Polyprotic Acids and Acid Base Salts by MikeJenny

VIEWS: 212 PAGES: 21

									Polyprotic Acids
 And Acid and Base Salts
Polyprotic Acids
   So far, we have only dealt with acids that
    can give up one proton.

   Most acids encountered in biological
    systems have multiple protons, depending
    on the pH of the solution.

   We call these „polyprotic‟ acids or bases.
Diprotic acids and bases

                                   
             
  H 2 A (aq)  H    (aq)      HA   (aq)   ; K a1
                                 2
    HA   -
         (aq)   
                 H   (aq)   A   (aq)   ; K a2

 How do we calculate the pH of a solution of:
            H2A, HA-, or A2- ?
Diprotic Acids
   Treat this like a weak monoprotic acid:
                                                                3
               
    H 2 A (aq)  H      (aq)    HA (aq) ; K a1  6.3x10
                                           -



                               -
            [H          ][ HA       (aq)   ]
                                                6.3x103
                 (aq)

                 [H 2 A (aq) ]

              H2A                          HA-              H+
       I     0.0750                        0                0
       C        -x                         +x               +x
       E    0.0750-x                        x               x
Diprotic Acids
                ( x)( x)
                           6.3x10 3
              0.0750 - x
         x 2  6.3x10 3 x  4.725 x10  4
         cannot ignore x in this case
                
         x  [H ]  [HA ]  0.019M
                           -


    [H 2 A]  0.0750M  0.019M  0.056M

                What about [A2-]?
  Diprotic Acids
                    
        H 2 A (aq)  H (aq)  HA - (aq) ; K a1  6.3x103
                             2
         HA - (aq)  H (aq)  A (aq) ; K a2  4.9 x1010
                                       2-
                     [H          ][ A          ]
                                                    4.9 x10 10
                          (aq)          (aq)

                       [HA - (aq) ]
 The amount of H+ from dissociation of HA- is insignificant
 relative to H+ from the dissociation of H2A, so total [H+] =
 0.019M, and HA- also equals 0.019 M

            4.9 x10 10[HA - (aq) ] (4.9 x10 10 )(0.019)
[A (aq) ] 
   2-
                      
                                                          4.9 x10 10 M
                  [H (aq) ]               (0.019)
Polyprotic Acids
   Here are three successive ionizations of phosphoric
    acid:

     H3PO4(s) + H2O(l)    H3O+(aq) + H2PO4−(aq)   Ka1= 7.25×10−3
     H2PO4−(aq)+ H2O(l)   H3O+(aq) + HPO42−(aq)   Ka2= 6.31×10−8
     HPO42−(aq)+ H2O(l)   H3O+(aq) + PO43−(aq)    Ka3= 3.98×10−13


   The first dissociation constants for phosphoric acid
    is much greater than the second, about 100,000
    times greater
   This means nearly all the H+ ions in the solution
    comes from the first step of dissociation.
Example
 Calculate the H+, H3PO4, H2PO4-, HPO42-,
 and PO43- concentrations at equilibrium in a
 0.10 M H3PO4 solution, for which Ka1 = 7.1 x
 10-3, Ka2 = 6.3 x 10-8, and Ka3 = 4.2 x 10-13.



  H3PO4(s) + H2O(l)    H3O+(aq) + H2PO4−(aq)   Ka1= 7.25×10−3
  H2PO4−(aq)+ H2O(l)   H3O+(aq) + HPO42−(aq)   Ka2= 6.31×10−8
  HPO42−(aq)+ H2O(l)   H3O+(aq) + PO43−(aq)    Ka3= 3.98×10−13
Example




    Ka =   x2            X = .023

         0.10 - x
 [H3PO4] = 0.10 - .023 = 0.077 M

 [H3O+] = [H2PO4-] = 0.023 M
Example
Substituting what we know about the H3O+ and
 H2PO4- ion concentrations into the second
 equilibrium expression gives:




[HPO42-] = 6.3 x 10-8
Example


   Substituting what we know about the
    concentrations of the H3O+ and HPO42- ions
    into this expression gives




                      PO43- = 1.2 x 10-18
Salts

   In general, salts are ionic compounds
    composed of metallic ions and nonmetallic
    ions
   Salts dissociate in water. Salt solutions are
    generally electrolytes.
   An electrolyte is a substance that ionizes or
    dissociates into ions when it dissolves in
    water (conducts electricity)
Salt + Water

   The reaction of a salt and water to form an acid and
    base is called hydrolysis.
               NaCl + H2O  NaOH + HCl
   This is the reverse of a neutralization reaction in
    which acid and bases react to form a salt and water.

   When acids and bases react, the relative strength of
    the conjugated acid-base pair in the salt determines
    the pH of its solution.
Adding a “salt” to water

   If the salt is from a strong acid or base, then
    nothing will happen (like adding table salt to
    water – no change in pH).
   If it is a conjugate of a weak acid or base,
    then the “salt” is itself also a weak base or
    acid. So it hydrolyzes and makes some H+ or
    OH-, which changes the pH.
Adding a “salt” to water

   If the salt is from a result of a strong acid and
    base then the pH is 7, for example KNO3.
   A salt formed between a strong acid and a
    conjugate of a weak base is an acid salt, for
    example NH4Cl, and the pH will be acidic.
   A salt formed between a conjugate of a weak
    acid and a strong base is a basic salt, for
    example NaCH3COO, and the pH will be
    basic.
Acid-base properties of salt
solutions: hydrolysis




 NaCl (aq)      NH4Cl (aq)     NaClO (aq)
Hydrolysis example

Which of the following salts, when added to
  water, would produce the most acidic
  solution?
a) KBr
b) NH4NO3
c) AlCl3
d) Na2HPO4
                The answer is B
Example #1
Will an aqueous solution that is 0.20 M NH4F
 be acidic, basic or neutral?
NH4+ is the conjugate of a weak base and F- is
 the conjugate of a weak acid, so how the two
 ions compare in their ability to affect the pH
 must be determined.
NH4+ } ka = 5.6 x 10-10    F- } kb = 1.5 x 10-11

The acid is stronger than the base so the
  solution will be slightly acidic.
Example #2
What is the pH of a 0.10 M solution of NaOCl?
   HOCl, ka = 3.0x10-8
1. What type of salt is this? Na+ (from a strong
   base so does not effect the pH) and OCl-
   (from the weak acid HOCl)
2. Write the equilibrium expression for the
   dissolved salt.
      OCl- + H2O < HOCl- + OH-
Example #2 cont.
       Kb = kw/ka = 3.3 x 10-7
3.   From ICE chart kb = x2/0.10
            x= 1.8 x 10-4 M,
       pOH = 3.74, pH =10.26
Example #3
What is the pH of a 0.20 M solution of
 hydrazinium chloride, N2H5Cl? Hydrazine,
 N2H4, is a weak base with kb = 1.7 x 10-6.

Answer: conj. Acid in solution write the acid
 dissociation and use the ICE chart to
 determine the H+ and the pH.
    x = 3.4 x 10-5, pH = 4.47

								
To top