15.8

	Chemistry 8th Edition / Chang
	Student Study Guide

Chapter 15: Acids and Bases

Index | 15.1 | 15.2 – 15.3 | 15.4 | 15.5 – 15.7 | 15.8 | 15.9 | 15.10 | 15.12 |

DIPROTIC AND POLYPROTIC ACIDS (15.8)

STUDY OBJECTIVES

Write ionic equations for the various stages of ionization of diprotic and polyprotic acids, and predict the relative extent of each stage.
Calculate the concentration of the related species present in a solution of a given polyprotic acid.

Stages of Ionization. The acids listed in Table 15.3 of the text are monoprotic acids; that is, they produce one proton per acid molecule. Acetic acid (CH₃COOH) contains four hydrogen atoms per molecule, but it is still monoprotic because only one of them is ionized in solution. The H atoms bonded to the C atom do not ionize because the electronegativity difference between C and H is not great enough to produce sufficient positive charge on the hydrogen atoms.

Acids that contain two ionizable H atoms are called diprotic acids. Those with three or more ionizable H atoms are called polyprotic acids. For example, sulfuric acid (H₂SO₄) is a diprotic acid, and phosphoric acid (H₃PO₄) with its three ionizable H atoms is a polyprotic acid. The ionization reactions occur stepwise; the conjugate base in the first step becomes the acid in the second step of ionization. For oxalic acid, H₂C₂O₄:

H₂C₂O₄ H⁺ + HC₂O₄^- K_a1 = 6.5 x 10^–2

HC₂O₄^- H⁺ + C₂O₄^- K_a₂ = 6.1 x 10^–5

This example is typical in that K_a₁ >> K_a₂. It is always more difficult to remove an H⁺ ion from a negative ion such as than from a neutral molecule such as H₂C₂O₄. Table 15.5 of the text lists acid ionization constants for several polyprotic acids.

Here are some important observations concerning polyprotic acids:

1.
2.
3.

EXAMPLE Solution Containing a Diprotic Acid

Calculate the concentrations of H₂A, HA^–, A^2–, and H₃O⁺ in a 1.0 M H₂A solution.

[H₂A]	= M
[H₃O⁺]	= M
[HA^-]	= M
[A^2-]	= x 10^M

Correct!

Click a Hint button for help.

First write the equations for the stepwise ionization of H₂A, given the K_a values shown.

H₂A H⁺ + HA^– K_a₁ = 1.3 x 10^–2
HA^– H⁺ + A^2– K_a₂ = 6.3 x 10^–8

•Calculation

Since K_a1 >> K_a2, we assume that essentially all the H⁺ comes from step 1. Let x be the concentration of H₂A ionized to reach equilibrium. We solve for x in the usual way.

Concentration H₂A H⁺ + HA^–

Initial (M) 1.00 0 0

Change (M) –x x x

Equilibrium (M) (1.00 – x) x x

For this acid K_a1 is quite large, and the percent ionization will be greater than 5%. Rearranging the preceding equation into quadratic form:

ax² + bx + c = 0
x² + (1.3 x 10^–2)x – 1.3 x 10^–2 = 0

where

a = 1, b = 1.3 x 10^–2, and c = –1.3 x 10^–2.

Substituting into the quadratic formula, and solving for x.

x = 0.11 M

Therefore, at equilibrium

[H⁺] = [HA^–] = 0.11 M

and

[H₂A] = 1.00 – 0.11 = 0.89 M

which corresponds to 11% ionization.

To determine the concentration of A^2–, we must consider the second stage of ionization:

HA^– H⁺ + A^2– K_a2 = 6.3 x 10^–8

Let y be the equilibrium concentration of A^2–, and note that x >> y.

Concentration HA^– H⁺ + A^2–

Initial (M) 0.11 0.11 0

Change (M) –y +y +y

Equilibrium (M) (0.11 – y) (0.11 + y) +y

Since K_a2 is very small, as for a typical weak acid, we can assume that less than 5% of the HA^– dissociates. Therefore,

0.11 M ± y 0.11 M

With this assumption

and

y = K_a2 = 6.3 x 10^–8 M

The concentration of A^2– is equal to K_a2.

[A^2–] = y = 6.3 x 10^–8 M

Checking the validity of the assumption that

0.11 M – (6.3 x 10^–8 M) 0.11 M

The percent ionization of step 2 is

•Comment

Because K_a₁ >> K_a₂ the species produced in step 1 will be present in much greater concentration than those from step 2. This can be seen by comparing the percent ionization of each step (11% vs. 0.000057%). Also, since H₂A is a weak acid, the concentration of undissociated H₂A should be greater than that of all ionic species. Therefore, we expect the following order of concentrations:

[H₂A] > [H⁺] [HA^–] > [A^2–]

OBJECTIVE CHECK

Complete the following questions to check your understanding of the material. Select the check button to see if you answered correctly.

Write chemical equations for the three stages of ionization of phosphoric acid, H₃PO₄.

Calculate the pH of a 0.50 M solution of phosphoric acid, H₃PO₄.

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H₂C₂O₄ H⁺ + HC₂O₄^-	K_a1 = 6.5 x 10^–2
HC₂O₄^- H⁺ + C₂O₄^-	K_a₂ = 6.1 x 10^–5

H₂A H⁺ + HA^–	K_a₁ = 1.3 x 10^–2
HA^– H⁺ + A^2–	K_a₂ = 6.3 x 10^–8


Concentration	H₂A	H⁺	+	HA^–

Initial (M)	1.00	0		0
Change (M)	–x	x		x

Equilibrium (M)	(1.00 – x)	x		x