4.4
4.4 CALCULATION
CALCULATION OF
OF ACIDITY
ACIDITY
4.4.1 Proton Balance Equation
4.4.2 Calculation the acidity of monoprotic acid (base)
4.4.3 Calculation the aacidity
cidity of amphiprotic substance
4.4.4 The acidity of other acid-base systems
4.4.1 Proton Balance Equation
4.4.1 Proton Balance Equation
Proton Balance Equation
1.
1.P
Also known as proton condition, which is a
mathematical equation to describe the relation
between the proton acceptors and proton donors.
It can be obtained either by zero level method or
from the mass and charge balance equations
alance Equation or electroneutrality
Charge B
Balance
relation, means that the charges on the positively
charged species equal to the charges on negatively
charged species
4.4.1 Proton Balance Equation
2. Zero level method
① set the zero level of proton: the species from
which the solution is made up.
② other species are compared with the zero
level, and classified as containing proton
excesses or proton deficiencies
③ the proton condition is obtained by balancing
all the proton excess on the left-hand side
against the proton deficiencies on the righthand side
4.4.1 Proton Balance Equation
E1: Write the PBEs for Na2CO3, H2C2O4 and
(NH4)2CO3 solutions.
For Na
Na22CO
CO33 solution
solution
(1) Zero level: ①CO32-; ②H2O.
(2) Possible changes:
CO32- + H2O = HCO3- +OH HCO3- + H2O = H2CO3 + OHH2O + H2O = H3O+ + OHHCO3-] + 2[
H 2CO3] = [[OH
OH -]
(3)PBE: [H+] + [[HCO
2[H
or [H+ ] = [[OH
OH -] - [[HCO
HCO3-] - 2[
H2CO3]
2[H
For example, in Na2CO3 solution, the zero level would be
①CO32-;
②H2O.
In NH4Cl solution, the zero level would be:
①NH4+;
②H 2O.
4.4.1 Proton Balance Equation
For
For H
H22C
C22O
O44 solution
solution
(1)Zero level: ①H2C2O 4; ②H 2O.
(2) Possible equilibrium
H2C2O4 = H+ + HC2O4HC2O4- = H+ + C2O42H2O + H2O = H 3O + + OH(3) PBE: [H +] = [OH-] + [HC2O 4-] + 2[C2O42-] .
1
4.4.1 Proton Balance Equation
4.4.2 Calculation the acidity of monoprotic acid (base)
For
For (NH
(NH44))22CO
CO33 solution
solution
(1) Zero level: ①NH4+; ②CO32-; ③H2O.
(2) equilibrium: NH4+ H+ + NH3
CO32- + H2O = HCO3- + OHHCO3- + H2O = H2CO3 + OH H2O + H2O = H 3O + + OHHCO3-] + 2[
H2CO3] = [[OH
OH-] + [[NH
NH3]
(3) PBE: [H +] + [[HCO
2[H
or [H +] = [[OH
OH-] + [[NH
NH3] - [HCO3-] - 2[
H2CO3]
2[H
Note
Note:: PBE can also be obtained from MBE and CBE
4.4.2 Calculation the acidity of monoprotic acid (base)
�
�
①If the acid is not too weak and dilute,i.e
i.e,, cK a≥10K w):
�
+
√
[H ] = K a[HA]
approximate
�
②If the acid is concentrate, but not strong, i.e., c/ K a≥ 105
�
�
[H+ ] = √ K a c , or pH = ((p
pK a + pc )/2
simplest
③ If the acid is concentrate, but not strong,
�
�
�
,, i.e., c/ K ≥105, cK
a
a < 10K w)
�
�
[H+] = √ K a c + K
w
1. For monoprotic weak acid
PBE of HA solution:
[H+] = [OH-] +[A-]
i.e., protons come from two sources
�
K a[HA]
[A-] =
[H+]
�
[OH-] = K w /[H +].
�
�
∴[H+] = K w /[H+] + K a[HA]/
[H+ ]
[HA]/[H
�
Precise formula
�
[H+ ] = √ K a[HA] + K
w
where: [HA] = δ1c , or [HA] = c - [H+].
4.4.2 Calculation the acidity of monoprotic acid (base)
E2: calculate the pH of 0.10 mol
mol··L-1 chloroacetic acid
�
solution. (K a= 1.4×10-3)
Solution:
�
�
∵cK a= 0.10× 1.4×10-3 > 10K
�
But ,
c /K a< 105
w
�
∴[H +] = √ K a[HA]= √ 1.4×10-3(0.10 - [H+ ])
= 1.1×10-2 mol
mol··L-1.
pH = 1.96.
approximate
approximate
4.4.2 Calculation the acidity of monoprotic acid (base)
4.4.2 Calculation the acidity of monoprotic acid (base)
mol
mol··L-1NH4Cl solution.
