Predicting Precipitate Formation 
The K_{sp} value is the product of the upper limit
of the
product of the concentration of the soluble ions. When the product of
the
concentration of the ions exceed the value of K_{sp}
they
cannot exist in equilibrium anymore. They will form a precipitate in
order
to reduce the concentrations of the ions in solution back to the
equilibrium
value.


To determine whether the K_{sp} is exceeded, just
substitute
the ion concentrations into an expression similar to the K_{sp}
expression and determine an experimental Trial Ion Product.
(T.I.P.)

If the T.I.P. exceeds the K_{sp} then a
precipitate forms.
If the T.I.P. is larger than the values of K_{sp} then the ions
are above the saturation line and the ions are in a supersaturated
situation.
The ions will precipitate until their individual concentrations
multiplied
together equal the value of K_{sp}.

If the T.I.P. does not exceed the value of K_{sp}
then the
ions are in an unsaturated situation, which means there aren't enough
ions
dissolved to form a precipitate.

If the T.I.P. equals the value of K_{sp} then the
solution
is saturated. The solution can hold no more dissolved ions without a
precipitate
forming.

Problem 
Does a ppt. of AgCl form when 1 mL of 0.1 mol/L AgNO_{3}
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
mL of
0.1 mol/L

Set up a ratio that states, if there are 0.1 moles of Ag^{+
}in
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
concentration)

The [Ag^{+}] will be this many moles of Ag^{+}
dissolved
in the volume of solution in this question. Therefore
[Ag^{+}] = 0.0001
mole
= 0.0001 mole = 1.0 x 10^{4 }mol/L
T.I.P. = [Ag^{+}][Cl^{}]
The K_{sp} for AgCl is 1.8 x 10^{10}
therefore T.I.P
> K_{sp} so a ppt. will form.

Followup Problem 
Does a ppt of AgI form when 10 mL of 0.1 mol/L AgNO_{3}
gets added to 90 mL of a solution containing 1.0 x 10^{10}
mol/L
of KI?
What is the highest concentration of I^{} ions
that can
exist in equilibrium with 0.01 mol/L Ag^{+} ions? How many
grams
of KI are present in 1 litre of a solution with the I^{} ion
concentration
found above?
K_{sp} = [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
}mol/L

If [I^{}] exceeds 8.3 x 10^{15} mol/L
then a ppt.
will form.
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
that
can exist in equilibrium with 0.001 mol/L Tl^{+} ions at 25^{o}C?
2. Calculate the [CrO_{4}^{2}] required to begin precipitation of these metal ions from solutions containing 1 g of the metal ions per L of solution. a) Ag^{+} b) Ca^{2+} c) Pb^{2+}
