Solubility Unit Plan

 

 

Notes- double click on the lesson number and download Power Point Viewer if you do not have it.                                                                                                

Worksheets                              Quiz

·        1. Solubility and Saturated Solutions.                               WS 1                                       1

·        2. Ion Concentration Calculations and Ionic Equations.   WS 2                                        2

·        3. Solubility to Ksp.                                                         WS 3                                        3

·        4. Ksp to Solubilty and Size of Ksp.                                 WS 4                                        4

·        5. Trial Ksp.                                                                     WS 5                                        5

·        6. Separating Ions                                                             WS 6                                        6

·        7. Common Ion Effect and                                                WS 7                                        7

·        8. Titrations and Max Ion Concentration                          WS 8    Web Review               8      Quizmebc

·        9. Review                                                                         Practice Test1

·        10. Review                                                                       Practice Test 2                                                                                                                 

·       11. Test

 

 

 

Text book                    Hebden                        Read Unit III

 

 

 

            The following workbook is designed to ensure that you can demonstrate your understanding of all aspects of the solubility unit. Ask yourself, “do I want to do well in this class?” If you are determined to be successful the minimum expectation that you should have for yourself is that you do all of these questions by the due dates given by your teacher. There are other things that you should do to prepare for the test at the end of the unit. Remember, what you put into this course is what you will get out. There is no substitute for consistent effort and hard work. If you can’t do a question, get some help before the end of the unit, you need to know, understand, and remember everything. Good luck! I know you can do well in this unit. Keep up the great work!

 

Web Site Address: http://www.wjmouat.com/

 

 

 

 

 

 

 

 

Chemistry 12                Solubility and Saturated Solutions                          WS #1

                       

1. Define and give units for solubility.           M,    g/100mL,       g/L

 

2. Describe the relationship between the rate of dissolving and the rate of crystallization when a small amount of solute is added to an unsaturated solution.

           

            Rate of dissolving is greater than the rate of crystallization

 

3. Describe the relationship between the rate of dissolving and the rate of crystallization when a small amount of solute is added to a saturated solution.

 

            Rate of dissolving equals the rate of crystallization

 

4. Describe the relationship between the rate of dissolving and the rate of crystallization when a small amount of solute is added to a supersaturated solution.

 

            Rate of dissolving is less than the rate of crystallization

 

5. Which of the above solutions would need to be prepared in order to determine the solubility of an ionic solution.

 

             Saturated

 

6. 2.65 g of Ba(OH)2 is dissolved in 70.0 mL of water to produce a saturated solution at 20 oC. Calculate the solubility in units of g/100 mL, g/L, and M.

 

            0.221M

 

            37.9g/L

           

            3.79g/100mL

 

7. A beaker containing 100. mL of saturated BaCO3 solution weighs 159.60 g. The beaker is evaporated to dryness and weighs 56.36 g. The empty beaker weighs 24.33 g. Calculate the solubility in units of g/100 mL, g/ L, and M.

 

 

              56.36 g                      32.03g     =      320g    x   1 mole        =     1.62 M                 Do not use the mass with water included!

            - 24.33g                       100 mL              L           197.3g                               

              32.03 g

 

           

 

 

8. Write dissociation equations to represent the equilibrium present for a saturated solution of each ionic compound. Write the solubility product (Ksp expression) for each of the equilibrium systems. The first one is done.

 

a) Al2(SO4)3                          ⇄        2Al3+               +         3SO42-                                             Ksp = [Al3+]2 [SO42-]3

b) FeCO3                                 ⇄   Fe2+                      +         CO32-                                              Ksp = [Fe2+] [CO32-]

c) Co2(SO4)3                        ⇄   2Co3+                   +         3SO42-                                              Ksp = [Co3+]2 [SO42-]3

d) Na3PO4                               ⇄    3Na+                    +         PO43-                                               Ksp = [Na+]3 [PO43-]

10. Write formula, complete ionic, and net ionic equations for each.

 

a) H2SO4(aq)         +      NaOH(aq)    →   

 

