Common Ion Effect

Homework:
p492 (9)
P492 (10,11,12)

Determine the molar solubility of Fe(OH)3 in a 0.300M Fe2(SO4)3 solution.

      Fe(OH)3(s)   <=>     Fe3+ (aq)      +       3 OH-(aq)

I       n/a                           0.600M*                    0
C     n/a                                 +x                       +3x
E     n/a                            0.600 + x                  +3x
                                  ~0.600 (small K)

initial Fe3+ = 0.600M - from the 0.300M Fe2(SO4)3,
where there are twice as many Fe as Fe2(SO4)3 units

Ksp = [Fe3+][OH-]3
2.6x10-39 = (0.600)(3x)3
2.6x10-39 = 0.600(27x3)
5.4x10-14 = x 
molar solubility of Fe(OH)3 in the 0.300M Fe2(SO4)3 solution
 is 5.4x10-14 mol/L

Molar Solubility to Ksp

Molar solubility is measured at llbm.

question: Determine the  Ksp of salt AB2 if the molar solubility is
 0.00015 mol/L

1.Write the llbm dissociation rxn.
2. Write the Ksp expression.
3. Set up the ICE chart.
4. Use molar solubility to determine [ion]llbm
5.Sub into Ksp expression.

    AB2 (s)     <=>    A+(aq)   +   2B-(aq)                        Ksp=[A+][B-]2
I    n/a                      0              0
C  n/a
E   n/a                0.00015*        2(0.00015)
* for every mole of A+ that is present at llbm, one mole of AB2 dissolved

Ksp = (0.00015)(0.00030)2
Ksp = 1.4x10-11 for AB2

Sometimes the molar solubility info can be 'hidden':
for example, a saturated solution contains 0.24g of salt per 500mL.
[Use molar mass, convert the given info to mol/L]

Homework:
p486(4)
p493(7,8,9,12)
 ...not all are molar solblty to Ksp, some are Ksp to molar solblty
What would you get if image 1 and  image 2 were combined.

[Ans: a molar solution.]

Ksp to Molar Solubility

Slightly soluble salts (see data table page 802)

Molar solubility of the slightly soluble salt identifies the degree
 to which the salt actually dissolves in water.
Expressed in mol/L
The concentration of the ions at llbm is related to the molar solubility of
the salt  by the mole relationship in the balanced llbm equation.

example Determine the molar solubility of iron (II) hydroxide.

1.Write the llbm dissociation reaction.
2. Write the Ksp expression.
3.Complete the ICE chart. (not considering solids - llbm reaction, solid concentration will not change)
4.Determine the value of "x"
5.State the molar solubility value.

            Fe(OH)2(s)        <=>         Fe2+ (aq)        +      2OH-(aq)

I            n/a                                    0                           0
C          n/a                                    +x                         +2x
E          n/a                                     x                           2x

     Ksp = [Fe3+][OH-]2
4.9x10-17 = (x)(2x)2
4.9x10-17 = 4x3
2.3x10-6 = x

The molar solubility of Fe(OH)2 is 2.3x10 -6 mol/L.
[According to the balanced llbm equation, for every mole of Fe2+ that shows up at llbm, one mole of Fe(OH)2 had to dissolve.]

Homework:
p486: 1,2,3
p493: 4,5,6

KSCN lab Follow-Up

The nice orange-red colours are now a thing of the past.

A few helpful hints:
Carefully determine the initial concentrations of SCN- and Fe3+

The initial [Fe3+] for test tubes 2 to 5 requires the use of c1V1=c2V2 twice.

When the determination of Kc is done, the results may vary quite a bit.
Generally they are in the 100 to 300 range.

Complete the follow-up to this lab in a timely fashion. Sooner = better.

If you were absent for the lab - see me for some "results" that could be used to work through the various calculations.

Next up:
Solubility (Have you completed the introduction to solubility?)

Equilibrium: The Movie?

If the teeter-totter represents an llbm reaction, what would be its K value?

Small K approximation

The basis of this lesson is the significance of the size of K as it relates to equilibrium position.

K = [product] / [reactant]

A large K value indicates that there is much more product than reactant.

A small K value indicates that there is much less product than reactant.

Since there is very little product actually produced, the change to the amount of reactant is minimal. So minimal that there is approximately no change to the amount of reactant at llbm compared to the amount of reactant initially. This approximation will simplify the calculation.

There is still product made. The amount of product is still represented by some multiple of 'X'.

Initially 0.200mol of A is placed into a 1.00L container, sealed and allowed to reach llbm.
Determine [B]llbm
Consider the generic equation:
    2A      -->        3B      +        2C      K = 3.50x10-5
I  0.200M             0                   0
C -2X                   +3X             +2X
E  0.200-2X*         3X                2X

but since K is small, 'X' is also very small
there 0.200-2X ~ 0.200M

K   =   [B]3[C]2 / [A]2

3.50x10-5 =  [3X]3[2X]2 / [0.200M]2
1.4x10-6 = 108X5
2.65x10-2 =X
0.079M = [B]llbm

Homework: - a few adjustments to the text info
p476 (7) - H2S <=> H2 + S  [ans=9.17X10-4 M]
p476 (8)  2HCl <=> H2 + Cl2 [H2/Cl2 llbm = 3.6x10-17M]
p481 (6)
p640 (15) [ans: O2 llbm = 1.9x10-3M]