Separation of Ions
383
whereas in this problem the [NH 3 ] is decreased as a result of precipitating
Mn(OH) 2 :
Mn
2+ + 2NH 3 + 2H 2 O -H> Mn(OH) 2 | + 2NH 4
+
Because we are precipitating 0.20 mole of Mn
2+ per liter, we will use up 0.40
mole of NH 3 per liter, leaving [NH 3 ] = 1.00 - 0.40 = 0.60 M. The [NH 4
+ ],
0.40 mole/liter of which is produced in the precipitation reaction, is our unknown.
Substituting these values as we did in the last problem, we obtain
fNH+]2 _ [Mg
2 +][NH 3 J
2 _ (0.20)(0.60)
2 _
[NHJ ~
0.0383
~
0.0383
~
L88
[NH 4
+ J = 1.37 M
Thus, 1.37 moles/liter is the final concentration of NHJ that must be in solution
if the Mg
2+ is to remain unprecipitated, but 0.40 mole/liter of this is produced
by the precipitation of Mn(OH) 2 , leaving only 0.97 mole/liter to be added in the
form of NH 4 C1.
Wtof NH 4 C1 needed = ( 0.97 ^!i) (0.500 liter) ( 53.5
\
liter /
V
mole/
= 25.9g
PROBLEM:
To what pH must a solution be buffered if you wish to permit the maximum
precipitation of Mn(OH) 2 with no precipitation of Mg(OH) 2 , if the solution is
0.200 M in both Mg
2+ and Mn
2+ ?
SOLUTION:
This problem presents the easy alternative for making the separation of Mg'
2+ and
Mn
2+ in the preceding two problems. The chemical equilibrium is based solely
on
Mg(OH) 2 *± Mg
2+ + 2OHThe maximal tolerable [OH~J that can exist without precipitating any of the
0.200 M Mg
2+ is
pOH = 5.12
pH = 14.00 - 5.12 = 8.88
If you use a buffer of pH = 8.88 that possesses high concentrations of a weak
base and its salt, the precipitation of Mn
2+ will not use up enough base to
significantly change the [base]/[salt] ratio (see p 356), the pH will remain
essentially unchanged, and no Mg
2+ will precipitate. The same buffer solution
could be used to dissolve the Mg(OH) 2 from the mixture of Mn(OH) 2 and
Mg(OH) 2 in the problem on p 381, because the Mg(OH) 2 will not use up enough