                        H2SO4 (aq)             +   2NaOH (aq)    →         2H2O (l)          +        Na2SO4 (aq) 

                        2H+(aq) + SO42-(aq) + 2Na+(aq) + 2OH-(aq)             →          2H2O (l)  +  2Na+(aq)    + SO42-(aq)

                        2H+(aq) + 2OH-(aq)            →   2H2O (l) 

 

b) Mg(NO3)2(aq)    +      Na2CO3(aq)    →

 

 Mg(NO3)2 (aq)             +   Na2CO3 (aq)   →                    MgCO3 (s)         +        2NaNO3 (aq)

              Mg2+(aq)      +   2NO3-(aq)         +  2Na+(aq) +   CO32-(aq) →  MgCO3 (s)  +     2Na+(aq) +       2NO3 - (aq)

                        Mg2+(aq)      +   CO32-(aq) →     MgCO3 (s) 

 

 c) Al(NO3)3(aq)    +      (NH4)3PO4(aq)    →

 

 Al(NO3)3 (aq)             +   (NH4)3PO4 (aq)   →              AlPO4 (s)           +        3NH4NO3 (aq)

              Al3+(aq)      +   3NO3-(aq)         +  3NH4+(aq) +   PO43-(aq) →  AlPO4 (s)+     3NH4+(aq) +       3NO3 - (aq)

                        Al3+(aq)      +   PO43-(aq) →     AlPO4 (s)

 

 d) H3PO4(aq)    +     Ca(OH)2(aq)     →

 

 2H3PO4(aq)    +     3Ca(OH)2(aq)     →  Ca3(PO4)2(s)    +   6HOH(l)

6H+ (aq)    + 2PO43-(aq)   +  3Ca2+(aq)   +   6OH-(aq)          →  Ca3(PO4)2(s)      +   6HOH(l)

6H+ (aq)    + 2PO43-(aq)   +  3Ca2+(aq)   +   6OH-(aq)          →  Ca3(PO4)2(s)      +   6HOH(l)

 

 

 

Chemistry 12     Solubility       WS #2                       Ion Concentration Calculations

 

1.         What is the concentration of each ion in a 10.5 M sodium silicate solution?

 

 Na2SiO3         ⇄        2Na+     +         SiO32-

 

10.5 M            21.0 M 10.5 M

 

            [Na+] = 21.0 M,          [SiO32-] = 10.5 M

 

 

2.         What is the concentration of each ion in the solution formed when 94.5 g of nickel (III) sulphate is dissolved into 850.0 mL of water?

 

 Molarity  =      94.5 g  x        1 mole   

                                                405.7g             =          0.2740

                                         0.850L

 

Ni2(SO4)3        ⇄        2Ni3+               +          3SO42-

 

0.2740                         0.548 M                      0.822 M

 

[Ni3+] = 0.548 M,        [SO42-] = 0.822 M

 

 

3.         If 3.78 L of 0.960 M sodium fluoride solution is added to 6.36 L of 0.550 M calcium nitrate solution, what is the resulting concentration of [Ca+2] and [F-]?

 

 

                                    NaF     ⇄        Na+                  +          F-                                             Ca(NO3)2        ⇄                    Ca2+                +          2NO3-

 

             3.78    x   0.960 M  =             0.358 M                      0.358M                                   6.36    x   0.550 M  =              0.345 M                      0.690M

            10.14                                                                                                                           10.14

 

 

 

 

            [Ca2+] = 0.345 M,       [F-] = 0.358 M

 

 

4.         What is the concentration of each ion in the solution formed when 94.78 g of iron (III) sulphate is dissolved into 550.0 mL of water?

