Equilibria Problems Equilibria Answers Problems


Equilibria Solutions


1) What is the hydronium concentration of 0.03M solution of C3H7COOH, butanoic acid?
      What is the percent ionization?  The Ka of butanoic acid is 1.52x10-5.

       C3H7COOH    +   H2O  <--->  C3H7COO-    +   H3O+

          Ka[C3H7COO-][H3O+ ]
                  [C3H7COOH]      

        1.52x10-5 [x][x] 
                            0.03M

        4.56 x 10-7 =  x2

        6.75 x 10-4 M =  x  =   [C3H7COO-]  =  [H3O+ ]    

        % ionization =       [H3O+ ]        =   6.75 x 10-4 M  x   100   =   2.25%
                               [C3H7COOH]                0.03M

 

2) What is the Ka of  0.15M citric acid, C4H6O(COOH)3 when [C4H6O(COOH)2COO-]
      is 0.011M?

        C4H6O(COOH)3  +   H2O  <--->  C4H6O(COOH)2COO-   +   H3O+

          Ka[C4H6O(COOH)2COO-][H3O+ ]
                             [C4H6O(COOH)3]

         Ka [0.011M][0.011M]
                            0.15M

         Ka =  7.44 x 10-4

 3) What is the Kb of  2.8M phosphate, PO43- when [OH-] is 0.198M?

      PO43- +  H2O  <--->  PO4H2-  +  OH-

        Kb[PO4H2-][OH- ]
                         [PO43- ]

         Kb [0.198M][0.198M]
                            2.8M

         Kb =  1.4 x 10-2       

4) What is the hydroxyl concentration of a 0.2M solution of ammonia, NH3
     What is the percent ionization?  The Kb of ammonia is 1.74 x 10-5

       NH3  +   H2O  <--->  OH-    + NH4+

          Kb = [OH-][NH4+
                       [NH3]      

        1.74 x 10-5 [x][x] 
                            0.2M

        3.48 x 10-6 =  x2

        1.87 x 10-3 M =  x  =   [OH-]  =  [NH4+ ]    

        % ionization =       [ OH-]        =   1.87 x 10-3 M  x   100   =   .93%
                                     [NH3]                0.2M

 

COMMON ION PROBLEMS

5)  What is the hydronium ion concentration, [H3O+], in a solution of 0.521M HKCO3 and 
      0.024 M K2CO3?  Ka of  HKCO3 is 4.37 x 10-7.  

       HKCO+   H2O <-->  1H3O+  KCO3-  (+ KCO3-  + KOH)         

       Ka = [H3O+][ KCO3- + 0.024M KCO3-]  
                            [HKCO3 ]

        4.37 x 10-7 = [X][X + 0.024M KCO3-]  
                                           0.521M

        4.37 x 10-7 = 0.024X  
                              0.521M

        2.28 x 10-7 = 0.024X

        9.49 x 10-6 = X =  [H3O+]

 

6)  What is the hydronium ion concentration, [H3O+], in a solution of  1.13M HBO3-2 and 
      0.02 M Na3BO3?  Ka of  HBO3-2 is 1.58 x 10-14

      H BO3-2 +   H2O <-->  1H3O+  BO3-3   (+ BO3-3  + Na3OH3)

      

        Ka = [H3O+][ BO3- 3+ 0.02 M BO3-3]  
                            [HBO3-2 ]

        1.58 x 10-14 = [X][X + 0.02 M BO3-3]  
                                           1.13 M

        1.58 x 10-14 = 0.02 X  
                               1.13 M

        1.79x 10-14 = 0.02 X

        8.93 x 10-13 = X =  [H3O+]

TITRATION PROBLEMS

7) How many mL of  0.15 M KOH are needed to neutralize 25.00 mL of  .20 M HCl?

     HCl + KOH <--> KCl + HOH

  [25.00 mL HCl][        1 L HCl][.20 mol HCl][1 mol KOH][1L KOH][1000 mL KOH] =
                           [1000mL HCl][1 L HCl      ][1 mol HCl   ][.15mol KOH][1L KOH   ]

  33.33 mL KOH required to neutralize HCl

8) If 28.25 mL of 0.50 M NaOH are needed to neutralize 50.00 mL of  HNO3, what
    is the concentration of HNO3

    HNO3 + NaOH <-->  NaNO3 + HOH

    [28.25 mL NaOH][   1 L  NaOH][.50 mol NaOH][1 mol HNO3]          [1000mL] =
                            [1000mL NaOH][1 L    NaOH][1 mol NaOH][50 mL HNO3][1L]

   0.283 M HNO3 is the concentration of the acid

 

Equilibria Problems