5 M
−4
8.33 × 10
(b) After adding 5.00 mL of base
c HCl =
25.00 × 0.1200 − 5.00 × 0.100
= 0.0833 M
25.00 + 5.00
[H3O+] = 0.0833 + [A-] ≈ 0.0833 M, pH = 1.08
Check this assumption:
[A - ] 1.00 × 10 −4
[A - ]
=
= 0.001200 M , [HA ] =
, cHA = [HA] + [A-]
[HA ]
0.0833
0.001200
[A - ]
0.0800 × 25.00
[HA ] =
+ [A - ] ≈ (8.34 × 10 2 )[A - ] =
= 0.0667 M
0.001200
30.00
[A-] = 8.0 × 10-5 M ( << 0.0833 M )
(c) After adding 29.00 mL of base
c HCl =
25.00 × 0.1200 − 29.00 × 0.100
= 1.85 × 10 −3 M
25.00 + 29.00
92
c HA =
25.00 × 0.0800
= 3.70 × 10 − 2 M
25.00 + 29.00
[H3O+] = cHCl + [A-] = 1.85×10-3 + [A-], cHA = [HA] + [A-] = 3.70×10-2 M
[H 3 O + ][A - ]
[HA ] =
,
1.00 × 10 −4
[A - ] =
[H 3 O + ][A - ]
+ [A - ] = 3.70 × 10 2
−4
1.00 × 10
3.70 × 10 −6
3.70 × 10 −6
+
−3
, [H 3 O ] = 1.85 × 10 +
[H 3 O + ] + 1.00 × 10 −4
[H 3 O + ] + 1.00 × 10 − 4
[H3O+]2 + (1.00 × 10-4) [H3O+] = (1.85 × 10-3) [H3O+] + 1.85 × 10-7 + 3.7 × 10-6
[H3O+]2 - (1.75 × 10-3) [H3O+] - 3.885 × 10-6 = 0
[H3O+] = 3.03 × 10-3 M,
pH = 2.52
3.03 × 10-3 – 1.85 × 10-3 = 1.18 × 10-3 ([H3O+] from HA)
1.85 × 10-3 ([H3O+] from HCl)
Fig 15-1 Curves for the titration of
strong acid/weak acid mixtures with
0.1000 M NaOH. Each titration is on
25.00 mL of a solution that is 0.1200 M
in HCl and 0.0800 M in HA.
15B Polyfunctional Acids and Bases
15B-1 The phosphoric acid system
H3PO4 + H2O ⇔ H2PO4- + H3O+
[H 3O + ][H 2 PO -4 ]
K a1 =
= 7.11 × 10 − 3
[H 3PO 4 ]
H2PO4- + H2O ⇔ HPO42- + H3O+
K a2 =
HPO42- + H2O ⇔ PO43- + H3O+
K a3 =
[H 3O + ][HPO 24- ]
[H 2 PO -4 ]
[H 3O + ][PO34- ]
[HPO 24 - ]
= 6.32 × 10 −8
= 4.5 × 10 −13
Ka1 > Ka2 > Ka3
93
H3PO4 + 3H2O ⇔ PO43- + 3H3O+
[H 3O + ]3[PO 34- ]
K a1K a2 K a3 =
[H 3PO 4 ]
= 7.11 × 10 − 3 × 6.32 × 10 −8 × 4.5 × 10 −13 = 2.0 × 10 − 22
15B-2 The carbon dioxide carbonic acid system
K hyd =
CO2(aq) + H2O ⇔ H2CO3
+
[ H 3 O + ][HCO 3- ]
= 1.5 × 10 − 4
K1 =
[H 2 CO 3 ]
-
H2CO3 + H2O ⇔ H3O + HCO3
-
+
[H 2 CO 3 ]
= 2.8 × 10 − 3
[CO 2 (aq)]
[ H 3 O + ][CO 32- ]
K2 =
= 4.69 × 10 −11
[HCO 3 ]
2-
HCO3 + H2O ⇔ H3O + CO3
---------------------------------------------------------------------------------------CO2(aq) + 2H2O ⇔ H3O+ + HCO3[ H 3 O + ][HCO 3- ]
K a1 =
= 2.8 × 10 −3 × 1.5 × 10 − 4 = 4.2 × 10 −7
[CO 2 (aq)]
-
+
[ H 3 O + ][CO 32- ]
K2 =
= 4.69 × 10 −11
[HCO 3 ]
2-
HCO3 + H2O ⇔ H3O + CO3
Ex. 15-3 Calculate the pH of a solution that is 0.02500 M CO2.