E4: Calculate the pH of 1.0×10-4 mol
mol··L-1HCN. K a=
�
6.2×10-10(pK a = 9.21)
E3: calculate the pH of 0.10
�
�
K b= 1.79×10-5 (pK b = 4.75).
�
�
�
∵pK w = pK a + pK b
�
∴ pK a = 14.00 - 4.75 = 9.25.
�
∵c/K a = 0.10/
10-9.25 > 105�
0.10/10
�
cK a = 0.10×10-9.25 > 10K w
�
∴pH = ((p
pK a + pc)/2 = (9.25 + 1.00)/2 = 5.12.
�
�
∵c/K a = 1.0×10-4/6.2×10-10 > 105;
�
�
cK a = 1.0×10-4×6.2×10-10 < 10K w.
�
�
∴[H+] = √ K a c + K w
= √ 1.0×10-4 × 6.2×10-10 + 1.0×10-14
= 2.7×10-7 mol
mol··L-1 ∴pH = 6.57
Using the simplest formula: [H+] = 2.5×10-7, pH = 6.60
Er= (2.5×10-7 - 2.7×10-7)/ 2.7 ×10-7 =7%
2
4.4.2 Calculation the acidity of monoprotic acid (base)
E5: Calculate the pH of 0.10 mol
mol··L-1NH3 aqueous
�
solution. pK b = 4.75.
4.4.3 Calculation the acidity of amphiprotic substance
[H+]For NaHA
NaHA,, PBE:
[H+ ] = [OH-] + [A2-] – [H 2A]
�
[OH-] = K w [H +]
�
K a2[HA-]
[A2-] =
�
∵c/K b = 0.10/10-4.75 > 105;
�
�
cK b = 0.10×
10-4.75 > 10K w
�
∴pOH = ((p
pK b + pc)/2 = (4.75 + 1.00)/2 = 2.88
�
pH = pK
w
[H2A] =
- pOH = 14.00 - 2.88 = 11.12
+
[H ] =
4.4.3 Calculation the acidity of amphiprotic substance
K a1θ (K θa2c + K Wθ )
K a1θ + c
�
�
�
If cK a2≥10K w, and c≥10K a1, then:
[H+ ] =
[ H + ] = K aθ1 K aθ2
Or
�
�
pH = (pK a1+ pK
a2)/2
4.4.3 Calculation the acidity of amphiprotic substance
E7: calculate the pH of 0.010 mol
mol··L-1 Na
2HPO 4
�
�
�
solution. pK a1 = 2.12
2.12,, pK a2 = 7.20
7.20,, pK a3 = 12.36
12.36..
[H+][HA-]
�
K
a1
K a1θ (K θa2[HA − ] + K Wθ )
K a1θ + [HA − ]
4.4.3 Calculation the acidity of amphiprotic substance
E6: calculate the pH of 0.10 mol
mol··L-1NaHCO3 aqueous
�
�
solution. pK a1 = 6.37, pK a2 = 10.32.
�
�
∵ cK a2 = 0.10
×10-10.32≥10K w,
0.10×
�
c≥10K a1. �
�
∴pH = (pK a1 + pK a2)/2
= (6.37 + 10.32)/2
= 8.34
4.4.4 The acidity of other acid-base systems
Na2HPO 4 is the second step dissociation product of H3PO 4
1. Strong but very dilute acid or base solution
If its concentration is close to 10-7 mol
mol··L-1, the
dissociation of water can not be neglected
K aθ2 (K aθ3 c + K Wθ )
K aθ2 + c
�
�
By using the simplest formula pH = (pK a2 + pK a3)/2
= (7.20 + 12.36)/2
= 9.78
Approximate formula, pH = 9.52
.
c + c 2 + 4K Wθ
2
2. Polyprotic acid (base)
Only the first step is to be considered, therefore the
calculation will be same as the monoprotic acid or
base.
[H + ]=
[H +] =
3
4.4.4 The acidity of other acid-base systems
E8: calculate the pH of saturated CO2 aqueous
�
�
solution. pK a1= 6.38; pK a2 = 10.25).
The solution
can be regarded as 0.04 mol L-1 H2CO3
�
�
As K a1 » K a2 , so treat it as monoprotic acid
�
pH = (pK a1+ pc)/2
= {6.38 - (lg0.04)}/2
= 3.89
4