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CHAPTER 17 - LECTURE 3:

17.5 SOLUBILITY EQUILIBRIA

Solubility Product

Solubility Rules:
1- All nitrates (NO₃^-) are soluble; all acetates (C₂H₃O₂^2-)
        are soluble.
2- All Cl^- Br^-, and I^- are soluble, except for Ag⁺, Hg₂²⁺, Pb²⁺.
3- All sulfates (SO₄^2-) are soluble, except for Ca²⁺, Sr²⁺,
        Ba²⁺, Hg₂²⁺, Pb²⁺.
4- All Group I and Group II and NH₄⁺ salts are soluble.
5- All S^2- are insoluble, except for Group I, NH₄⁺,Ca²⁺,
        Sr²⁺, Ba²⁺
6- All CO₃^2- are insoluble, except for Group I, NH₄⁺
7- All PO₄^3- are insoluble, except for Group I, NH₄⁺
8- All OH^- are insoluble, except for Group I, NH₄⁺,Ca²⁺,
        Sr²⁺, Ba²⁺
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

    CaC₂O₄(s) + H₂O -----> Ca²⁺(aq)+ C₂O₄^2-(aq)
                                 <-----
         K_sp = [Ca²⁺] [C₂O₄^2-]

This is the equilibrium constant for a slightly soluble
(or nearly insoluble) ionic compound

PbI₂(s) + H₂O -----> Pb²⁺(aq)+ 2I^-(aq); K_sp = [Pb²⁺] [I^-]²
<-----

AgCl(s) -----> Ag⁺¹(aq) + Cl^-1(aq); K_sp = [Ag⁺¹] [Cl^-1]
<-----

Hg₂Cl₂(s) -----> Hg₂²⁺(aq)+ 2Cl^-1(aq);K_sp = [Hg₂²⁺] [Cl^-1]²
<-----

Pb₃(AsO₄)₂(s) -----> 3Pb²⁺(aq)+ 2AsO₄^-3(aq)
<-----

K_sp = [Pb²⁺]³ [AsO₄^-3]²
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Molar Solubility and Solubility

1.000 L of a solution saturated with CaC₂O₄ is evaporated
to dryness, giving 0.0061g residue of CaC₂O₄. Calculate the
solubility product constant K_spat 25^oC.
CaC₂O₄(s) -----> Ca²⁺+ C₂O₄^2-
_<------

Let S = [Ca²⁺] = [C₂O₄^2-]
         K_sp= [Ca²⁺] [C₂O₄^2-]
         S = 0.0061g CaC₂O₄/(128g/mol) = 4.8 x 10^-5mol/L
        K_sp= (4.8 x 10^-5mol/L)(4.8 x 10^-5mol/L) = 2.3 x 10^-9
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

K_spfor BaCrO₄ is 1.2 x 10^-10.Find the molar solubility, S.
BaCrO₄(s) -----> Ba²⁺+ CrO₄^2-
_<-----

         Let S = [Ba²⁺] = [CrO₄^2-]
         K_sp = [Ba²⁺] [CrO₄^2-]
         1.2 x 10^-10= (S)(S) = S²
S = (1.2 x 10^-10)^1/2= 1.1 x 10^-5M
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Find K_spfor Pb₃(AsO₄)₂. The molar solubility, S, is
3.33 x 10^-8 M.

Pb₃(AsO₄)₂(s) -----> 3Pb²⁺+ 2AsO₄^3-
_<-----

K_sp = [Pb²⁺]³ [AsO₄^3-]²
2S = [AsO₄^3-]; 3S = [Pb²⁺]

K_sp= (3 x 3.33 x 10^-8M)³(2 x 3.33 x 10^-8M)²
K_sp = (1.00 x 10^-21)(4.45 x 10^-15) = 4.45 x 10^-36
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

The K_sp for Cu(OH)₂is 2.6 x 10^-19. Calculate the molar
solubility S.

           Cu(OH)₂----> Cu²⁺(aq) + 2OH^-(aq)
                          <----
        Let S = [Cu²⁺];    therefore, [OH^-] = 2S

K_sp = [Cu²⁺][OH^-]²= (S)(2S)² = 4 S³

2.6 x 10^-19= 4 S³

(2.6 x 10^-19/4)^1/3= S = 4.02 x 10^-7 M
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

The K_sp for Mg₃(AsO₄)₂is 2.0 x 10^-20. Calculate the molar
solubility S.

Mg₃(AsO₄)₂----> 3Mg²⁺(aq) + 2AsO₄^-3(aq)
<----

Let 3S = [Mg²⁺]; therefore, [AsO₄^-3] = 2S

K_sp = [Mg²⁺]³[AsO₄^-3]²= (3S)³(2S)² = (27S³)(4S²)
K_sp = 108 S⁵

2.0 x 10^-20= 108 S⁵

(2.0 x 10^-20/108)^1/5= S = 4.50 x 10^-5 M
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

The K_sp for Mg(OH)₂is 1.8 x 10^-11. Calculate the molar
solubility S, and the pH of the solution.

