|Predicting Precipitate Formation|
|The Ksp value is the product of the upper limit
product of the concentration of the soluble ions. When the product of
concentration of the ions exceed the value of Ksp
cannot exist in equilibrium anymore. They will form a precipitate in
to reduce the concentrations of the ions in solution back to the
|To determine whether the Ksp is exceeded, just
the ion concentrations into an expression similar to the Ksp
expression and determine an experimental Trial Ion Product.
|If the T.I.P. exceeds the Ksp then a
If the T.I.P. is larger than the values of Ksp then the ions
are above the saturation line and the ions are in a super-saturated
The ions will precipitate until their individual concentrations
together equal the value of Ksp.
|If the T.I.P. does not exceed the value of Ksp
ions are in an unsaturated situation, which means there aren't enough
dissolved to form a precipitate.
|If the T.I.P. equals the value of Ksp then the
is saturated. The solution can hold no more dissolved ions without a
|Does a ppt. of AgCl form when 1 mL of 0.1 mol/L AgNO3
is added to a beaker containing 1 L of tap water with a Cl-
ion concentration of 1.0 x 10-5 mol/L?
[Cl-] = 1.0 x 10-5 mol/L
[Ag+] calculation. Start with the given info, 1
|Set up a ratio that states, if there are 0.1 moles of Ag+
1000 mL how many moles will be in 1 mL?
0.1 mol = x
1000mL 1 mL
x = 0.0001 moles of Ag+ are present.
(This is how many moles there are, this is not the
|The [Ag+] will be this many moles of Ag+
in the volume of solution in this question. Therefore
[Ag+] = 0.0001
= 0.0001 mole = 1.0 x 10-4 mol/L
T.I.P. = [Ag+][Cl-]
The Ksp for AgCl is 1.8 x 10-10
> Ksp so a ppt. will form.
|Does a ppt of AgI form when 10 mL of 0.1 mol/L AgNO3
gets added to 90 mL of a solution containing 1.0 x 10-10
What is the highest concentration of I- ions
exist in equilibrium with 0.01 mol/L Ag+ ions? How many
of KI are present in 1 litre of a solution with the I- ion
Ksp = [Ag+][I-]
8.3 x 10-17 = (1.0 x 10-2)[I-]
There [I-] = 8.3 x 10-17 =
8.3 x 10-15
|If [I-] exceeds 8.3 x 10-15 mol/L
then a ppt.
K+ + I- ------> KI
If we have 8.3 x 10-15 moles of I- we also have 8.3 x 10-15 moles of K+ and hence we also have 8.3 x 10-15 moles of KI.
g = moles x molecular mass
|Follow up Problems|
|1. What is the highest concentration of Cl- ions
can exist in equilibrium with 0.001 mol/L Tl+ ions at 25oC?
2. Calculate the [CrO42-] required to begin precipitation of these metal ions from solutions containing 1 g of the metal ions per L of solution.
a) Ag+ b) Ca2+ c) Pb2+