Ka
HB (aq) ⇄
H+ (aq)
+ B- (aq)
Ka = [ H+] [ B-]
[HB]
STRONG ACID: 100% dissociation
Ka ~ infinity since [large]
~0
WEAK ACID: small % dissociation
Ka usually << 1 since [small]
large
Eq:
Eq:
HCl H+ + Cl~0
large large
CH3COOH H+ + CH3COOlarge
small
small
Types of Ka Calculation problems
UNIDENTIFIED ACID:
Ka not known
Arithmetic problem:
start with [ H+]e or [OH-]e known
start with pH, pOH, % dissociation
IDENTIFIED ACID:
Ka known
Algebra Problem with weak acid:
Assume for now: monoprotic acids: HB
Simple Arithmetic problem with strong acid
% dissociation = [ H+] e x 100
[HB]i
oxy acids of chlorine and their
Ka
HClO
hypochlorous acid
5.6 x 10-8
HClO2
chlorous acid
1 x 10-2
HClO3
chloric acid
large
HClO4
perchloric acid
very large
-As the central chlorine atom becomes more positive, then the electrons shared between
the H-O bond are pulled closer toward the Cl atom.
Therefore the H-O bond becomes more polar.
SEE P. 409
The strength of oxyacids is determined by the number of O's attached to the oxyacid.
(more O's = stronger)
H2SO5
persulphuric acid
very very strong
H2SO4
sulphuric acid
very strong
H2SO3
sulphurous acid
1.7 x 10-2
H2SO2
hyposulphurous acid
very weak
The more positive the S becomes, the more easily the H is removed.
Sample Ka calcs:
UNIDENTIFIED ACID PROBLEMS
1. 0.100 M HB (aq) has a pH of 4.00.
Find Ka, % dissociation, and possible identity.
2. 0.0250 M HB (aq) has pOH = 8.25.
Find [ H+] , [OH-] , pH, Ka, identity, % dissocation.
3. 0.00450 M HX (aq) is 5.0 % dissociated.
Find pH, Ka, identity.
4. Find identity and % dissociation if 0.046 M HB (aq) has pH = 4.82.
5. 0.00800 M HB (aq) has pOH = 9.75.
Find pH, [ H+] , [OH-] , Ka, identity, % dissocation.
Sample Ka calcs:
IDENTIFIED ACID PROBLEMS
1. Find pH of 0.250 M H2O2 (aq). Ka = 2.4 x 10 -12
2. Find pH of 0.048 M HBr (aq).
3. Find pH, pOH, [ H+] , [OH-] and % dissocation for 0.250 M HCO3- (aq).
4. Find % dissocation for 0.96 M HNO2 (aq).
NOW YOU CAN COMPLETE WORKSHEET 4.3
Wkst 4.3: Ka Problems
+
1- A 0.0100 M solution of HB(aq) has [H ]e = 8.0 x 10-4 M. Determine the
pH, % dissociation, pOH, [OH ], Ka and the possible identity of the acid
from your Ka table.
+
-
2- Determine the pH, pOH, [H ], [OH ], and % dissociation of 0.50 M
H2O2(aq).
3- Find the pH, pOH of 0.0750 M HNO3(aq).
4- If a 0.0010 M solution of an unidentified acid HB(aq) has a pH of 5.00,
find its Ka, % dissociation and possible identity.
5- A 0.100 M solution of acid HX(aq) is 2.0% dissociated. Find the pH, pOH,
Ka and possible identity of the acid.
6- Find the pH and the % dissociation of a 0.044 M solution of HBrO(aq). Ka
= 2.5 x 10-9
7- Find the pH and pOH for 0.0065 M HI(aq).
8- An unknown acid HB(aq), in a 0.0045 M solution, has a pOH of 9.50.
Find the Ka,
% dissociation and possible identity of the acid.
9- A 0.20 M solution of HAsO2(aq) is 5.5 x 10-3% dissociated. Find its Ka.
Wkst 4.3: Ka Problems
-
1- pH = 3.10, % dissociation = 8.0 %, [OH ] = 1.3 x 10-11 M, pOH = 10.90,
Ka = 7.0 x 10-5 (similar to benzoic acid)
+
-
2- [H ] = 1.1 x 10-6 M, pH = 5.96, % dissociation = 2.2 x 10-4 %, [OH ] =
9.1 x 10-9 M,
pOH = 8.04
3- pH = 1.120, pOH = 12.880
4- % dissociation = 1.0 %, Ka = 1.0 x 10-7 (similar to hydrogen sulphite ion)
5- pH = 2.70, pOH = 11.30, Ka = 4.1 x 10-5 (similar to hydrogen oxalate)
6- pH = 5.00, % dissociation = 2.3 x 10-2 %
7- pH = 2.19, pOH = 11.81
8- % dissociation = 0.71 %, Ka = 2.3 x 10-7 (similar to hydrogen sulphite or
monohydrogen citrate ions)
9- Ka = 6.05 x 10-10
Kn
Kn =
Ka (forward reaction)
Ka (reverse reaction)
(this is a division or really, just a RATIO)
Example: HC2O4- + CN- ⇄ HCN + C2O42a) identify the acids:
HC2O4- and HCN
b) identify the acid strength by looking up the Ka values on page 6 of
the data booklet
HC2O4- : 6.4 x 10-5 and HCN : 4.9 x 10-10
- HC2O4- is the stronger acid (higher Ka value)
c) predict whether the forward or reverse reaction is favoured:
HC2O4- dissociating is the forward reaction and HCN dissociating is the reverse reaction
- since the Ka of the hydrogen oxalate ion is higher,
I predict that the forward reaction is favoured.
d) back up your answer by calculating Kn
(which is just a ratio of the Ka values of the two acids)
Kn = Ka (forward reaction)
Ka (reverse reaction)
= 6.4 x 10-5
4.9 x 10-10
= 1.3 x 105
OR
the acid on the reactants side
the acid on the products side
Since Kn >>1, my previous prediction is correct.
The forward reaction is favoured.
That is, the products are favoured.
NOW WORKSHEET 4.4 CAN BE COMPLETED
Wkst 4.4: Acid vs. Acid Competition
For each of the following neutralisation reactions, identify: the acids, their
relative strengths and whether reactants or products are favoured. After
predicting the winner, offer numerical proof by calculating Kn.
-
a)
HSO4 (aq)
b)
-
)
+
SO4-2(aq)

