Documents
Resources
Learning Center
Upload
Plans & pricing Sign in
Sign Out

IONIC EQUILIBRIUM INVOLVING WEAK ELECTROLYTES

VIEWS: 51 PAGES: 13

									             IONIC EQUILIBRIUM INVOLVING WEAK ELECTROLYTES
                                                                        SECTION I

                                          KCl                K+ + Cl-

                                          HCl + H2O                     H3O+ + Cl-

                                          HCl                H+ + Cl-

   1.      H2A + H2O                         H3O+ + HA-

   2.      HA- + H2O                         H3O + A=

   1a.            H2A                        H+ + HA-

   2a.            HA-                        H+ + A=

   3.             H2A                        2H+ + A=

   3a.     H2O + HOAc                                H3O+ + OAc-


   4.      K=
              H O OAc 
                    3
                                            


                  H 2 O HOAc

   5.      K[H2O] =
                            H O OAc 
                                    3
                                                       


                                         HOAc
           K[H2O] = Ka


   6.      Ka =
                  H O OAc 
                        3
                                                


                             HOAc

   7.      Ka =
                  H OAc 
                                        


                     HOAc
Example 1

    Calculate the Ka for HOAc if .1 M solution of HOAc is 1.34% ionizable.

Solution

                                                            HOAc           H+ + OAc-

Since HOAc is 1.34% ionizable

                                                                                       1.34
Concentration of [H+] which is equal to the concentration of [OAc-] =                        .1 = 1.34  10-3 m.l.
                                                                                       100
                                     1.34
The concentration of HOAc = .1 -           .1 = .099
                                     100


                 Ka =
                      134  10 134  10  = 1.8  10
                        .       3
                                   .         3
                                                                   -5
                                .099

Example 2

    Kb for NH4OH is 1.8  10-5. Calculate the concentration of NH4+ and the OH- and the % ionization if
.5 molar NH4OH solution.

Solution

                                     NH4OH                 NH4+ + OH-

     The concentration of NH4+ = the concentration of OH-. Let “x” be the concentration of NH4+. Then
“x” is the concentration of OH- and the concentration of NH4OH is .5 - x.


                                     Kb =
                                          NH OH 
                                                  4
                                                              


                                               NH 4 OH 
                                                       x. x
                                     1.8  10-5 =
                                                      .5  x

x is very small compared to .5. It can be deleted from the denominator:

                                                      x2
                                     1.8  10-5 =
                                                      .5
                                     .9  10-5 = x2

                                     9  10-6= x2

                                        9  10 6  x

                 3  10-3 = x = NH4+ = OH-

                  concentration of NH 4 + or concentration of OH -
% ionization =                                                           100
                           concentration of original base

                 3  10 3
% ionization =              100 = .6%
                    .5
                                            SECTION II

  1.   Calculate the concentration of each ion in the following solutions.

       (a)   .05 M HCl

       (b)   1.2 M NaOH

       (c)   .1 M NaOAc




  2.   Calculate the H+ and the % ionization in a .1 M HCN (Ka HCN 4  10-10).




  3.   .1 M NH4OH solution is 1.34% ionizable. Calculate the Kb for NH4OH.




  4.   Calculate the H+ and the % ionization in a 1 M HOAc (Ka 1.8  10-5).




ANSWERS FOR SECTION II:

  1.   (a)      [H+] = .05 [Cl-] = .05

       (b)      [Na+] = 1.2 [OH-] = 1.2

       (c)      [Na+] = .1 M [OAc-] = .1

  2.   [H+] = 6.3  10-6

       % ionization = 6.3  10-3%

  3.   1.85  10-5

  4.   [H+] = 4.24  10-3

       % ionization = .424%
                                                   SECTION III
                                                   (Common Ion)

                    PbCl2           Pb+2 + 2Cl-

                    HCl            H+ + Cl-

If HCl is added, the solubility of PbCl2 will be reduced.

                    HOAc            H+ + OAc-

Example 1

           Calculate the [H+] in a .1 M HOAc which contains .1 M NaOAc (Ka HOAc 1.8  10-5).

                    HOAc            H+ + OAc-

Solution

           HOAc is a weak electrolyte. The Ka expression is given by Equation 7 of Section 1.


                             H   OAc  
                                        
                    Ki =
                                 HOAc
           Assume the concentration of H+ = x. The concentration of OAc- from OHAc = x. NaOAc is a
           strong electrolyte; thus the concentration of the OAc- = .1, and the total concentration of OAc- is .1
           + x.


