Incorporating the Common Ion Effect
Example
Calculate
the molar solubility of solid calcium fluoride at 25°C
in a solution containing 0.010 M calcium nitrate.
Answer
First, let's construct an ICE table for the equilibrium between solid calcium fluoride and its aqueous ions.
In a similar question, in the last lesson, we put 0 M in for the in [ ] for both ions because the calcium fluoride was dissolving in pure water. Today, however, the calcium fluoride is dissolving in a solution of 0.010 M calcium nitrate. This is where the "common ion" in the common ion effect comes into play. There is already Ca2+(aq) in the solution in which the solid will be dissolving, so the in [Ca2+(aq)] is 0.010M.
The remainder of filling in the ICE table proceeds the same as the last lesson.
In a similar question, in the last lesson, we put 0 M in for the in [ ] for both ions because the calcium fluoride was dissolving in pure water. Today, however, the calcium fluoride is dissolving in a solution of 0.010 M calcium nitrate. This is where the "common ion" in the common ion effect comes into play. There is already Ca2+(aq) in the solution in which the solid will be dissolving, so the in [Ca2+(aq)] is 0.010M.
The remainder of filling in the ICE table proceeds the same as the last lesson.
CaF2(s) ↔ Ca2+(aq) + 2F-(aq)
in
[ ] ---- 0.010 0
Δ [
] ---- +x +2x
eq
[ ] ---- 0.010 + x 2x
Create the equilibrium expression. Plug in the Ksp value (from the chart). Plug in the eq [ ] for each ion. Solve, bearing in mind that because the Ksp value is so tiny, the value of x must be tiny as well and thus, can be deleted from the (0.010 + x) term. This is useful because, instead of trying to determine the roots for a cubic equation, we now have an easily solved equation.
Ksp = [Ca2+][F-]2
3.9 x 10-11 = (0.010 + x)(2x)2
3.9 x 10-11 = (0.010)4x2
x2 = 9.8 x
10-10
x = 3.1 x
10-5
∴ 3.1 x
10-5 mol of calcium fluoride dissolves per liter of 0.010 M calcium
nitrate solution.
Criteria for Precipitation or
Dissolution
Example
Will
a precipitate form when 0.100 L of 3.0 x 10-3
M lead (II) nitrate is added to 0.400 L of 5.0 x
10-3 M sodium sulfate?
Answer
First, let's make like a grade 11 and predict some products and their solubilities (using the solubility rules):
Pb(NO3)2 + Na2SO4 →
2NaNO3(aq) + PbSO4(s)
Since PbSO4 is most likely to precipitate (although it will depend on the exact conditions), we will exclusively focus on it as we proceed. Set up the equilibrium relationship between the solid and its aqueous ions and look up the Ksp from the chart:
PbSO4(s) ↔
Pb2+(aq) + SO42-(aq) Ksp = 1.6 x 10-8
We will use c1V1
= c2V2 (which I have rearranged as c2
= c1V1/V2) to determine the in [ ] for both aqueous ions. Notice that V2 = 0.500L because the total volume after combining the two solutions would be 0.100 L + 0.400 L:
[Pb2+] = (3.0 x10-3 M)(0.100 L)/0.500 L
[Pb2+] = (3.0 x10-3 M)(0.100 L)/0.500 L
= 6.0 x 10-4
M
[
SO42-] = (5.0 x 10-3 M)(0.400 L)/0.500 L
= 4.0 x
10-3 M
Next, we will find the ratio of [P]:[R], or [dissolved]:[undissolved], using the reaction quotient:
Next, we will find the ratio of [P]:[R], or [dissolved]:[undissolved], using the reaction quotient:
Q
= [Pb2+][ SO42-]
= (6.0 x 10-4)(
4.0 x 10-3)
= 2.4 x
10-6
Since
Ksp = 1.6 x 10-8, the [P]/[R] must change from 2.4 x
10-6 to1.6 x 10-8 (so the value must get smaller), for the
system to proceed from its initial state to the one required at equilibrium:
Initial
[P]/[R] = 2.4 x
10-6/1 ↠↠↠↠ Equilibrium [P]/[R] = 1.6 x 10-8/1
Thus,
within this ratio, [P] must decrease. Therefore,
the reaction will proceed to the left (←) to produce more reactants to reach
equilibrium. Notice that the reactant side is where we find the solid precipitate.
∴ Q > Ksp, the equilibrium
will shift to the left, so precipitation will occur
HINT: I really like this kind of question.
Homework #38-41
I take you through the answer to HW #38 in this video and HW #39 in this video.
The Equilibrium Review can be found here.