HNO2(aq)
+
CO3-2(aq)

HF(aq)
+
HO2 (aq)

NH3(aq)
+
H2SO4(aq)
H2PO4 (aq)

HPO4-2(aq)
+
HS (aq)
HBr(aq)

Br (aq)
+
CH3COOH(aq)
+
C6H5COO (aq)
HCO3 (aq)
+
NO2 (aq)
c)
H2O2(aq)
+
F (aq)
d)
HSO4 (aq)
+
e)
S-2(aq)
+
f)
CH3COO (aq)
+
-
-
-
C6H5COOH(aq
-
-
NH4
+
(aq)
-

-
-
-
Wkst 4.4: Acid vs. Acid Competition
For each of the following neutralisation reactions, identify: the acids, their
relative strengths and whether reactants or products are favoured. After
predicting the winner, offer numerical proof by calculating Kn.
a)
-
HSO4 (aq)
Acid
+
C6H5COO
(aq)

C6H5COOH(aq
)
+
SO4-2(aq)
+
CO3-2(aq)
+
HO2
+
H2SO4(aq)
Acid
+
HS (aq)
Acid
+
CH3COOH(aq)
Acid
Acid
Kn = 1.8 x 102; products are favoured.
b)
-
HCO3 (aq)
Acid
+
NO2
(aq)

HNO2(aq)
Acid
Kn = 1.2 x 10-7; reactants are favoured.
c)
H2O2(aq)
Acid
+
F
(aq)

HF(aq)
Acid
(aq)
Kn = 6.9 x 10-9; reactants are favoured.
d)
HSO4
(aq)
+
+
NH4 (aq)
Acid

NH3(aq)
Kn = 0; reactants are favoured.
e)
S-2(aq)
+
-
H2PO4 (aq)
Acid

HPO4-2(aq)
-
Kn = 4.8 x 105; products are favoured.
f)
CH3COO
(aq)
+
HBr(aq)
Acid

Br
(aq)
Kn = ∞ ; products are favoured.