Acid dissociation constants are used to calculate the hydrogen ion concentration
in the solution of an acid.
A. Hydrogen Ion Concentration in Solutions
of Strong Acids
Strong acids with one ionizable hydrogen are completely ionized in aqueous solution;
therefore, the hydrogen ion concentration of these solutions is equal to the
molar concentration of the acid.
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Example: What is the hydrogen ion concentration in 1.0 M HCl? Solution Hydrochloric acid is a strong acid that is completely ionized in water. HCl Therefore, in a solution prepared by adding 1.0 mol HCl to enough water to make 1.0 L solution, the concentration of H+ is 1.0 M, that of Cl- is 1.0 M and that of undissociated acid is 0. |
B. Hydrogen Ion Concentration in Solutions
of Weak Acids
The hydrogen ion concentration of an aqueous solution of a weak acid depends
on the value of its acid dissociation constant and is always less than the concentration
of the weak acid. The hydrogen ion concentration can be calculated using the
value of Ka and the molar concentration of the weak acid.
Acetic acid is a weak acid that ionizes according to the equation
HC2H3O2
H+ + C2H3O2-
Its acid dissociation constant is
| Ka = | [H+][C2H3O2-]
[HC2H3O2] |
= 1.8 X 10-5 |
The hydrogen ion concentration of a 1.0 M acetic acid solution can be calculated as follows: The solution contains 1.0 mol acetic acid in 1.0 L solution. Because acetic acid is a weak electrolyte, only a small fraction of the molecules ionize to hydrogen and acetate ions; most remain as un-ionized acetic acid molecules. Let x stand for the number of moles of acetic acid that ionize. If x moles ionize, then 1.0 - x moles of acetic acid remain un-ionized. For x moles of acetic acid that ionize, x moles of H+ and x moles of C2H3O2- are formed. The resulting concentrations of acetic acid, hydrogen ion, and acetate ion at equilibrium are
| HC2H3O2
1.0 - x |
|
H+ x |
+ |
C2H3O2- x |
Substituting these values into the expression for the acid dissociation constant gives:
| Ka = | [H+][C2H3O2-]
[HC2H3O2] |
= | (x)(x)
1.0 - x |
= 1.8 X 10-5 |
The tiny value of the acid dissociation constant suggests that the amount
of acid dissociated is very small (less than 0.01 M). Using the rules for significant
figures in addition and subtraction Section 2.2C), we know that the quantity
1.0 - 0.01 expressed to two significant figures is 1.0. If x has a value less
than 0.01, it is appropriate to disregard x in the expression 1.0 - x and change
the Ka expression to:
| Ka = | x2
1.0 |
= 1.8 X 10-5 |
Solving this equation gives:
x2 = 1.8 X 10-5 = 18 X 10-6
x = 4.2 X 10-3 = [H+] = [C2H3O2-]
These values are shown in Table 12.5.
| In calculation | Calculated value | |
|---|---|---|
| acetic acid molecules | 1.0 - x | 1.0 M |
| hydrogen ions | x | 4.2 X 10-3 M |
| acetate ions | x | 4.2 X 10-3 M |
The hydrogen ion concentration in 1.0 M acetic solution is, then, 4.2 X 10-3 M, or 0.0042 M. The number 0.0042 is not significant when subtracted from 1.0, so our simplification of the original equation was justified. If the acid is very dilute, for example 10-3 M, or if it is one with a large dissociation constant, such as 10-2, this simplification would not be valid. These calculations emphasize the difference between strong and weak acids. The hydrogen ion concentration of a 1.0 M solution of a strong acid is 1.0 M (see Example 12.7). The hydrogen ion concentration of a 1.0 M solution of a weak acid can be calculated from the acid dissociation constant of the weak acid and is much less than 1.0 M.
