Themes > Science > Chemistry > Inorganic Chemistry > Acids and Bases > Acids and Bases Index > Polyprotic acids


A polyprotic acid is one that has multiple ionizable protons, such as H2SO4 or H3PO4. Each proton has its own equilibrium constant Ka. For example, for a diprotic acid H2A,

  • H2A(aq) < = > H+(aq) + HA-(aq)          Ka1 = [H+][HA-]/[H2A]
  • HA-(aq) < = > H+(aq) + A-2(aq)          Ka2 = [H+][A-2]/[HA-]
In general, Ka1 >> Ka2 >> Ka3. By the law of combining reactions, you can compute the K for the total ionization of the acid. If you add the above equations, you get
H2A (aq) < = > 2H+)(aq) + A-(aq)          Ktotal = Ka1*Ka2

When computing the pH of a solution of a polyprotic acid, since Ka1 is almost always much greater than the other K's, you can treat the acid as a monoprotic acid.

There is one other useful relationship to note for a polyprotic acid. For a diprotic acid, the same amount of HA- and H+ is formed. For the second ionization

HA-(aq) < = > H+(aq) + A-2(aq)          Ka2 = [H+][A-2]/[HA-]

Since Ka2 is much smaller than Ka1, the amount of hydrogen ion generated by the second reaction is very small, and the amount of H+ is about the same as the amount generated by the first reaction. Thus, since the first proton reaction gives us [H+] = [HA-], we can rewrite the above equilibrium expression for Ka2 as

Ka2 = [A-2]
Therefore, we can easily compute the amount of [A-] in the solution. (What is the case if it is a triprotic acid?)

Example: What is the pH of a solution of 0.100 M carbonic acid, H2CO3? What is the concentration of the carbonate ion CO3-2? Ka1 = 4.4*10-7, Ka2 = 4.7*10-11

Solution: For the pH, only the first proton matters. Thus, the reaction is

  • H2CO3(aq) < = > H+(aq) + HCO3-(aq)
  • Ka1 = [H+][HCO3-]/[H2CO3]
This is solved in the usual fashion for equilibrium concentrations. If we say that the amount of H+ formed is x, then
[H2CO3] (M) [H+] (M) [HCO3-] (M)
Initial 0.100 0.00 0.00
Change -x +x +x
Equilibrium 0.100-x x x
 
Ka1 = [H+][HCO3-]/[H2CO3]
4.4*10-7 = (x)(x)/(0.100-x)      x << 0.100
4.4*10-7 = x2/0.100
x = 2.1*10-4
Thus, the hydrogen ion concentration is 2.1*10-4 M and the pH is 3.7

For the second part, use the relation between [A-2] and Ka2 for a diprotic acid. [CO3-2] = 4.7*10-11 M


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