 

 

 [Fe3+] = 0.8619 M,    [SO42-] = 1.293 M

 

 

5. If the [F-]  = 0.200 M, calculate the number of grams AlF3 that would be dissolved in 2.00 L of water.

 

 

                        AlF3                 ⇄        Al3+     +          3F-

 

                        0.06667M                   0.06667M       0.200M

 

            2.00L  x  0.06667 mole   x     84.0 g     =    11.2g

                                    L                     mole

 

 

6. If the [SO42-]  = 0.200 M in 2.0 L of Al2(SO4)3, determine the [Al3+] and the molarity of the solution.

 

                        Al2(SO4)3             ⇄        2Al3+               +         3SO42-  

 

                        0.067 M                      0.13 M            0.20 M

 

 

 

Dissociation Equations     Write a dissociation equation for any chemical which dissociate when dissolved in water:

 

            1.         HCl (aq)                                        ⇄         H+                   +          Cl-

 

            2.         C6H12O6 (s)                     ⇄             C6H12O6(aq)                             (molecular compounds do not dissociate)

 

            3.         Na2S (s)                                        ⇄             2 Na+ (aq)                 +         S2- (aq)   

 

            4.         Al(CH3COO)3 (s)                ⇄        Al3+ (aq) +          3CH3COO- (aq)

 

       5.         MgBr2 (s)                                                        ⇄             Mg2+ (aq)                  +          2 Br- (aq)

 

        6.               Na2CO3 (s)                                                    ⇄             2 Na+ (aq)                 +          CO32- (aq)

 

            7.         C12H22O11 (s)                             ⇄        C12H22O11 (aq)                        (molecular compounds do not dissociate)

 

            8.         K3PO4 (s)                      ⇄             3 K+ (aq)                    +          PO43- (aq)

 

            9.         CH3OH (l)                     ⇄             CH3OH (aq)                             (molecular compounds do not dissociate)

 

 Net Ionic Equations

 

Write chemical equations, total ionic equations and net ionic equations for each reaction. The first one is done for you. (assume that all reactions occur):

 

1.         Magnesium metal is placed in hydrochloric acid

                                       

                        Mg (s)            +   2 HCl   (aq)           →          MgCl2 (aq)                +         H2 (g)

                        Mg (s)    +  2 H+ (aq) +  2 Cl- (aq)   →       Mg2+ (aq)  +   2Cl- (aq)  +  H2 (g)

                        Mg (s)           +  2 H+ (aq)                 → Mg2+ (aq)       +    H2 (g)

 

2.         Zinc metal is placed in silver nitrate solution

 

Zn (s)             +   2Ag NO3 (aq)  → Ag (s)     +            2 Zn(NO3)2 (aq)

            Zn(s)     +    2Ag+(aq)    +  2 NO3-(aq)   →                     2Ag(s)      +  Zn+2 (aq) +   +  2NO3-(aq)

Zn(s)                +       2Ag+ (aq)        →    Zn+2 (aq)      +          2Ag(s)              

 

3.         Barium chloride solution is added to lead (II) nitrate solution.

 

BaCl2 (aq)             +   Pb(NO3)2 (aq)   →         PbCl2 (s)           +        Ba(NO3)2 (aq)

             

Ba2+(aq)            +  2Cl-(aq) +   Pb2+(aq) +   2NO3-(aq)     →  PbCl2 (s) +         Ba2+ (aq)      +       2NO3 - (aq)

                        Pb2+ (aq)        +      2Cl- (aq)                →     PbCl2 (s)

 

4.         Sulphuric acid is added to Strontium hydroxide solution.

 

H2SO4 (aq)             +   Sr(OH)2 (aq)    →         2H2O (l)         +        SrSO4 (s)

                       

2H+(aq) + SO42-(aq) + Sr2+(aq) + 2OH-(aq)             →          2H2O (l)  +  SrSO4 (s)

                        2H+(aq) + SO42-(aq) + Sr2+(aq) + 2OH-(aq)            →   2H2O (l)  +  SrSO4 (s)

5.         Sodium carbonate solution is added to nickel (III) nitrate solution.

 

 3Na2CO3(aq)          +   2Ni(NO3)3 (aq)    → Ni2(CO3)3 (s) +          6NaNO3 (aq)