                    1.8  10-5 =
                                    x .1  x
                                    .1
           The value of x compared to .1 is very small. It can be neglected. Hence:


                    1.8  10-5 =
                                    x .1
                                      .1
                    H+ = x = 1.8  10-5

Reaction 1:         HOAc            H+ + OAc-

Reaction 2:         NaOAc          Na+ + OAc-

                    H+ + OH            H2O

Example 2

           Calculate the [OH-] in a .2 M NH3 solution which also contains .5 M NH4Cl. Kb NH3 is
           1.8  10-5.
Solution

           NH4OH is a weak electrolyte.

                    NH3 + H2O               NH4 + + OH-


                    Kb =
                         NH  OH 
                               4
                                           


                                NH 3 
                    NH4Cl          NH4+ + Cl-

           Assume the concentration of OH- = x.

       Concentration of NH4+ from NH4OH is also x. But in addition to NH4+ from NH4OH, we have
    +
NH4 from NH4Cl which is .5 M. The total NH4+ = .5 + x.

           Substitute the values obtained from the ionization expression for NH4OH


                    1.8  10-5 =
                                   . 5  x  x
                                        . 2
           x is very small compared to .5; thus it can be neglected.


                    1.8  10-5 =
                                   . 5  x
                                       .2

                    .36  10-5 = .5x

                    .72  10-5 = x

                    7.2  10-6 = x

                    x = 7.2  10-6
                                           Section IV

  1.   Calculate the concentration of [H+] in a .1 M HCN solution which also contains .2 M NaCN
       (Ka HCN 4  10-10).




  2.   Calculate the [H+] in a .2 M HOAc solution which also contains .2 M NaOAc
       (Ka HOAc 1.8  10-5).




  3.   How many m/l of hydrogen ions are contained in a .1 M HOAc solution which contains .2 Mole of
       NaOAc? (Ka HOAc 1.8  10-5 ).




ANSWERS TO SECTION IV:

  1.   [H+] = 2  10-10 mole/liter

  2.   [H+] = 1.8  10-5 mole/liter

  3.   [H+] = 9  10-6 mole/liter
                                                     Section V
                                                (Ionization of Water)

Reaction 3:           H2O + H2O           H3O+ + OH-

Reaction 4:           H2O            H+ + OH-


Eq. 8      Ke =
                   H OH 
                                


                       H 2 O

           Ke[H2O] = [H+][OH-]

Eq. 9      Kw = [H+][OH-]

           Kw is the ion product constant for water. Kw is a constant at 25C and has a value of 1  10-14.

Eq. 10 [H+][OH-] = 1 10-14

           It follows that in an aqueous solution at 25C.

Eq. 11 pH + pOH = 14

           In pure water the [H+] = [OH-] = 1 10-7.

           For example, calculate the [OH-] in a .01 molar HCl.

Solution

           [H+][OH-] = 1 10-14

           [1  10-2][OH-] = 1  10-14

                    1  10 14
           OH- =
                    1  10 2

                  = 1  10-12
                                                  Section VI
                                                        (pH)

Eq. 12 pH = -log[H+]

Eq. 13 pOH = -log[OH-]

Example 1

           Calculate the pH of a solution in which [H+] is 1  10-3 m/l.

Solution

           Substitute the value given in Equation 12.

                    pH = -log[H+]

                       = -log[1  10-3]

                       = -[log 1 + log 10-3]

                       = -[-3 log 10]

                       = -[-3] = 3

Example 2

           Calculate the pH of .0055 molar HCl.

Solution

           HCl is completely ionzied; thus the H+ is .0055 or 5.5  10-3. Substitute the value in Equation 12.

                    pH = -log[5.5  10-3

                       = [log 5.5 + log 10-3]

                       = -[.74 - 3]

                       = -[-2.26]

                       = 2.26

           If the pOH is required, remember that:

                       pH + pOH = 14

                       pOH = 14 - 2.26

                       pOH = 11.74
Example 3

           Calculate the pH of .1 M HOAc. Ka 1.8 10-5.

Solution

           HOAc is a weak electrolyte. Ki expression is given by Equation 7.