Mg(OH)₂----> Mg²⁺(aq) + 2OH^-1(aq)
<----

Let S = [Mg²⁺]; therefore, [OH^-] = 2S

K_sp = [Mg²⁺][OH^-1]²= (S)(2S)²

1.8 x 10^-11= 4 S³

(1.8 x 10^-11/4)^1/3= S = 1.65 x 10^-4 M

[OH^-1] = 2 x 1.65 x 10^-4 M = 3.30 x 10^-4 M

Therefore, pOH = 3.48; and pH = 14.00 - 3.48 = 10.52
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Predicting Precipitation Reactions

A solution has 0.100 M Ag⁺ (from AgNO₃) and 0.100 M Hg₂²⁺
        [from Hg₂(NO₃)₂].
When Cl^- is added, both AgCl (K_sp = 1.8 x 10^-10) and Hg₂Cl₂
        (K_sp = 1.3 x 10^-18) will precipitate from solution. What
        [Cl^-] is necessary to begin the precipitation of each salt?
        Which salt precipitates first?

For AgCl: K_sp = 1.8 x 10^-10= [Ag⁺][Cl^-]; and [Ag⁺] = 0.100M
Therefore, [Cl^-] = 1.8 x 10^-10/0.100M = 1.8 x 10^-9M

For Hg₂Cl₂: K_sp = 1.8 x 10^-18= [Hg₂⁺²][Cl^-]²;
and [Hg₂⁺²] = 0.100M
Therefore, [Cl^-] = (1.8 x 10^-18/0.100M)^1/2 = 3.6 x 10^-9M

Therefore, when Cl^- is added to the mixture of these two ions,
AgCl will precipitate first when the [Cl^-] = 1.8 x 10^-9M,
and then Hg₂Cl₂will precipitate when [Cl^-] = 3.6 x 10^-9M.
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

K_sp for Zn(OH)₂ = 5.0 x 10^-17. Will a solution made by mixing
50.0 mL of 5.0 x 10^-3 M ZnCl₂ solution and 100.0 mL of
5.0 x 10^-6 M NaOH precipitate?

Zn(OH)₂ -----> Zn²⁺(aq) + 2OH^-(aq)
^<-----
     5.0 x 10^-17= [Zn⁺²][OH^-]² <>^{Solution #1:}
    [Zn²⁺] = 0.050 L x 5.0 x 10^-3 M/0.150 L = 1.67 x 10^-3 M
^{Solution #2:}
     [OH^-1] = 0.100 L x 5.0 x 10^-6 M/0.150 L = 3.33 x 10^-6M

Now, calculate Q and compare to K_sp
Q = [Zn⁺²][OH^-]²= (1.67 x 10^-3 M)(3.33 x 10^-6M)²
Q = 1.86 x 10^-14
Since Q > K_sp, therefore, precipitation occurs. AND,
    precipitation continues until Q = K_sp
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

17.6 The Common Ion Effect and Solubility

a) Calculate the molar solubility of PbCrO₄(s) in water.
Given: K_sp for PbCrO₄= 2.8 x 10^-13

K_sp = [Pb²⁺][CrO₄^-2] = (S)(S) = 2.8 x 10^-13

S = 5.29 x 10^-7M

b) Calculate the molar solubility of PbCrO₄(s) in a solution
      containing 0.100 M Na₂CrO₄(note: the common ion CrO₄^2-
      is coming from 2 different sources: from PbCrO₄(s) and
      from the soluble salt 0.100 M Na₂CrO₄).

[Pb²⁺][CrO₄^-2] = (S)(S) = 2.8 x 10^-13, where S = [Pb²⁺]

2.8 x 10^-13 = (S)(0.100 + S) = (S)(0.100)
We assume S is small as compared to 0.100M

Therefore, S = 2.8 x 10^-12M, which is small compared to
0.100M

     Therefore, adding the common ion CrO₄^-2from the soluble
      salt Na₂CrO₄, reduces the solubility from 5.29 x 10^-7M to
      2.8 x 10^-12M, an order of 5 magnitudes!!
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

2nd Ex.: AgCl ----> Ag⁺ + Cl^-
<----

K_sp = [Ag⁺][Cl^-] = 1.0 x 10^-10= (S)(S) = S²

S = 1.0 x 10^-5M in water!!