cCO 2 = 0.02500 = [CO 2 (aq)] + [H 2 CO 3 ] + [HCO 3- ] + [CO 32- ]
[CO 2 (aq)] >> [H 2 CO 3 ] + [HCO 3- ] + [CO 32- ], [CO 2 (aq)] ≈ cCO 2 = 0.02500 M
[H3O+] = [HCO3-] + 2[CO32-] + [OH-],
From charge-balance
2[CO32-] + [OH-] << [HCO3-]
→ [H3O+] ≈ [HCO3-]
[H 3O + ]2
K a1 =
= 4.2 × 10 −7 , [ H 3 O + ] = 0.02500 × 4.2 × 10 -7 = 1.02 × 10 − 4 M
[CO 2 (aq)]
pH = -log(1.02 × 10-4 ) = 3.99
15C Buffer Solutions Involving Polyprotic Acids
Ex. 15-4 Calculate the [H3O+] for a buffer solution that is 2.00 M in H3PO4 and
1.50M in KH2PO4.
-
+
H3PO4 + H2O ⇔ H2PO4 + H3O
[H 3O + ][H 2 PO -4 ]
K a1 =
= 7.11 × 10 − 3
[H 3PO 4 ]
94
H2PO4- + H2O ⇔ HPO42- + H3O+
K a2 =
[H 3O + ][HPO 24- ]
[H 2 PO -4 ]
= 6.32 × 10 −8
Assume: [HPO42-] and [PO43-] << [H2PO4-] and [H3PO4]
[H 3 PO 4 ] ≈ cH3PO 4 = 2.00 M,
[H 2 PO -4 ] ≈ cH PO- = 1.50 M
2
4
7.11×10 -3 × 2.00
[H 3 O ] =
= 9.48 × 10 −3 M
1.50
Check the assumption
+
[H 3 O + ][HPO 24- ]
[H 2 PO -4 ]
9.48 ×10 −3 [HPO 24- ]
=
= 6.34 ×10 −8
1.50
[HPO42-] = 1.00 × 10-5 M and [PO43-] < [HPO42-]
the assumption is valid
*For a buffer prepared from NaHA and Na2A
HA- + H2O ⇔ H2A + OH- is disregarded
[H2A] << [HA-] and [A2-]
Ex. 15-5 Calculate the [H3O+] for a buffer solution that is 0.0500 M in potassium
hydrogen phthalate (KHP) and0.150M in potassium phthalate (K2P).
-
HP + H2O ⇔
+
H3O + P
2-
K a2 =
[H 3 O + ][P 2- ]
-
[HP ]
= 3.91×10 −6
Assume: [H2P] is negligible
[HP - ] ≈ cKHP = 0.0500 M,
[P 2- ] ≈ cK 2P = 0.150 M
3.91×10 -6 × 0.0500
[H 3 O ] =
= 1.30 × 10 −6 M
0.150
Check the assumption
+
(1.30 × 10 -6 )(0.0500)
= K a1 = 61.12 × 10 −3
[H 2 P]
[H2P] = 6 × 10-5 M and [H2P] << [HP-] and [P2-],
the assumption is valid
15D Calculation of the pH of Solution of NaHA
HA- + H2O ⇔ A2- + H3O+
HA- + H2O ⇔ H2A + OH-
K a2 =
K b2
[H 3 O + ][A 2- ]
[HA - ]
K w [H 2 A][OH - ]
=
=
K a1
[HA - ]
For mass balance: cNaHA = [H2A] + [HA-] + [A2-]
95
For charge balance: [Na+] + [H3O+] = [OH-] + [HA-] + 2[A2-] = cNaHA+ [H3O+]
[H2A] + [HA-] + [A2-]+ [H3O+] = [OH-] + [HA-] + 2[A2-]
[H3O+] = [A2-] + [OH-]- [H2A]
[H 3 O + ][HA - ]
K a2 [HA - ]
2[H 2 A] =
, [A ] =
K a1
[H 3 O + ]
[H 3 O + ] =
K a2 [HA - ]
[H 3 O + ][HA - ]
Kw
+
−
K a1
[H 3 O + ] [H 3 O + ]
( × [H3O+] )
⎞
[H 3 O + ]2 [HA - ]
+ 2 ⎛ [HA ]
+ 1⎟⎟ = K a2 [HA - ] + K w
[H 3 O ] = K a2 [HA ] + K w −
→ [H 3 O ] ⎜⎜
K a1
⎝ K a1
⎠
+ 2
-
K a2 [HA - ] + K w
[H 3 O ] =
[HA - ]
1+
K a1
+
if CNaHA/Ka1 >> 1 and
→
+
→ [H 3 O ] =
K a2 C NaHA + K w
C
1 + NaHA
K a1
Ka2CNaHA >> Kw
[H3O + ] ≅ K a1K a 2
Ex. 15-6 Calculate the [H3O+] of a 1.00 × 10-3 M Na2HPO4 solution.
