To see how a ladder diagram is constructed, we will use the acid–base equilibrium between HF and F–
HF(aq) + H2O(l)
t
H3O+(aq) + F–(aq)for which the acid dissociation constant is
Taking the log of both sides and multiplying through by –1 gives – log – log [ ] – log[ ]
[ ]
,
–
Ka HF H O F
= 3 + HF Ka HF H O F
, HF
[ ][ –] [ ]
= 3 + Q =
= ×
+ +
0 5
0 5 4
3
3 2
32 2
. ( . )
[ ( ) ]
[ ][ ]
Ag NH
Ag NH
eq eq
β Q = ( . )[+ ( ) ]+
( . )[ ] ( . ) [ ] 0 5
0 5 0 5
3 2
2 32
Ag NH
Ag NH
eq
eq eq
Figure 6.4
Ladder diagram for HF, showing areas of predominance for HF and F–.
Finally, replacing the negative log terms with p-functions and rearranging leaves us with
6.31 Examining equation 6.31 tells us a great deal about the relationship between pH and the relative amounts of F–and HF at equilibrium. If the concentrations of F–and HF are equal, then equation 6.31 reduces to
pH = pKa,HF= –log(Ka,HF) = –log(6.8×10–4) = 3.17
For concentrations of F–greater than that of HF, the log term in equation 6.31 is positive and
pH > pKa,HF or pH > 3.17
This is a reasonable result since we expect the concentration of hydrofluoric acid’s conjugate base, F–, to increase as the pH increases. Similar reasoning shows that the concentration of HF exceeds that of F–when
pH < pKa,HF or pH < 3.17
Now we are ready to construct the ladder diagram for HF (Figure 6.4).
The ladder diagram consists of a vertical scale of pH values oriented so that smaller (more acidic) pH levels are at the bottom and larger (more basic) pH levels are at the top. A horizontal line is drawn at a pH equal to pKa,HF. This line, or step, separates the solution into regions where each of the two conjugate forms of HF predominate. By referring to the ladder diagram, we see that at a pH of 2.5 hydrofluoric acid will exist predominately as HF. If we add sufficient base to the solution such that the pH increases to 4.5, the predominate form be- comes F–.
Figure 6.5 shows a second ladder diagram containing information about HF/F–and NH4+/NH3. From this ladder diagram we see that if the pH is less than 3.17, the predominate species are HF and NH4+. For pH’s between 3.17 and 9.24 the predominate species are F– and NH4+, whereas above a pH of 9.24 the predominate species are F–and NH3.
Ladder diagrams are particularly useful for evaluating the reactivity of acids and bases. An acid and a base cannot coexist if their respective areas of predominance do not overlap. If we mix together solutions of NH3and HF, the reaction
HF(aq) + NH3(aq)
t
NH4+(aq) + F–(aq) 6.32occurs because the predominance areas for HF and NH3do not overlap. Be- fore continuing, let us show that this conclusion is reasonable by calculating the equilibrium constant for reaction 6.32. To do so we need the following three reactions and their equilibrium constants.
HF H O H O F
NH H O OH NH
H O OH H O
a
b
w
( ) ( ) ( ) ( ) .
( ) ( ) ( ) ( ) .
( ) ( ) ( )
.
– –
– –
–
–
aq aq aq K
aq aq aq K
aq aq K
K
+ + = ×
+ + = ×
+ = =
×
+
+ +
2 3 4
3 2 4 5
3 2 14
6 8 10 1 75 10
2 1 1
1 00 10 l
l
l
t t
t
pH p F
a HF
= K +log[ ] [ ]
–
Chapter 6 Equilibrium Chemistry
151
pH = pKa,HF = 3.17
HF F–
pH
pH = pKa,HF = 3.17
HF NH3
NH4+ F–
pH
pH = pKa,NH3 = 9.24
Figure 6.5
Ladder diagram for HF and NH3.
152
Modern Analytical ChemistryAdding together these reactions gives us reaction 6.32, for which the equilibrium constant is
Since the equilibrium constant is significantly greater than 1, the reaction’s equilib- rium position lies far to the right. This conclusion is general and applies to all lad- der diagrams. The following example shows how we can use the ladder diagram in Figure 6.5 to evaluate the composition of any solution prepared by mixing together solutions of HF and NH3.
EXAMPLE
6.7
Predict the pH and composition of a solution prepared by adding 0.090 mol of HF to 0.040 mol of NH3.
SOLUTION
Since HF is present in excess and the reaction between HF and NH3is favorable, the NH3will react to form NH4+. At equilibrium, essentially no NH3
remains and
Moles NH4+= 0.040 mol Converting NH3to NH4+consumes 0.040 mol of HF; thus
Moles HF = 0.090 – 0.040 = 0.050 mol Moles F–= 0.040 mol
According to the ladder diagram for this system (see Figure 6.5), a pH of 3.17 results when there is an equal amount of HF and F–. Since we have more HF than F–, the pH will be slightly less than 3.17. Similar reasoning will show you that mixing together 0.090 mol of NH3and 0.040 mol of HF will result in a solution whose pH is slightly larger than 9.24.
If the areas of predominance for an acid and a base overlap each other, then practically no reaction occurs. For example, if we mix together solutions of NaF and NH4Cl, we expect that there will be no significant change in the moles of F– and NH4+. Furthermore, the pH of the mixture must be between 3.17 and 9.24. Because F–and NH4+can coexist over a range of pHs we cannot be more specific in estimat- ing the solution’s pH.
The ladder diagram for HF/F– also can be used to evaluate the effect of pH on other equilibria that include either HF or F–. For example, the solubility of CaF2
CaF2(s)
t
Ca2+(aq) + 2F–(aq)is affected by pH because F–is a weak base. Using Le Châtelier’s principle, if F–is converted to HF, the solubility of CaF2will increase. To minimize the solubility of CaF2we want to control the solution’s pH so that F–is the predominate species.
From the ladder diagram we see that maintaining a pH of more than 3.17 ensures that solubility losses are minimal.
K K K
= K = × ×
× = ×
a b w
( . )( . )
( . ) .
– –
–
6 8 10 1 75 10
1 00 10 1 19 10
4 5
14
6
Figure 6.6
Ladder diagram for metal–ligand complexes of Cd2+and NH3.
Chapter 6 Equilibrium Chemistry
153
Cd(NH3)32+
Cd(NH3)22+
Cd(NH3)2+
Cd2+
log K1 = 2.55
log K4 = 0.84 log K3 = 1.34 log K2 = 2.01
Cd(NH3)42+
p NH3