Saturday, October 2, 2010

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-, HPO42-, and PO43- are all present.

1) H3PO4 (aq) + H2O (l) ⇌ H3O+ (aq) + H2PO4- (aq)
Ka1 = [H3O+][H2PO4-] / [H3PO4]

2) H2PO4- (aq) + H2O (l) ⇌ H3O+ (aq) + HPO42- (aq)
Ka2 = [H3O+][HPO42-] / [H2PO4-]

3) HPO42- (aq) + H2O (l) ⇌ H3O+ (aq) + PO43- (aq)
Ka3 = [H3O+][PO43-] / [HPO42-]

Ka1, the first ionization constant, is much larger than the second one. For a triprotic acid, Ka2, the second ionization constant is much larger than the third one.

Example:

For 1.25 M of H3PO4 at 25oC, find pH, [H3PO4], [H2PO4-], [HPO42-], [PO43-], and [H3O+].

(Ka for H3PO4 = 6.9 x 10-3, Ka for H2PO4- = 6.2 x 10-8, Ka for HPO42- = 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) + HPO42- (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
[HPO42-] = x = 6.2 x 10-8
[H3O+] = 0.0929M - 6.2 x 10-8 = 0.0929M

We know that Ka2 is supposed to be much smaller than Ka1. 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:

HPO42- (aq) + H2O (l) ⇌ H3O+ (aq) + PO43- (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, [PO43-] = x = 3.20 x 10-19 M

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

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