                        6Na+(aq) +3CO32-(aq) +2Ni3+(aq) +6NO3-(aq)     →     Ni2(CO3)3 (s) + 6Na+(aq) +6NO3-(aq)

                        3CO32-(aq)            +    2Ni3+(aq)   → Ni2(CO3)3 (s)

 6.        Aqueous chlorine is added to sodium bromide solution.

 

Cl2 (aq)                   +   2NaBr (aq)     →   2NaCl (aq)      +      Br2 (aq)

                        Cl2 (aq)    +   2Na+ (aq) +  2Br- (aq)    →  2Na+ (aq) +  2Cl- (aq)  +  Br2 (aq)

                        Cl2 (aq)                  +   2Br- (aq)       →  2Cl- (aq)     +      Br2 (aq)

 7.        Nitric acid is added to aluminum hydroxide solution.

 

 2HNO3 (aq)           +   Sr(OH)2 (aq)   →    2H2O (l)                +      Sr(NO3)2 (aq)

                        2H+(aq) + 2NO3-(aq) +  Sr2+(aq) + 2OH-(aq)   →       3H2O (l) + Sr2+(aq) + 2NO3-(aq)

                                           H+(aq)   +   OH-(aq)          →                              H2O (l)

 

 

   

Chem 12         WS #3                         Solubility to  Ksp

 

 

The Ksp is a measure of the solubility of an ionic salt. The larger the value of the Ksp, the greater is the solubility of the salt. You can only calculate a Ksp if the solution is saturated. Only saturated salt solutions are in equilibrium. You can calculate the Ksp from the solubility of a salt, since the solubility represents the concentration required to saturate a solution.

 

1. Calculate the Ksp for CaCl2 if 200.0 g of CaCl2 are required to saturate 100.0 mL of solution.

 

 

Molarity         =          200 g  x  1 mole                                 

                                                  111.1 g          =          18.001 M

                                          0.100L

 

CaCl2              ⇄        Ca2+    +          2Cl-

 

18.0 M            18.0 M 36.0M

 

 

Ksp  =  [Ca2+][Cl-]2

 

Ksp  =  [18.0][36.0]2

 

Ksp = 2.33 x 104

 

  

2. Calculate the Ksp for AlCl3 if 100.0 g is required to saturate 150.0 mL of a solution.

 

 

 Ksp = 1.679 x 104

 

 

3. The solubility of SrF2 is 2.83 x 10-5 M. Calculate the Ksp.

 

 

 Ksp = 9.07  x 10-14

 

 

4. The solubility of GaBr3 is 15.8 g per 100 mL. Calculate the Ksp.

 

 Molarity        =          15.8 g  x  1 mole

                                                   309.4 g         =     0.51066 M                                  

                                         0.100 L

 

 

                        GaBr3             ⇄        Ga3+    +          3Br-

 

            0.51066 M                  0.51066 M      1.532 M

 

 

            Ksp  =  [Ga3+][Br-]3

 

Ksp  =  [0.51066][1.532]3

 

Ksp = 1.83

 

 

5.         The solubility of Ag2SO4 is 1.33 x 10-7g per 100 mL. Calculate the Ksp.

 

 

            Ksp = 3.10 x 10-25

 

 

6.         If 2.9 x 10-3 Ca(OH)2 g is needed to saturate 250 mL of solution, what is the Ksp.

 

 

            Ksp =1.5 x 10-11                                 

 

 

7.       At a certain temperature, a 40.00 mL sample of a saturated solution of barium hydroxide, is neutralized by 29.10 mL of 0.300 M HCl. Calculate the Ksp of          Ba(OH)2.