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Example: Calculate the hydrogen ion concentration in 0.10 M ascorbic acid, C6H8O6, a weak acid. the Ka for ascorbic acid is 8.0x10-5. Solution In 0.10 M ascorbic acid, the equilibrim equation is
The Ka for this equilibrium is
Let [H+] = x; then [C6H7O6-] also equals x and [C6H8O6] = 0.10 - x Substituting these values into the expression for the acid dissociation constant gives:
Assuming, as before that [H+] is so much less than 0.10 M as to be insignificant we rewrite the equation as
solving for x, we get: x2 = 8.0x10-6 x = 2.8x10-3 Thus, in 0.10 M ascorbic acid, [H+] = 2.8x10-3 M. |
C. Changing the Hydrogen Ion Concentration
in Solutions of Weak Acids; The Common-Ion Effect
We know that the solution of a weak acid contains the equilibrium
HA
and, as long as the equilibrium exists, everything is in balance so that
| Ka = | [H+][A-]
[HA] |
If a substance is added to the solution that changes one of the concentrations, the system goes out of equilibrium. Then either more molecules dissociate or more ions combine until concentrations again fit the equilibrium constant expression. When this set of concentrations is established, an equilibrium is again present, the opposing rates are equal, and the concentrations become constant.
It is possible to calculate the new equilibrium concentrations after a change in one concentration. Suppose we have one liter of a solution containing one mole of acetic acid and one mole of sodium acetate. The acetic acid is present as an equilibrium mixture if acetic acid molecules, hydrogen ions, and acetate ions. The sodium acetate is present only as ions (recall from Chapter 7, Section 7.5C, that salts are completely ionized in solution). The acetate ions from both the acetic acid and the sodium acetate participate in the acetic acid equilibrium:
H+ + C2H3O2- Ka
= 1.8 X 10-5
Although sodium ions are also present in solution, they do not participate in the equilibrium, they do not appear in the equation for the equilibrium, and they play no role in determining the concentrations of those substances whose formulas do appear in the equilibrium expression. To calculate the hydrogen ion concentration in this solution, we begin as in previous problems when only the weak acid was present. If x equals the concentration of ionized acetic acid molecules, the concentrations at equilibrium are:
| 1.0 - x = | the concentration of un-ionized acetic acid molecules |
| x = | the concentration of hydrogen ions |
| 1.0 + x = | the concentration of acetate ions (the concentration of sodium acetate in the solution plus the acetate ions from the ionization of acetic acid) |
In the equilibrium equation, the concentrations are:
| HC2H3O2 1.0 - x |
|
H+ x |
+ |
C2H3O2- 1.0 + x |
As in the case of a solution containing only acetic acid, we predict that the concentration of ionized acetic acid, and therefore of hydrogen ions, is very small and not significant when added to or subtracted from 1.0 M. Given this assumption, 1.0 + x is approximately equal to 1.0, and 1.0 - x is also approximately equal to 1.0. Thus, the concentrations can be expressed as
| HC2H3O2
1.0 |
|
H+ x |
+ |
C2H3O2- 1.0 |
Substituting these values into the expression for the acid dissociation constant for acetic acid and solving gives:
| Ka = | [H+][C2H3O2-]
[C2H3O2] |
= | (x)(1.0)
(1.0) |
= 1.8 X 10-5 | |
| x = | 1.8 X 10-5 | ||||
| [H+] = | 1.8 X 10-5 M | ||||
The addition of sodium acetate has decreased the hydrogen ion concentration from 4.2 X 10-3 M in 0.1 M acetic acid solution to 1.8 X 10-5 M in a 1.0 M acetic acid solution that is also 1.0 M in acetate ion. This decrease is tremendous.
These calculations have shown that an ionic equilibrium such as the ionization of a weak acid can be shifted by the addition of another ionic substance (such as salt) that contains one of the ions present in the equilibrium. This effect is known as the common-ion effect because it is caused by the addition of an ion common to both substances.
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Example: Calculate the hydrogen ion concentration of a solution that is 1.0 M in acetic acd and 0.20 M in sodium acetate. Solution If we let x equal the moles of acetic acid that ionize, the concentration at equilibrium will be:
We can drop x from the concentration of acetic acid and acetate ions because, as has been shown before, x is not significant when added to or subtracted from a number as large as 1.0 M or 0.20 M. The equilibrium concentrations become:
Substituting these values into the expression for the acid dissociation constant gives:
or 0.20x = 1.8x10-5 x = 9.0x10-5 A solution that is 1.0 M in acetic acid and 0.20 M in sodium acetate has a hydrogen ion concentration of 9.0x10-5 M. |
H+ +
Cl-