Page 188 - Modern physical chemistry

180 Equilibria in Condensed Phases

whence

Example 8.6

Calculate m H + in 0.0100 m ammonium formate.
Anunonium formate consists of ammonium ions and formate ions. These undergo the
reactions

H+ + OH- ~ H 20,

on dissolving in water. Here A- represents the formate ion. Since the ammonium hydrox-
ide and the formic acid formed do not react further, we have

Furthermore, electrical neutrality is maintained; so

Because the excess hydrogen ions and hydroxide ions combine, a small difference in
the first two reactions causes a large difference in m H+ and m OH -:

m H + ,;,m OH -'
But if enough ammonium formate is added to water, m NH4 + is large with respect to m H+
and m A - is large with respect to m OH -' Then the electrical neutrality condition reduces to

Substituting this into the equation for the overall molality yields

mNH40H == mBA'
For the ionization of ammonium hydroxide, we write

mNH40H
and for the ionization of formic acid, we write

mA-mW = 1.772 x 10-4 m,
mBA
assuming each Qy = 1. Dividing the last equation by the previous one and canceling the
molalities that are nearly equal yields

mw = 1.772 x 10-4 = 1.013 x 10.
mow 1. 75 x 10- 5