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Percent Ionization in Weak Acids

Author: John Hutchinson

Table 1 shows that the pH of 0.1M acid solutions varies from one weak acid to another. If we dissolve 0.1 moles of acid in a 1.0L solution, the fraction of those acid molecules which will ionize varies from weak acid to weak acid. For a few weak acids, using the data in table 1 we calculate the percentage of ionized acid molecules in 0.1M acid solutions in table 2.

Table 2: Percent Ionization of 0.1M Acid Solutions
Acid [H3O+] (M) % ionization
HNO26.210-36.2%
HCN710-60.007%
HIO110-60.001%
HF8.310-38.3%
HOCN5.510-35.5%
HClO22.810-228.2%
CH3COOH (acetic acid)1.310-31.3%
CH3CH2COOH (propionic acid)1.110-31.1%

We might be tempted to conclude from table 2 that we can characterize the strength of each acid by the percent ionization of acid molecules in solution. However, before doing so, we observe the pH of a single acid, nitrous acid, in solution as a function of the concentration of the acid.

HNO2(aq) + H2O(l) H3O+(aq) + NO2-(aq) [2]

In this case, "concentration of the acid" refers to the number of moles of acid that we dissolved per liter of water. Our observations are listed in table 3, which gives [H3O+], pH, and percent ionization as a function of nitrous acid concentration.

Table 3: % Ionization of Nitrous Acid
c0 (M) [H3O+] pH % Ionization
0.501.710-21.83.3%
0.201.010-22.05.1%
0.107.010-32.27.0%
0.0504.810-32.39.7%
0.0202.910-32.514.7%
0.0102.010-32.720.0%
0.0051.310-32.926.7%
0.0014.910-43.349.1%
0.00053.010-43.560.8%

Surprisingly, perhaps, the percent ionization varies considerably as a function of the concentration of the nitrous acid. We recall that this means that the fraction of molecules which ionize, according to equation 2, depends on how many acid molecules there are per liter of solution. Since some but not all of the acid molecules are ionized, this means that nitrous acid molecules are present in solution at the same time as the negative nitrite ions and the positive hydrogen ions. Recalling our observation of equilibrium in gas phase reactions, we can conclude that equation 2 achieves equilibrium for each concentration of the nitrous acid.

Since we know that gas phase reactions come to equilibrium under conditions determined by the equilibrium constant, we might speculate that the same is true of reactions in aqueous solution, including acid ionization. We therefore define an analogy to the gas phase reaction equilibrium constant. In this case, we would not be interested in the pressures of the components, since the reactants and products are all in solution. Instead, we try a function composed of the equilibrium concentrations:

[3]

The concentrations at equilibrium can be calculated from the data in table 3 for nitrous acid. [H3O+] is listed and [NO2-]=[H3O+]. Furthermore, if c0 is the initial concentration of the acid defined by the number of moles of acid dissolved in solution per liter of solution, then [HA]=c0-[H3O+]. Note that the contribution of [H2O(l)] to the value of the function K is simply a constant. This is because the "concentration" of water in the solution is simply the molar density of water, n(H2O)/V = 55.5M, which is not affected by the presence or absence of solute. All of the relevant concentrations, along with the function in equation 3 are calculated and tabulated in table 4.

Table 4: Equilibrium Concentrations and K for Nitrous Acid
c0 (M) [H3O+] [NO2-] [HNO2] K
0.50 1.710-2 1.710-2 0.48 1.010-5
0.20 1.010-2 1.010-2 0.19 9.910-6
0.10 7.010-3 7.010-3 9.310-2 9.610-6
0.050 4.810-3 4.810-3 4.510-2 9.410-6
0.020 2.910-3 2.910-3 4.510-2 9.410-6
0.010 2.010-3 2.010-3 8.010-3 8.910-6
0.005 1.310-3 1.310-3 3.610-3 8.810-6
0.001 4.910-4 4.910-4 5.110-4 8.510-6
0.00053.010-4 3.010-4 2.010-4 8.510-6

We note that the function K in equation 3 is approximately, though only approximately, the same for all conditions analyzed in table 3. Variation of the concentration by a factor of 1000 produces a change in K of only 10% to 15%. Hence, we can regard the function K as a constant which approximately describes the acid ionization equilibrium for nitrous acid. By convention, chemists omit the constant concentration of water from the equilibrium expression, resulting in the acid ionization equilibrium constant, Ka, defined as:

Ka=[H3O+][NO2-]/[HNO2] [4]

From an average of the data in table 4, we can calculate that, at 25C for nitrous acid, Ka=510-4. Acid ionization constants for the other weak acids in table 2 are listed in table 5 (a more complete table of ionization constants can be found in the appendix).

Table 5: Weak Acid Ionization Constants
Acid Ka pKa
HNO2 510-4 3.3
HCN 4.910-10 9.3
HIO 2.310-11 10.6
HF 6.810-4 3.17
HOCN 3.510-4 3.4
HClO2 1.110-2 2.0
CH3COOH (acetic acid) 1.710-5 4.8
CH3CH2COOH (propionic acid) 1.410-5 4.9

We make two final notes about the results in table 5. First, it is clear the larger the value of Ka, the stronger the acid. That is, when Ka is a larger number, the percent ionization of the acid is larger, and vice versa. Second, the values of Ka very over many orders of magnitude. As such, it is often convenient to define the quanity pKa, analogous to pH, for purposes of comparing acid strengths:

pKa=-(logKa) [5]

The value of pKa for each acid is also listed in table 5. Note that a small value of pKa implies a large value of Ka and thus a stronger acid. Weaker acids have larger values of pKa. Ka and pKa thus give a simple quantitative comparison of the strength of weak acids.




Last Update: 2011-04-07