 

          2HCl               +          Ba(OH)2         →        BaCl2  +          2HOH

          0.02910 L                   0.0400 L

          0.300 M                      ? M

 

          Molarity Ba(OH)2     =          0.02910 L HCl            x          0.300 moles    x          1 mole Ba(OH)2

                                                                                                          1 L                              2 moles HCl

                                                                                              0.0400 L

 

 

                                              =          0.109 M

 

 

          Ba(OH)2         ⇄        Ba2+                 +          2OH-

         

          0.1091 M                    0.1091 M                    0.2183 m

 

          Ksp      =          [Ba2+][OH-]2

 

          Ksp      =          [0.1091][0.2183]2

 

 

                      =          5.20   x  10-3

 

Calculate the concentrations of all ions in each solution.

 

8.         0.50 M Al2(SO4)3(aq)

 

      Al2(SO4)3   ⇄        2Al3+               +          3SO42-

 

      0.50 M                  1.0 M                          1.5 M

 

      [Al3+] = 1.0M                                 [SO42-] = 1.5M

 

 

 

9.   25.7g (NH4)3PO4 (aq) in 250mL H2O.

 

 

 [NH4+] = 2.07M        [PO43-] = 0.690M

 

 

 

10. 210g CoCl2 • 6H2O in 800mL H2O.

 

 

 [Co2+] = 1.10M        [Cl-] = 2.20M

 

 

Chemistry 12                           Ksp to  Solubility                                            WS # 4            

 

Calculate the solubility in M and g/L for each. Use the Ksp values found in your chart.

1)  BaCO3

 

BaCO3(s)         ⇄        Ba2+     +          CO32-

 

      x                x                      x                       

 

 

      ksp            =          [Ba2+][ CO32-]

 

      ksp            =          x2

 

      2.6  x  10-9 =          x2

     

      5.099  x  10-5 M

 

      5.099 x  10-5  mole   x   197.3 g    =

                   L                      1 mole

     

      1.0 x 10-2  g/L

 

 

 

2)   Fe(OH)2

 

 

 

2.1 x 10-4  g/L

 

 

 

3)   PbCl2

 

 

 4.0 g/L

 

 

 

4)   How many grams of Mg(OH)2 are required to completely saturate 1.5  L of solution?

 

 

            Mg(OH)2                    ⇄        Mg2+    +          2OH-

 

            x                                              x                      2x

 

 

            Ksp                              =          [Mg2+][OH-]2

 

            Ksp                              =          [x][2x]2

 

            Ksp                              =          4x3

 

            5.6   x  10-12                =          4x3

 

            1.119  x  10-4 M          =          x

 

 

            1.5 L   x   1.119 x  10-4 mole   x   58.3 g        =          9.8 x 10-3 g                 

                                        1 L                    1 mole

 

           

           

Review

 

1.   If 200 g of MgCl2 is required to saturate 1.5 L of solution at 20 oC, calculate the Ksp.

 

 Ksp = 11      

 

 

2.     If the Ksp for Al2O3 is 2.8 x 10-8, calculate [Al3+] and [O-2] in  mol/L.

 

 [Al+3] = 2.4  x  10-2 M     [O-2] = 3.6  x  10-2 M

 

 

 Trial Ksp                   Worksheet 5

 

1. Will a precipitate form if 200ml 0.00020M Ca(NO3)2  is mixed 300ml of 0.00030M  Na2C03?  

 

            CaCO3                       ⇄                    Ca2+                +                      CO32-

 

                                                            200  x  0.00020 M                  300  x  0.00030 M

                                                            500                                          500

 

                                                            0.000080 M                            0.00018 M

 

                                    Trial Ksp        =          [0.000080][0.00018]

                                               

                                    Trial Ksp        =          1.4 x  10-8        >          Ksp(5.0   x   10-9)

 

                                                        Therefore a precipitate forms!

 

 

2. Will a precipitate form if 25.0ml of .0020M  Pb(NO3)2  is mixed with  25.0ml of .040M  NaBr.

 

 

 Trial ksp = 4.0 x  10-7  no ppt

 

 

3. Will a precipitate form if equal volumes of 0.00020M Ca(NO3)2  is mixed with  0.00030M  Na2C03?  

 

 Note: When equal volumes are mixed the dilution factor is ½ for each ion.