Ka2 = 6.32 × 10-8 and Ka3 = 4.5 × 10-13
CNa2HPO4/Ka2 = (1.00 × 10-3)/(6.32 × 10-8) >> 1
and Ka3CNa2HPO4 = 4.5 × 10-13 × 1.00 × 10-3 < Kw
K a3C Na 2 HPO 4 + K w
4.5 × 10 −13 × 10 −3 + 10 −14 = 8.1 × 10-10
=
[H 3O ] =
C Na 2 HPO 4
10 −3
1+
K a2
6.32 × 10 −8
+
Ex. 15-7 Find the [H3O+] of a 0.0100 M NaH2PO4 solution.
Ka1= 7.11 × 10-3 and Ka2= 6.32 × 10-8
CNaH2PO4/Ka1 = 0.0100/7.11 × 10-3
and Ka2CNaH2PO4 = 6.32 × 10-8 × 0.0100 >> Kw
[H3O + ] =
K a2C NaH 2 PO 4 + K w
6.32 × 10 −8 × 10 − 2 = 1.62 × 10-5
=
C NaH 2 PO 4
10 − 2
1+
1+
K a1
7.11× 10−3
96
Ex. 15-8 Calculate the [H3O+] of a 0.100 M NaHCO3 solution.
CO2 + 2H2O ⇔ H3O+ + HCO3K a1 =
-
+
[ H 3 O + ][HCO 3- ]
= 2.8 × 10 −3 × 1.5 × 10 − 4 = 4.2 × 10 −7
[CO 2 (aq)]
2-
HCO3 + H2O ⇔ H3O + CO3
K2 =
[ H 3 O + ][CO 32- ]
= 4.69 × 10 −11
[HCO 3 ]
∵ CNaHA/Ka1 >> 1 and Ka2CNaHA >> Kw
[H 3 O + ] = 4.2 × 10 −7 × 4.69 × 10 −11 = 4.4 × 10 −9 M
15E Titration Curves for Polyfunctional Acids
Fig. 15-2 Titration of 20.00 mL of
0.1000 M H2A with 0.1000 M
NaOH. For H2A, Ka1 = 1.00×
10-3 and Ka2 = 1.00 × 10-7. The
method of pH calculation is
shown for several points and
regions on the titration curve.
Volume of 0.100 M NaOH, mL
Ex. 15-9. Construct a curve for the titration of 25.00 mL of 0.1000 M maleic acid,
HOOC-CH=CH-COOH, with 0.1000 M NaOH.
H2M + H2O ⇔ H3O+ + HM-
Ka1 = 1.3 × 10-2
HM- + H2O ⇔ H3O+ + M2Ka1/Ka2 is large (2 × 104)
Ka2 = 5.9 × 10-7
Initial pH
[H3O+] ≈ [HM-], [H2M] + [HM-] ≈ 0.1000
[H2M] = 0.1000 - [HM-] = 0.1000 - [H3O+]
K a1 = 1.3 × 10
−2
[H 3 O + ] 2
=
0.1000 − [H 3 O + ]
[H3O+]2 + 1.3 × 10-2 [H3O+] - 1.3 × 10-3 = 0
[H3O+] = 3.01 × 10-2, pH = 2 - log 3.01 = 1.52
97
First Buffer Region: addition of 5.00 mL of base
CNaHM ≈ [HM-] = 5.00 × 0.1000/30.00 = 1.67 × 10-2 M
CH2M ≈ [H2M] = (25.00 - 5.00) × 0.1000/30.00 = 6.67 × 10-2 M
[H3O+] << CH2M or CHM- is not valid
[HM-] = 1.67 × 10-2 + [H3O+] - [OH-]
[H2M] = 6.67 × 10-2 - [H3O+] + [OH-]
K a1 =
[H 3O + ](1.67 × 10 − 2 + [H 3O + ])
6.67 × 10 − 2 − [H 3O + ]
= 1.3 × 10 − 2
[H3O+]2 + 2.97 ×10-2 [H3O+] - 8.67 ×10-4 = 0
[H3O+] = 1.81 ×10-2
pH = -log 1.81 × 10-2 = 1.74
Just Prior to First Equivalence Point
CNaHM ≈ [HM-] = 24.90 × 0.1000/49.90 = 4.99 × 10-2 M
CH2M ≈ [H2M] = (25.00 – 24.90) × 0.1000/49.90 = 2.00 × 10-4 M
Mass balance: CH2M + CNaHM = [H2M] + [HM-] + [M2-]
