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By the end of this section, you will be able to:
  • Assess the relative strengths of acids and bases according to their ionization constants
  • Rationalize trends in acid–base strength in relation to molecular structure
  • Carry out equilibrium calculations for weak acid–base systems

We can rank the strengths of acids by the extent to which they ionize in aqueous solution. The reaction of an acid with water is given by the general expression:

HA ( a q ) + H 2 O ( l ) H 3 O + ( a q ) + A ( a q )

Water is the base that reacts with the acid HA, A is the conjugate base of the acid HA, and the hydronium ion is the conjugate acid of water. A strong acid yields 100% (or very nearly so) of H 3 O + and A when the acid ionizes in water; [link] lists several strong acids. A weak acid gives small amounts of H 3 O + and A .

This table has seven rows and two columns. The first row is a header row, and it labels each column, “6 Strong Acids,” and, “6 Strong Bases.” Under the “6 Strong Acids” column are the following: H C l O subscript 4 perchloric acid; H C l hydrochloric acid; H B r hydrobromic acid; H I hydroiodic acid; H N O subscript 3 nitric acid; H subscript 2 S O subscript 4 sulfuric acid. Under the “6 Strong Bases” column are the following: L i O H lithium hydroxide; N a O H sodium hydroxide; K O H potassium hydroxide; C a ( O H ) subscript 2 calcium hydroxide; S r ( O H ) subscript 2 strontium hydroxide; B a ( O H ) subscript 2 barium hydroxide.
Some of the common strong acids and bases are listed here.

The relative strengths of acids may be determined by measuring their equilibrium constants in aqueous solutions. In solutions of the same concentration, stronger acids ionize to a greater extent, and so yield higher concentrations of hydronium ions than do weaker acids. The equilibrium constant for an acid is called the acid-ionization constant, K a . For the reaction of an acid HA:

HA ( a q ) + H 2 O ( l ) H 3 O + ( a q ) + A ( a q ) ,

we write the equation for the ionization constant as:

K a = [ H 3 O + ] [ A ] [HA]

where the concentrations are those at equilibrium. Although water is a reactant in the reaction, it is the solvent as well, so we do not include [H 2 O] in the equation. The larger the K a of an acid, the larger the concentration of H 3 O + and A relative to the concentration of the nonionized acid, HA. Thus a stronger acid has a larger ionization constant than does a weaker acid. The ionization constants increase as the strengths of the acids increase. (A table of ionization constants of weak acids appears in Appendix H , with a partial listing in [link] .)

The following data on acid-ionization constants indicate the order of acid strength CH 3 CO 2 H<HNO 2 < HSO 4 :

CH 3 CO 2 H ( a q ) + H 2 O ( l ) H 3 O + ( a q ) + CH 3 CO 2 ( a q ) K a = 1.8 × 10 −5
HNO 2 ( a q ) + H 2 O ( l ) H 3 O + ( a q ) + NO 2 ( a q ) K a = 4.6 × 10 −4
HSO 4 ( a q ) + H 2 O ( a q ) H 3 O + ( a q ) + SO 4 2− ( a q ) K a = 1.2 × 10 −2

Another measure of the strength of an acid is its percent ionization. The percent ionization    of a weak acid is the ratio of the concentration of the ionized acid to the initial acid concentration, times 100:

% ionization = [ H 3 O + ] eq [ HA] 0 × 100

Because the ratio includes the initial concentration, the percent ionization for a solution of a given weak acid varies depending on the original concentration of the acid, and actually decreases with increasing acid concentration.

Calculation of percent ionization from ph

Calculate the percent ionization of a 0.125- M solution of nitrous acid (a weak acid), with a pH of 2.09.

Solution

The percent ionization for an acid is:

[ H 3 O + ] eq [ HNO 2 ] 0 × 100

The chemical equation for the dissociation of the nitrous acid is: HNO 2 ( a q ) + H 2 O ( l ) NO 2 ( a q ) + H 3 O + ( a q ) . Since 10 −pH = [ H 3 O + ] , we find that 10 −2.09 = 8.1 × 10 −3 M , so that percent ionization is:

8.1 × 10 −3 0.125 × 100 = 6.5 %

Remember, the logarithm 2.09 indicates a hydronium ion concentration with only two significant figures.

Check your learning

Calculate the percent ionization of a 0.10- M solution of acetic acid with a pH of 2.89.

Answer:

1.3% ionized

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Source:  OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9
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