Chapter 17 Ionic Equilibria III - Solubility Product Principle:
           (from Dr. Raynor)

Solubility Rules:
    1-  All nitrates (NO3-) are soluble; all acetates (C2H3O22-) are soluble.
    2-  All Cl- Br-, and I- are soluble, except for Ag+, Hg22+, Pb2+.
     3-  All sulfates (SO42-) are soluble, except for Ca2+, Sr2+, Ba2+,
            Hg22+, Pb2+.
    4-  All Group I and Group II and NH4+ salts are soluble.
    5-  All S2- are insoluble, except for Group I, NH4+,Ca2+, Sr2+, Ba2+
    6-  All CO32- are insoluble, except for Group I, NH4+
    7-  All PO43- 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)  + C2O42-(aq)
                                   <-----
         Ksp =  [Ca2+] [C2O42-]

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

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

    AgCl(s)  ----->  Ag+1(aq) + Cl-1(aq)                    Ksp =  [Ag+1] [Cl-1]
                   <-----
 

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

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

             Ksp =  [Pb2+]3 [AsO4-3]2
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

1.000 L of a solution saturated with CaC2O4 is evaporated to dryness,
giving  0.0061g residue of CaC2O4.  Calculate the solubility product
constant Ksp at 25oC.
        CaC2O4(s) -----> Ca2+ + C2O42-
                                <------

           Let S = [Ca2+] = [C2O42-]
        Ksp= [Ca2+] [C2O42-]
            S = 0.0061g CaC2O4/(128g/mol) = 4.8 x 10-5 mol/L
        Ksp= (4.8 x 10-5 mol/L)(4.8 x 10-5 mol/L) = 2.3 x 10-9
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

     Ksp for BaCrO4 is 1.2 x 10-10.  Find the molar solubility, S.
        BaCrO4(s) -----> Ba2+ + CrO42-
                                <------

             Let S = [Ba2+] = [CrO42-]
         Ksp =  [Ba2+] [CrO42-]
        1.2 x 10-10  = (S)(S) = S2
                               S = (1.2 x 10-10)1/2 = 1.1 x 10-5 M
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

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

        Pb3(AsO4)2(s) -----> 3Pb2+ + 2AsO43-
                                      <------

         Ksp =  [Pb2+]3 [AsO43-]2

           2S = [AsO43-]; 3S = [Pb2+]

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

The  Ksp 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

                Ksp = [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  Ksp 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

       Ksp = [Mg2+]3[AsO4-3]2 = (3S)3(2S)2 = (27S3)(4S2) = 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  Ksp 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

                Ksp = [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
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Common-Ion Effect:

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

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

           S = 5.29 x 10-7  M
 

  b)  Calculate the molar solubility of PbCrO4(s) in a solution
        containing 0.100 M Na2CrO4  (note:  the common ion CrO42-
        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 10-12  M, which is small compared to 0.100M

       Therefore, adding the common ion CrO4-2  from the soluble
        salt Na2CrO4, reduces the solubility from 5.29 x 10-7M to
        2.8 x 10-12 M.
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

2nd Ex.:  AgCl ----> Ag+ + Cl-
                        <----

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

           S = 1.0 x 10-5 M  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
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Predicting Precipitate Formation:

A solution has 0.100 M Ag+ (from AgNO3) and 0.100 M Hg22+
        [from Hg2(NO3)2].
When Cl- is added, both AgCl (Ksp = 1.8 x 10-10) and Hg2Cl2
        (Ksp = 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:    Ksp = 1.8 x 10-10 = [Ag+][Cl-];  and [Ag+] = 0.100M
                    [Cl-] = 1.8 x 10-10 /0.100M = 1.8 x 10-9 M

For Hg2Cl2:  Ksp = 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 10-9 M

Therefore, AgCl will precipitate first, when the [Cl-] = 1.8 x 10-9 M,
    and then  Hg2Cl2will precipitate when [Cl-] = 3.6 x 10-9 M.
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

 Ksp 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

         [Zn2+] = 0.050 L x 5.0 x 10-3 M/0.150 L = 1.67 x 10-3 M
        [OH-1] = 0.100 L x 5.0 x 10-6 M/0.150 L = 3.33 x 10-6 M

            Q = [Zn+2][OH-]2 = (1.67 x 10-3 M)(3.33 x 10-6 M)2
                = 1.86 x 10-14
    Since Q > Ksp, therefore, precipitation occurs.  AND, precipitation
        continues until Q = Ksp
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

 Ksp 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 acidic!!

    However, [OH-] coming from the water = 1.0 x 10-7M, and 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!!
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

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?  Ksp for Cu(OH)2 is
    1.6 x 10-19; Kb for aqueous NH3 is 1.8 x 10-5

    2 different equilibria are involved:

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

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

    Let's calculate the [OH-] from the weak base equilibrium:
         Kb = [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 Qsp = [Cu2+][OH-]2 = (0.050)(0.00164)2
                                           = 1.34 x 10-7

 And Qsp > Ksp.  Therefore, Cu(OH)2 will precipitate until Qsp = Ksp
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

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)
                               <----

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

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

    Or:    Kdissociation = [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)
                                              <-----

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

  A Problem:
    Calculate [Cu2+] left in solution when concentrated NH3 is added
    to 0.050 M CuCl2 to give final [NH3] = 0.100 M.
               Kf = [Cu(NH3)42+]/[Cu2+][NH3]4= 5.0 x 1012

               Kd = [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

                  Kf = [Cu(NH3)42+]/[Cu2+][NH3]4= 5.0 x 1012
                  Kf = 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

BACK HOME