16.2: Polyprotic Acids

Polyprotic Acids:

-Polyprotic Acid: an acid that yields two or more H₃O⁺ per molecule.

Example:

H₃PO₄ is a triprotic acid, it can lose three hydrogens to yield three H₃O⁺ in aqueous solution. But in that solution, H₃PO₄, H₂PO₄^-, HPO₄^2-, and PO₄^3- are all present.

1) H₃PO₄ (aq) + H₂O (l) ⇌ H₃O⁺ (aq) + H₂PO₄^- (aq)
K_a1 = [H₃O⁺][H₂PO₄^-] / [H₃PO₄]

2) H₂PO₄^- (aq) + H₂O (l) ⇌ H₃O⁺ (aq) + HPO₄^2- (aq)
K_a2 = [H₃O⁺][HPO₄^2-] / [H₂PO₄^-]

3) HPO₄^2- (aq) + H₂O (l) ⇌ H₃O⁺ (aq) + PO₄^3- (aq)
K_a3 = [H₃O⁺][PO₄^3-] / [HPO₄^2-]

K_a1, the first ionization constant, is much larger than the second one. For a triprotic acid, K_a2, the second ionization constant is much larger than the third one.

Example:

For 1.25 M of H₃PO₄ at 25^oC, find pH, [H₃PO₄], [H₂PO₄^-], [HPO₄^2-], [PO₄^3-], and [H₃O⁺].

(K_a for H₃PO₄ = 6.9 x 10^-3, K_a for H₂PO₄^- = 6.2 x 10^-8, K_a for HPO₄^2- = 4.8 x 10^-13)

Since this reaction takes place in parts, lets look at the first part and see what information we get from it. The first reaction is:

H₃PO₄ (aq) + H₂O (l) ⇌ H₃O⁺ (aq) + H₂PO₄^- (aq)



(x²) / (1.25 - x) = 6.9 x 10^-3

Approximation method holds, so:

(x²) / (1.25) = 6.9 x 10^-3
x = 0.0929

According to the table, [H₃O⁺] = x = 0.0929
and pH = -log[H₃O⁺] = 1.03
[H₂PO₄^-] = x = 0.0929, but it is used in the next reaction, so we don't know the final concentration yet.

The second reaction is:

H₂PO₄^- (aq) + H₂O (l) ⇌ H₃O⁺ (aq) + HPO₄^2- (aq)




Notice that the initial concentration of H₂PO₄^- is the 0.0929M we had left over from the first reaction. And the initial concentration of H₃O⁺ is also 0.0929M, which was yielded in the first reaction also.

(0.0929 + x)(x) / (0.0929 - x) = 6.2 x 10^-8

Approximation method holds, so:

(0.0929)(x) / (0.0929) = 6.2 x 10^-8
x = 6.2 x 10^-8

According to the table, [H₂PO₄^-] = 0.0929M - 6.2 x 10^-8 = 0.0929M
[HPO₄^2-] = x = 6.2 x 10^-8
[H₃O⁺] = 0.0929M - 6.2 x 10^-8 = 0.0929M

We know that K_a2 is supposed to be much smaller than K_a1. When we subtract this very small value from 0.0929, the difference is nearly negligible, and it rounds back to 0.0929. So really, we found these values after the first reaction.

The third reaction is:

HPO₄^2- (aq) + H₂O (l) ⇌ H₃O⁺ (aq) + PO₄^3- (aq)



4.8 x 10^-13 = (0.0929 + x)(x) / (6.2 x 10^-8 - x)

Approximation method holds, so:

4.8 x 10^-13 = (0.0929)(x) / (6.2 x 10^-8)
x = 3.20 x 10^-19 M
According to the table, [PO₄^3-] = x = 3.20 x 10^-19 M

(Note that there is no K_a for strong acids, because they ionize 100% in water).