Page 169 - Modern Analytical Chemistry

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152 Modern Analytical Chemistry

Adding together these reactions gives us reaction 6.32, for which the equilibrium
constant is

a
KK b (.68 ´10 –4 )( .175 ´ 10 –5 ) 6
K = = –14 = .119 ´ 10
K w (.100 ´10 )
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 NH 3 .

EXAMPLE 6.7

Predict the pH and composition of a solution prepared by adding 0.090 mol of
HF to 0.040 mol of NH 3.
SOLUTION

Since HF is present in excess and the reaction between HF and NH 3 is
+
favorable, the NH 3 will react to form NH 4 . At equilibrium, essentially no NH 3
remains and
+
Moles NH 4 = 0.040 mol
+
Converting NH 3 to NH 4 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 NH 3 and 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
–
NH 4 Cl, we expect that there will be no significant change in the moles of F and
+
NH 4 . Furthermore, the pH of the mixture must be between 3.17 and 9.24. Because
+
–
F and NH 4 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
CaF 2
–
2+
CaF 2 (s) t Ca (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 CaF 2 will increase. To minimize the solubility of
–
CaF 2 we 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.