Solubility In the Presence of a Common Ion
375
If we lets = the molar solubility of PbSO 4 , we see that [Pb
2+ ] = s, and [SO|~] = 5
because, for every mole of PbSO 4 that dissolves, there is produced one mole each
of Pb
2+ and SO|~. Substituting these values in the K<, v expression, we obtain
K sv = [Pb
2+ ][SO 4
2 -] = (s)(.s) = s- = 1.66 x 108
5 = 1.29 x 10-" moles/liter of PbSO 4
gofPbStyiOOml = (\.29 x 1(T
4 j~\ (0.100 liter) ^303.3
= 3.91 x 10~
3 g
PROBLEM:
What is the solubility of Ag 2 CrO 4 in grams/100 ml of water?
SOLUTION:
The chemical reaction is
Ag 2 Cr0 4 <=* 2Ag
+
If we let s = the molar solubility of Ag 2 CrO 4 , we see that [Ag
+ ] = 2s, and
[CrO 4 ~] = s because, for every mole of Ag 2 CrO 4 that dissolves, there are produced 2 moles of Ag
+ and 1 mole of CrO|~. Substituting these values in the K sv
expression, we have
KSP = [Ag
+ ]
2 [CrO 4
2 -J = (2sY(s) = 4s
3 = 1.12 x 10~
12
/ 1.12 x 10-'V
, CA
,. . moles , . „ _
s =
= 6.54 x 10-= —
of Ag 2 CrO 4
\
4
/
liter
gof AgzCrCVlOO ml = ^6.54 x 10^
!
^
i
) (0.100 liter) (331.8
= 2.17 x ID"
3 g
SOLUBILITY IN THE PRESENCE OF A COMMON ION
If there is introduced into the solution from some other source an ion that is in
common with an ion of the insoluble solid, the chemical equilibrium is shifted to
the left, and the solubility of that solid will be greatly decreased from what it is
in pure water. This is called the "common-ion effect." This effect is important
in gravimetric analysis, where one wishes to precipitate essentially all of the ion
being analyzed for, by adding an excess of the "common-ion" precipitating
reagent. There is a practical limit to the excess, however, which involves such
factors as purity of precipitate and possibility of complex formation. You can
calculate the solubility under a variety of conditions, as illustrated in the following problem.
375
If we lets = the molar solubility of PbSO 4 , we see that [Pb
2+ ] = s, and [SO|~] = 5
because, for every mole of PbSO 4 that dissolves, there is produced one mole each
of Pb
2+ and SO|~. Substituting these values in the K<, v expression, we obtain
K sv = [Pb
2+ ][SO 4
2 -] = (s)(.s) = s- = 1.66 x 108
5 = 1.29 x 10-" moles/liter of PbSO 4
gofPbStyiOOml = (\.29 x 1(T
4 j~\ (0.100 liter) ^303.3
= 3.91 x 10~
3 g
PROBLEM:
What is the solubility of Ag 2 CrO 4 in grams/100 ml of water?
SOLUTION:
The chemical reaction is
Ag 2 Cr0 4 <=* 2Ag
+
If we let s = the molar solubility of Ag 2 CrO 4 , we see that [Ag
+ ] = 2s, and
[CrO 4 ~] = s because, for every mole of Ag 2 CrO 4 that dissolves, there are produced 2 moles of Ag
+ and 1 mole of CrO|~. Substituting these values in the K sv
expression, we have
KSP = [Ag
+ ]
2 [CrO 4
2 -J = (2sY(s) = 4s
3 = 1.12 x 10~
12
/ 1.12 x 10-'V
, CA
,. . moles , . „ _
s =
= 6.54 x 10-= —
of Ag 2 CrO 4
\
4
/
liter
gof AgzCrCVlOO ml = ^6.54 x 10^
!
^
i
) (0.100 liter) (331.8
= 2.17 x ID"
3 g
SOLUBILITY IN THE PRESENCE OF A COMMON ION
If there is introduced into the solution from some other source an ion that is in
common with an ion of the insoluble solid, the chemical equilibrium is shifted to
the left, and the solubility of that solid will be greatly decreased from what it is
in pure water. This is called the "common-ion effect." This effect is important
in gravimetric analysis, where one wishes to precipitate essentially all of the ion
being analyzed for, by adding an excess of the "common-ion" precipitating
reagent. There is a practical limit to the excess, however, which involves such
factors as purity of precipitate and possibility of complex formation. You can
calculate the solubility under a variety of conditions, as illustrated in the following problem.