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When mixing chemicals, it is often useful to know if a precipitate will form.
This is done by computing an ion product P, which is analogous to the reaction
quotient Q. P has the same form as Ksp
and determines if a given solution will precipitate or not
- If P > Ksp, the ion concentration is higher than the
solution can absorb and the ions will form a precipitate.
- If P < Ksp, the ion concentration is lower than the solution
can absorb and the ions not will form a precipitate.
- If P = Ksp, the system is exactly saturated with ions and any
extra amount of ion added will cause precipitation.
One thing to be
careful of when computing the ion product is that you are most likely mixing two
solutions. This will change the total volume of the mixture and thus all the
concentrations will change.
Example: You mix 100 ml of 0.010 M Pb(NO3)2 and
100 ml of 0.010 M KF. Will a precipitate of PbF2 form? Ksp
for PbF2 is 7.1*10-7
Example: We need to compute P and compare it to Ksp. We
must be careful though- the solution volume changes when we mix the solutions,
so we must adjust the concentrations of both ions. First, write out the
ionization expressions
- Pb(NO3)2 - > Pb+2(aq) +
2NO3-(aq)
- KF -> K+(aq) + F-(aq)
We start with 100
ml of 0.010 M lead nitrate. This means we have 0.0010 moles of lead nitrate,
which forms 0.0010 moles of Pb+2. We also have 100 ml of 0.010 M KF,
0.0010 moles of KF, which generates 0.0010 moles of F-. Our total
solution volume is 100ml+100ml =200 ml, so the concentrations of the species are
- 0.0010 mol Pb+2/ 0.200 L = 0.0050 M Pb+2
- 0.0010 mol F-/ 0.200 L = 0.0050 M F-
Now we
can compute P. It has the same form as Ksp
- PbF2(s) < = > Pb+2(aq) + F-(aq)
- P = [Pb+2][F-]2
Simply plug in the needed values and compare to Ksp
- P = [Pb+2][F-]2 =
(0.0050)(0.0050)2
- P = 1.3*10-7
- P < Ksp, so the mixture does not precipitate
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