Charge balance : [H3O+] + [Na+] = [HM-] + 2[M2-] + [OH-]
CNaHM = [HM-] + 2[M2-] – [H3O+] → [H3O+] = CH2M + [M2-] – [H2M]
+
[H 3 O ] = c H 2 M
[HM-] ≈ 4.99 × 10-2 M,
[H 3 O + ][HM - ]
+
−
K a1
[H 3 O + ]
K a 2 [HM - ]
CH2M = 2.00 × 10-4 M
Ka1 = 1.3 × 10-2
Ka2 = 5.9 × 10-7
⎛ [HM - ] ⎞
⎟⎟ − cH 2M [H 3 O + ] − K a 2 [HM - ] = 0
[H 3 O + ]2 ⎜⎜1 +
K a1 ⎠
⎝
[H3O+]2 + 2.00 ×10-4 [H3O+] – 2.94 ×10-8 = 0
[H3O+] = 1.014 ×10-4 pH = 3.99
at adding 24.99 mL of titrant, [H3O+] = 8.01 ×10-5
pH = 4.10
First Equivalence Point
[HM-] ≈ CNaHM = 25.00 × 0.1000/50.00 = 5.00 × 10-2 M
K a2 C NaHM
5.9 × 10 −7 × 5.00 × 10 −2
=
[H 3 O ] =
1 + C NaHM / K a1
1 + (5.00 × 10 − 2 ) /(1.3 × 10 − 2 )
+
= 7.80 × 10-5
pH = -log (7.80 × 10-5) = 4.11
Just After the First Equivalence Point
CHM- = [25.00-(25.01-25.00)] × 0.1000/50.01 = 0.04997 M
CM2-= (25.01 - 25.00) × 0.1000/50.01 = 1.996 × 10-5 M
98
Mass balance: CHM- + CM2- = [H2M]+[HM-]+[M2-]=0.04997+1.996×10-5=0.049999 M
Charge balance : [H3O+] + [Na+] = [HM-] + 2[M2-] + [OH-]
[Na+] = 25.01 × 0.1000/50.01 = 0.05001 M
[H3O+] = [M2-] – [H2M] – (0.05001 – 0.049999)
K a 2 [HM - ] [H 3 O + ][HM - ]
[H 3 O ] =
−
− 1.9996 × 10 −5
+
K a1
[H 3 O ]
+
1.9996 ×10 −5 ± (1.9996 ×10 −5 ) 2 − 4 × 4.8438 × (−2.948 ×10 −8 )
=
= 7.40 × 10 −5 M
2 × 4.8438
pH = 4.13
Second Buffer Region : addition of 25.50 mL of NaOH
[M2-] ≈ CNa2M ≈ (25.50 - 25.00) × 0.1000/50.50 = 0.050/50.50 M
[HM-] ≈ CNaHM ≈ [25.00-(25.50-25.00)] × 0.1000/50.50 = 2.45/50.50 M
[H3O+] = 5.9 × 10-7 (2.45/50.50) /(0.050/50.50) = 2.89 × 10-5 M
[H3O+] << CNa2M and CNaHM is valid and pH = -log 2.89 × 10-5 = 4.54
Just Prior to Second Equivalence Point
at adding 49.90 and 49.99 mL of titrant, M2- >> HMat adding 49.90 mL CHM- = 1.335 × 10-4 and CM2- = 0.3324
M2- + H2O ⇔ HM- + OHK b1 =
K w [OH − ][HM - ] [OH − ](1.335 × 10 −4 + [OH - ]) 1.00 × 10 −14
=
=
=
= 1.69 × 10 −8
2−7
K a2
[M ]
(0.03324 − [OH ])
5.9 × 10
[OH-]2 + (1.335 × 10-4 + Kb1) [OH-] – 0.03324 Kb1 = 0
[OH-] = 4.10 × 10-6 M → pOH = 5.39 → pH = 14 – 5.39 = 8.61
at adding 49.90 mL: [OH-] = 1.80 × 10-5 M → pH = 9.26
Second Equivalence Point: addition of 50.00 mL of NaOH
[Na2M] = 0.0333 M
M2- + H2O ⇔ OH- + HM[OH-] ≈ [HM-],
K w [OH - ][HM - ]
K b1 =
=
= 1.69 × 10 −8
2
Ka2
[M ]
[M2-] = 0.0333 - [OH-] ≅ 0.0333
[OH-]2/0.0333 = 1.69 × 10-8, [OH-] = 2.38 × 10-5
pH = 14.00 – (- log 2.38 × 10-5) = 9.38
