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14.3 Relative strengths of acids and bases  (Page 4/18)

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Acetic acid, CH 3 CO 2 H, is a weak acid. When we add acetic acid to water, it ionizes to a small extent according to the equation:

CH 3 CO 2 H ( a q ) + H 2 O ( l ) ⇌ H 3 O + ( a q ) + CH 3 CO 2 − ( a q ) ,

giving an equilibrium mixture with most of the acid present in the nonionized (molecular) form. This equilibrium, like other equilibria, is dynamic; acetic acid molecules donate hydrogen ions to water molecules and form hydronium ions and acetate ions at the same rate that hydronium ions donate hydrogen ions to acetate ions to reform acetic acid molecules and water molecules. We can tell by measuring the pH of an aqueous solution of known concentration that only a fraction of the weak acid is ionized at any moment ( [link] ). The remaining weak acid is present in the nonionized form.

For acetic acid, at equilibrium:

K a = [ H 3 O + ] [ CH 3 CO 2 − ] [ CH 3 CO 2 H ] = 1.8 × 10 −5
This image shows two bottles containing clear colorless solutions. Each bottle contains a single p H indicator strip. The strip in the bottle on the left is red, and a similar red strip is placed on a filter paper circle in front of the bottle on surface on which the bottles are resting. Similarly, the second bottle on the right contains and orange strip and an orange strip is placed in front of it on a filter paper circle. Between the two bottles is a pack of p Hydrion papers with a p H color scale on its cover.
pH paper indicates that a 0.l- M solution of HCl (beaker on left) has a pH of 1. The acid is fully ionized and [ H 3 O + ] = 0.1 M . A 0.1- M solution of CH 3 CO 2 H (beaker on right) is a pH of 3 ( [ H 3 O + ] = 0.001 M ) because the weak acid CH 3 CO 2 H is only partially ionized. In this solution, [ H 3 O + ] <[CH 3 CO 2 H]. (credit: modification of work by Sahar Atwa)
Ionization Constants of Some Weak Acids
Ionization Reaction K a at 25 °C
HSO 4 − + H 2 O ⇌ H 3 O + + SO 4 2− 1.2 × 10 −2
HF + H 2 O ⇌ H 3 O + + F − 3.5 × 10 −4
HNO 2 + H 2 O ⇌ H 3 O + + NO 2 − 4.6 × 10 −4
HCNO + H 2 O ⇌ H 3 O + + NCO − 2 × 10 −4
HCO 2 H + H 2 O ⇌ H 3 O + + HCO 2 − 1.8 × 10 −4
CH 3 CO 2 H + H 2 O ⇌ H 3 O + + CH 3 CO 2 − 1.8 × 10 −5
HCIO + H 2 O ⇌ H 3 O + + CIO − 2.9 × 10 −8
HBrO + H 2 O ⇌ H 3 O + + BrO − 2.8 × 10 −9
HCN + H 2 O ⇌ H 3 O + + CN − 4.9 × 10 −10

[link] gives the ionization constants for several weak acids; additional ionization constants can be found in Appendix H .

At equilibrium, a solution of a weak base in water is a mixture of the nonionized base, the conjugate acid of the weak base, and hydroxide ion with the nonionized base present in the greatest concentration. Thus, a weak base increases the hydroxide ion concentration in an aqueous solution (but not as much as the same amount of a strong base).

For example, a solution of the weak base trimethylamine, (CH 3 ) 3 N, in water reacts according to the equation:

( CH 3 ) 3 N ( a q ) + H 2 O ( l ) ⇌ ( CH 3 ) 3 NH + ( a q ) + OH − ( a q ) ,

giving an equilibrium mixture with most of the base present as the nonionized amine. This equilibrium is analogous to that described for weak acids.

We can confirm by measuring the pH of an aqueous solution of a weak base of known concentration that only a fraction of the base reacts with water ( [link] ). The remaining weak base is present as the unreacted form. The equilibrium constant for the ionization of a weak base, K b , is called the ionization constant of the weak base, and is equal to the reaction quotient when the reaction is at equilibrium. For trimethylamine, at equilibrium:

K b = [ ( CH 3 ) 3 NH + ] [ OH − ] [ ( CH 3 ) 3 N ]
This photo shows two glass containers filled with a transparent liquid. In between the containers is a p H strip indicator guide. There are p H strips placed in front of each glass container. The liquid in the container on the left appears to have a p H of 10 or 11. The liquid in the container on the right appears to have a p H of about 13 or 14.
pH paper indicates that a 0.1- M solution of NH 3 (left) is weakly basic. The solution has a pOH of 3 ([OH − ] = 0.001 M ) because the weak base NH 3 only partially reacts with water. A 0.1- M solution of NaOH (right) has a pOH of 1 because NaOH is a strong base. (credit: modification of work by Sahar Atwa)
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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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