Imagine adding NaCl (soluble salt) to this equilibrium
    system. This shifts the equilibrium to the left, since we are
    adding the common ion Cl^-.
How much [Ag⁺] will there be in a solution which is 0.10 M
        NaCl above AgCl(s)?
        1.0 x 10^-10= [Ag⁺][Cl^-] = [Ag⁺](0.10 M)
        [Ag⁺] = 1.0 x 10^-9 M in 0.10 M NaCl

How much [Ag⁺] will there be in a solution which is 3.0 M
        NaCl above AgCl(s)?
        1.0 x 10^-10= [Ag⁺][Cl^-] = [Ag⁺](3.0 M)
        [Ag⁺] = 3.3 x 10^-11 M in 3.0 M NaCl
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

K_sp for Al(OH)₃ = 5.0 x 10^-33. a) What is its molar solubility?
                                                     b) What is its solubility in g/L?
                                                     c) What is the pH of the solution?
Al(OH)₃ -----> Al³⁺(aq) + 3OH^-(aq)
^<-----

      Let S = [Al³⁺];    therefore, 3S = [OH^-]

5.0 x 10^-33= [Al⁺³][OH^-]³= (S)(3S)³= 27S⁴

a)S = 3.69 x 10^-9 M = molar solubility

b) 3.69 x 10^-9 M x 77.98 g/mol = 2.88 x 10^-7 g/L
= gram solubility

   c) [OH^-] = 3S = 3 x 3.69 x 10^-9 M = 1.11 x 10^-8M
        Therefore, pOH = -log (1.11 x 10^-8M) = 7.96
        Therefore, pH = 6.04 -------IMPOSSIBLE!!! This means
        that the solution is acidic!!

However, [OH^-] coming from the water = 1.0 x 10^-7M, and
this cannot be ignored.

Total [OH^-] = 1.0 x 10^-7M (from water)
                    + 1.11 x 10^-8M [from Al(OH)₃]
                     = 1.11 x10^-7M
     Therefore, pOH = -log (1.11 x 10^-7M) = 6.95
     Therefore, pH = 7.04      This is a basic solution!!
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

17.7 Complex Ion Equilibria and Solubility

Simultaneous Equilibria:

Will Cu(OH)₂ precipitate if 100.0 mL of 0.050 M Cu(NO₃)₂
    are added to 100.0 mL of 0.150 M aqueous ammonia?
    K_sp for Cu(OH)₂ is 1.6 x 10^-19; K_b for aqueous NH₃ is
    1.8 x 10^-5

2 different equilibria are involved:

Cu(OH)₂----> Cu²⁺(aq) + 2OH^-(aq)
^<----
K_sp= [Cu²⁺][OH^-]²= 1.6 x 10^-19

NH₃+ H₂O ----> NH₄⁺(aq) + OH^-(aq)
^<----
K_b= [NH₄⁺][OH^-]/[NH₃] = 1.8 x 10^-5

Let's calculate the [OH^-] from the weak base equilibrium:
K_b= [NH₄⁺][OH^-]/[NH₃] = 1.8 x 10^-5= (x)(x)/0.150-x
x = 0.00164 M = [OH^-]

Since copper nitrate is soluble, [Cu²⁺] = 0.050 M

We can calculate Q_sp = [Cu²⁺][OH^-]²= (0.050)(0.00164)²
= 1.34 x 10^-7

And Q_sp > K_sp. Therefore, Cu(OH)₂will precipitate until
Q_sp = K_sp
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Complex Ion Formation:

AgCl(s) + 2 NH₃(aq) ----> Ag(NH₃)₂⁺(aq) + Cl^-(aq)
<----

This can be thought of as:
AgCl(s) ----> Ag⁺(aq) + Cl^-(aq)
<----

and
Ag⁺(aq) + 2 NH₃(aq) ----> Ag(NH₃)₂⁺(aq)
<----

And we can write:
K_formation= [Ag(NH₃)₂⁺(aq)]/[Ag⁺(aq)][NH₃(aq)]²
K_formation = 1.7 x 10⁷

Or: K_dissociation= [Ag⁺(aq)][NH₃(aq)]²/[Ag(NH₃)₂⁺(aq)]
= 1/1.7 x 10⁷= 5.88 x 10^-8

Another example:
Cu(NH₃)₄²⁺(aq) -----> Cu²⁺(aq) + 4 NH₃(aq)
<-----

K_diss= [Cu(NH₃)₄²⁺]/[Cu²⁺][NH₃]⁴= 2.0 x 10^-13
K_form= [Cu²⁺][NH₃]⁴/[Cu(NH₃)₄²⁺] = 1/2.0 x 10^-13
K_form= 5.0 x 10¹²

A Problem:
Calculate [Cu²⁺] left in solution when concentrated NH₃
is added to 0.050 M CuCl₂to give final [NH₃] = 0.100 M.
K_f= [Cu(NH₃)₄²⁺]/[Cu²⁺][NH₃]⁴= 5.0 x 10¹²

K_d= [Cu²⁺][NH₃]⁴/[Cu(NH₃)₄²⁺] = 2.0 x 10^-13

[Cu(NH₃)₄²⁺] [Cu²⁺] [NH₃]

    Init []                0.00                    0.050M       0.00
    D[]                  +0.050M             -0.050M     +0.100M
    Equil []              0.050M                x                 0.100M

K_f= [Cu(NH₃)₄²⁺]/[Cu²⁺][NH₃]⁴ = 5.0 x 10¹²
K_f= 0.050M/(x)(0.100)⁴= 5.0 x 10¹²

x = (0.050)/(5.0 x 10¹²)(0.100)⁴
x = 1.0 x 10^-10M = [Cu²⁺] left in solution

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