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Solubility Product and Precipitation
PROBLEM:
What is the solubility of Ag 2 CrO 4 in 0.100 M AgNO 3 ?
SOLUTION:
Just as in the preceding problem, the chemical equilibrium is
Ag 2 CrO 4 <=s 2Ag
+ + CrOfAgain, we let s = the molar solubility of Ag 2 CrO 4 , but in this case the Ag
+ concentration is determined primarily by the 0. 100 mole/liter of AgNO 3 , to which there is
added 2s moles/liter of Ag
+ from the Ag 2 CrO 4 to give a total [Ag
+ J = (0. 100 + 2s)
moles/liter. The only source of CrO|~ is the Ag 2 CrO 4 , so [CrO|~] = . y moles/liter.
Substituting these values into the K sl , expression, we have
K sv = [Ag+HCrOM = (0.100 + 2s)'
2 (s) = 1.12 x 1Q-'
2
In this problem, s will be even smaller than in the preceding problem, because the
solubility is decreased by the common ion; as a consequence, 2s can be neglected
compared to 0. 100. This gives
(0. 100)
2 (.s) = 1.12 x 10-'
2
s = 1.12 x 10-'°
- of Ag 2 Cr0 4
You can see that, compared to water solubility, the solubility has been decreased
by a factor of more than 100,000.
CRITERION FOR PRECIPITATION
A K^p value represents the product of concentrations that exist in a saturated
solution. If you try to put into solution a combination of ions whose concentrations exceed the solubility product, you will get a precipitate because you have
exceeded the quantities needed to saturate the solution. The ion product (Q) is
an expression that looks like the A" sp expression, but it uses the actual concentrations you propose to have in solution, not the equilibrium concentrations. If the
value of Q is larger than K st> , you will get a precipitate; if it is smaller than K sv ,
no precipitate will form; and if Q = K sv , the solution will be just saturated. The
prediction of precipitation is illustrated in the following problems.
PROBLEM:
If 1.00 mg of Na 2 CrO 4 is added to 200 ml of 2.00 x lO"
3 M AgNO 3 , will a precipitate form?
SOLUTION:
We must find out whether the proposed concentrations of [Ag
+ ] and [CrO|~J will
give a value of Q larger than A" S|J . The proposed concentrations are
Solubility Product and Precipitation
PROBLEM:
What is the solubility of Ag 2 CrO 4 in 0.100 M AgNO 3 ?
SOLUTION:
Just as in the preceding problem, the chemical equilibrium is
Ag 2 CrO 4 <=s 2Ag
+ + CrOfAgain, we let s = the molar solubility of Ag 2 CrO 4 , but in this case the Ag
+ concentration is determined primarily by the 0. 100 mole/liter of AgNO 3 , to which there is
added 2s moles/liter of Ag
+ from the Ag 2 CrO 4 to give a total [Ag
+ J = (0. 100 + 2s)
moles/liter. The only source of CrO|~ is the Ag 2 CrO 4 , so [CrO|~] = . y moles/liter.
Substituting these values into the K sl , expression, we have
K sv = [Ag+HCrOM = (0.100 + 2s)'
2 (s) = 1.12 x 1Q-'
2
In this problem, s will be even smaller than in the preceding problem, because the
solubility is decreased by the common ion; as a consequence, 2s can be neglected
compared to 0. 100. This gives
(0. 100)
2 (.s) = 1.12 x 10-'
2
s = 1.12 x 10-'°
- of Ag 2 Cr0 4
You can see that, compared to water solubility, the solubility has been decreased
by a factor of more than 100,000.
CRITERION FOR PRECIPITATION
A K^p value represents the product of concentrations that exist in a saturated
solution. If you try to put into solution a combination of ions whose concentrations exceed the solubility product, you will get a precipitate because you have
exceeded the quantities needed to saturate the solution. The ion product (Q) is
an expression that looks like the A" sp expression, but it uses the actual concentrations you propose to have in solution, not the equilibrium concentrations. If the
value of Q is larger than K st> , you will get a precipitate; if it is smaller than K sv ,
no precipitate will form; and if Q = K sv , the solution will be just saturated. The
prediction of precipitation is illustrated in the following problems.
PROBLEM:
If 1.00 mg of Na 2 CrO 4 is added to 200 ml of 2.00 x lO"
3 M AgNO 3 , will a precipitate form?
SOLUTION:
We must find out whether the proposed concentrations of [Ag
+ ] and [CrO|~J will
give a value of Q larger than A" S|J . The proposed concentrations are