 

Trial ksp = 1.5 x  10-8  ppt   > Ksp and there is a precipitate

 

 

Ksp

4. Co(OH)2      Solubility = 3.0x10-3 g/L    Ksp=?

 

ksp = 1.3 x  10-13   

 

 

5. Ag2C2O4    Solubility = 8.3x10-4 M        Ksp=?  

 

 

 ksp = 2.3 x  10-9      

 

 

Solubility

6. SrF2     Solubility in (M) = ?

 

 

  1.0  x  10-3 M  

 

 

7. Cu(IO3)2    Solubility (g/L) = ?

 

 

 1.1 g/L

 

 

 Separation Positive Ions:  Work from top to bottom of solubility chart!!              WS # 7

1.         Ag+      Mg2+    Ba2+

i) Add:            NaCl(aq)           Filter Out:       AgCl(s)                         Net Ionic equation:  Ag+     +  Cl- ------> AgCl(s)

ii) Add:           Na2SO4(aq)       Filter Out:       BaSO4(s)                      Net Ionic equation:  Ba2+     +  SO4-2 ------> BaSO4(s)

iii) Add:          NaOH(aq)         Filter Out:       Mg(OH)2(s)                 Net Ionic equation:  Mg2++2OH-------> Mg(OH)2(s

2.         Pb2+     Ba2+     Sr2+

i) Add:            NaCl(aq)           Filter Out:       PbCl2(s)                        Net Ionic equation:  Pb+2     +  2Cl- ------> PbCl2(s) 

ii) Add:           NaOH(aq)         Filter Out:       Ba(OH)2(s)                  Net Ionic equation:  Ba2++2OH-------> BaOH)2(s)

iii) Add:          Na3PO4(aq)       Filter Out:       Sr3(PO4)2(s)                 Net Ionic equation:  3Sr2++2PO4-3------> Sr3(PO4)2(s)

3.         Cu+      Ca2+     Sr2+

i) Add:            NaCl(aq)           Filter Out:       CuCl(s)                         Net Ionic equation:  Cu+     +  Cl- ------> CuCl(s)        

ii) Add:           NaOH(aq)         Filter Out:       Ca(OH)2(s)                  Net Ionic equation:  Ca2++2OH-------> Ca(OH)2(s)

iii) Add:          Na3PO4(aq)       Filter Out:       Sr3(PO4)2(s)                 Net Ionic equation:  3Sr2++2PO4-3------> Sr3(PO4)2(s)

4.         Be2+     Sr2+      Ag+

i) Add:            NaCl(aq)           Filter Out:       AgCl(s)                         Net Ionic equation:  Ag+     +  Cl- ------> AgCl(s)

ii) Add:           Na2SO4(aq)       Filter Out:       SrSO4(s)                       Net Ionic equation:  Sr2+     +  SO4-2 ------> SrSO4(s)

iii) Add:          NaOH(aq)         Filter Out:       Be(OH)2(s)                  Net Ionic equation:  Be2++2OH-------> Be(OH)2(s)

5.         Be2+     Ca2+     Pb2+

 

i) Add:            NaCl(aq)           Filter Out:       PbCl2(s)                        Net Ionic equation:  Pb+2     +  2Cl- ------> PbCl2(s)

ii) Add:           Na2SO4(aq)       Filter Out:       CaSO4(s)                      Net Ionic equation:  Ca2+     +  SO4-2 ------> CaSO4(s)

 

iii) Add:          NaOH(aq)         Filter Out:       Be(OH)2(s)                  Net Ionic equation:  Be2++2OH-------> Be(OH)2(s)

 

]

6. Calculate the Ksp for CaCl2, if 50.0 g is required to saturate 25.0 mL of water.

 

50.0 g   x   1 mole

                111.1 g        =    18.0 M                  Ksp  =  4x3      =          4(18.0)3           =            2.33  x  104

   0.0250 L

 

 

7. Calculate the molar solubility of Mg(OH)2.