99
pH Just Beyond the Second Equivalence Point: add 50.01 mL of NaOH
cM2- =25.00 × 0.1/75.01 =0.03333 M
excess [OH-] = 1.00 × 0.01/75.01 = 1.3333 × 10-5 M
[M2-] = cM2- – [HM-] = 0.0333 – [HM-],
K b1
[OH-] = 1.3333 × 10-5 + [HM-]
[OH - ][HM - ] [HM - ](1.3333 ×10 −5 + [HM - ])
=
=
[M 2- ]
0.03333 − [HM - ]
[HM-]2 + (1.33 × 10-5 + Kb1) [HM-] – 0.03333 Kb1 = 0
[HM-] = 1.807 × 10-5 M
[OH-] = 1.3333 × 10-5 + 1.807 × 10-5 = 3.14 × 10-5 M
pOH = 4.50 and pH = 14 – pOH = 9.50
add 50.10 mL of NaOH pH = 10.14
pH Beyond the Second Equivalence Point: add 51.00 mL of NaOH
[OH-] = 1.00 × 0.1000/76.0 = 1.32 × 10-3
pH = 14.00 - (- log 1.32 × 10-3) = 11.12
Volume of 0.1000 M NaOH, mL
Volume of 0.1000 M NaOH, mL
Fig. 15-3 Titration curve for Fig. 15-4 Curve for the
25.00 mL of 0.1000 M titration of 0.1000 M H3PO4
maleic acid, H2M, with
(A), 0.1000M oxalic acid
0.1000 M NaOH.
(B), and 0.1000 M
H2SO4(C).
Volume of 0.1000 M HCl, mL
Fig. 15-5 Curve for the
titration of 25.00 mL of
0.1000M Na2CO3 with
0.1000 M HCl.
15F Titration Curves for Polyfunctional Bases
CO3 + H2O ⇔ HCO3 + OH
K w 1.00 × 10 −14
−4
=
2
.
13
×
10
K b1 =
=
K a 2 4.69 × 10 −11
HCO3- + H2O ⇔ H2CO3 + OH-
K w 1.00 × 10 −14
−11
K b2 =
=
=
6
.
7
×
10
K a1 1.5 × 10 − 4
2-
-
-
100
15G Titration Curves for Amphiprotic Species
In NaH2PO4 solution: can be titrate with a standard base solution
K a2 = 6.32 × 10 −8
H2PO4- + H2O ⇔ HPO42- + H3O+
-
-
H2PO4 + H2O ⇔ OH + H3PO4
K b3
K w 1.00 × 10 −14
=
=
= 1.41× 10 −12
−3
K a1 7.11 × 10
In Na2HPO4 solution: can be titrate with a standard acid solution
K a3 = 4.5 ×10 −13
HPO42- + H2O ⇔ PO43- + H3O+
2-
-
K b2 =
-
HPO4 + H2O ⇔ OH + H2PO4
K w 1.00 × 10 −14
=
= 1.58 × 10 −7
−8
K a2 6.32 × 10
15H The Composition of Solutions of a Polyprotic Acid as a
Function of pH
α0 =
[H 2 M]
CT
[HM - ]
α1 =
CT
α2 =
α0 + α1 + α2 = 1
CT = [H2M] + [HM-] + [M2-]
α0 =
[H 3O + ]2
[H 3O + ]2 + K a1[H 3O + ] + K a1K a2
α2 =
[M 2 - ]
CT
α1 =
K a1[H 3O + ]
[H 3O + ]2 + K a1[H 3O + ] + K a1K a2
K a1K a2
[H3O + ]2 + K a1[H3O + ] + K a1K a2
Fig 15-6 Composition of HM
Fig. 15-7
solution as a function of pH.
Fig. 15-7 Titration of 25.00 mL of 0.1000 M maleic acid with 0.1000 M NaOH. The
solid curves are plots of alpha values as a function of volume. The broken
curve is a plot of pH as a function of volume.
101
102