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AcidBase: Polyprotic Acids

Skills to develop

Polyprotic Acids

Some Polyprotic Acids
FormulaName
H2S Hydrogen sulfide
H2SO4 Sulfuric acid
H2SO3 Sulfurous acid
H3PO4 Phosphoric acid
H2C2O4Oxalic acid
H2CO3 Carbonic acid
H2C3H2O4 Malonic acid
Some polyprotic acids are given in the table on the right here. Knowing their names and familar with their properties (ionization for example) is an asset for you.

Polyprotic acids contain more than one mole ionizable hydronium ions per mole of acids. They ionize to give more than one H+ ions per molecule. Possible forms of three polyprotic acids are given below after their dissociation into H+ ions. 

  H2S,    HS-,    S2-

  H2SO4,  HSO4-,  SO42-

  H3PO4,  H2PO4-, HPO42-,  PO43-
These acids ionize in several stages, giving out one proton at each stage. The acidity constants for these acids may be written as K1, K2, K3, ...

Consider H2S, 

     H2S  =  H+  +  HS-

             [H+] [HS-]
     K1  =  ----------
               [H2S]
and 
     HS-  =  H+  +  S2-

             [H+] [S2-]
     K2  =  -----------
               [HS-]
Obviously, for the overall ionization reaction, 
     H2S  =  2 H+  +  S2-

              [H+]2 [S2-]
     Koverall = -----------
                [H2S]

          =  K1 K2
Confirm the above obvious result on a sheet of paper to satisfy yourself.

For polyprotic acids, the following is always true:

      K1 > K2 > K3 > ... For most acids, K1/K2 = 1E5 or 100000, and K2/K3 = 1E5, but oxalic acid is different. For oxalic acid, K1 = 5.6E-2, and K2 = 5.4E-4. The two acidic groups are separated by a C-C bond in oxalic acid.

Examples 1

Calculate the overall equilibrium constant for oxalic acid H2C2O4 = 2 H+ + C2O42-.
K1 = 5.6E-2
K2 = 5.4E-5 Solution
The calculation is straight forward: Koverall = K1 K2
      = 3.0E-6

Example 2

A solution is acidified with HCl so that its pH is 1.0, and is saturated with H2S at 298 K. What is the sulfide S2- ion concentration in this solution? At 298 K, a saturated H2S solution has [H2S] = 0.10 M and
Koverall = 1E-20 for H2S

Solution


               [H+]2 [S2-]
    Koverall = ----------- 
                 [H2S]
Thus, 
                      0.1
     [S2-]  =  1E-20  -----
                     (0.1)2

              =  1E-19 F

Example 3

Sulfuric acid is a strong acid, and the pKa2 of HSO4- is 1.92. What is the pH of a 0.100 M NaHSO4 solution?

Solution
The salt is completely ionized in its solution.

NaHSO4 ® Na+ + HSO4- The anion further ionizes. If [H+] = x, then the equilibrium concentrations of various species are:   HSO4- = H+ + SO42-         Ka2 = 10-1.92 = 0.0120
0.100 - x   x,       x
Ka2 = x2 / (0.100-x) = 0.0120
[H+] = x
    = {-0.120 + (0.0122 + 4*0.00120)1/2} / 2
    = 0.0292 M
 Thus,
pH = - log 0.0292 = 1.54

Discussion
Is the NaHSO4 salt solution acidic?

Although no concentration is stated, such a solution is acidic because of the acidity of HSO4-.

For H2SO4, pKa2 = 1.92
For H3SO4, pKa1 = 2.12; pKa2 = 7.21; pKa3 = 12.67
Which of the following solutions are acidic, basic, or neutral?
Na2SO4
NaH2PO4
Na2HPO4
Na3PO4
NaNO3
 Work out the answer please, some of these will appear on the examinations. If the concentration of a salt solution is given, you may be required to evaluate the pH or pOH of the solution.

Example 4

What is the pH of a solution containing 0.500 M NaHSO4 and 0.300 M Na2SO4?

Solution
The 0.500 M solution of NaHSO4 supplies 0.500 M HSO4- as an acid, and similarly, the solution also contains 0.300 M SO42-.

Using the equation:

pH = pKa - log [salt]/[acid]
    = 1.92 + log (0.300/0.500)
    = 1.70

Confidence Building Questions

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