                        Ka =
                             H OAc 
                                             


                                   HOAc
                        H+ = x OAc- = x

                                       x. x
                        1.8  10-5 =
                                        .1

                        1.8  10-6 = x2

                        1.34  10-3 = x

therefore,              H+ = 1.34  10-3

therefore,              pH = -log[1.34  10-3]

                            = -[.11 - 3]

                            = -[-2.89]

                            = 2.89

                                     IMPORTANT POINTS TO REMEMBER

   1.      The pH of a neutral solution is 7; at this point the [H+] = [OH-] = 1  10-7 m/l.

   2.      If the pH of solution is less than 7, then the solution is acidic and the [H +] is larger than 1  10-7.

   3.      If the pH of a solution is more than 7, then the solution is basic and the [H +] is less than 1  10-7.
                                         Section VIII

  1.   Calculate the pH of .1 M HCl.




  2.   Calculate the pH of .1 M HOAc (Ka HOAc 1.8  10-5)




  3.   Calculate the pH, pOH of .0005 M HCl.




  4.   Calculate the pH of .05 M HCN (Ka 4  10-10)




  5.   Calculate the pOH of .5 M HOA (Ka HOAc 1.8  10-5)



ANSWERS TO SECTION VIII:

  1.   pH = 1

  2.   pH = 2.89

  3.   pH = 3.3

       pH = 10.7

  4.   pH = 5.35

  5.   pOH = 11.48
                                                                    Section IX
                                                                    (Hydrolysis)


Reaction 5:             NH4Cl + H2O                           NH4OH + HCl

Reaction 6:             NaOAc + H2O                           NaOH + HOAc

Reaction 7:             NH4+ + HOH                            NH4OH + H+

                  NH 4 OH  H  
                 NH 4   HOH
Eq. 14 Ka =




        Ka[HOH] = Kh

                  NH 4 OH  H  
                         NH 
Eq. 15 Kh =
                                         
                                     4


                    Kw
Eq. 16 Kh =
                    Kb

Eq. 9   [H+][OH-] = Kw



From the hydrolysis of the NH4+
                 NH4+ + HOH                                   NH4OH + H+

                                              NH 4 OH  H  
                                                   NH 
Eq. 17 Therefore, Kh =
                                                          
                                                      4


        Substitute the value of [H+] from Eq. 9 into Eq. 17 and obtain Eq. 18.

                  NH 4 OH  Kw
                    NH  OH 
Eq. 18 Kh =
                                              
                         4



               NH 4 OH                 
                                               1
         NH  OH 
but
                                             k1
                4


         Kw
i.e.        = Kh
         Kb
Example 1

           Calculate the pH of a solution which is .1 M NaOAc. (Ka HOAc 1.8  10-5)

Solution

                    OAc- + HOH                        HOAc + OH-

           The hydrolysis of OAc- indicates an increase in the OH-; hence, the solution should be basic.

           Assume the concentration of HOAc = x. Then the concentration of OH- = x.


                         Kw  HOAc OH
                            
                                                               
                                                                    
                                                           
                    Kh =
                         Ka     OAc 


                    1  10 14         x. x
                              5
                                   
                    18  10
                     .                  .1

                    10  10 16
                                    x2
                    18  10 5
                     .

                    5.5  10-11 = x2

                    55  10-12 = x

                    7.5  10-6 = x

                    pOH = -log[OH-]

                         = -log[7.5  10-6]

                         = [.88 - 6]

                        = 5.12
therefore, pH = 14 - 5.12 = 8.88
Example 2

           Calculate the pH of a .1 M NaCN solution. Ka HCN 4  10-10.

Solution

                    CN- + HOH                         HCN + OH-

          From the reaction between CN- and water, it can be seen that there is an increase in the OH-, and
therefore the solution will be basic. Let x be the concentration of HCN; then the concentration of OH - is x.
That is, [HCN] = [OH-] = x.


            Kw  HCN  OH
               
                                          
                                                  
                                      
Eq. 19 Kh =
            Ka     CN 
       10  10 15       x. x
                     
        4  10 10        .1

       2.5  10-6 = x2

       1.58  10-3 = x

       pOH = -log[1.58  10-3]

             = [.2 - 3]

             = 2.8

       pH = 14 - 2.8 = 11.2


                                             Section X

  1.   Calculate the [OH-], [H+], and the pH in .2 M (NaCN). (Ka HCN = 4  10-10)




  2.   Calculate the [H+] and the pOH in .2 M NH4Cl. (Kb NH4OH = 1.8  10-5)




  3.   Calculate the [H+] and the pH of .5 M NaOAc. (Ka HOAc 1.8  10-5)




ANSWERS FOR SECTION X:

  1.   [OH-] = 2.24  10-3       2.   [H+] = 1.05  10-5         3.   [H+] = 6  10-10
       [H+] = 4.46  10-12            pOH = 9.02                      pH = 9.22
       pH = 11.35

								
To top