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Observation 2: percent ionization in weak acids

[link] shows that the pH of 0.1 M acid solutions varies from one weak acid to another. If we dissolve 0.1 moles of acid in a 1.0 L 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 [link] we calculate the percentage of ionized acid molecules in 0.1 M acid solutions in [link] .

Percent ionization of 0.1 m acid solutions
Acid [H 3 O + ](M) % ionization
HNO 2 6.2 × 10 –3 6.2%
HCN 7 × 10 –6 0.007%
HIO 1 × 10 –6 0.001%
HF 5.5 × 10 –3 5.5%
HOCN 5.5 × 10 –3 5.5%
HClO 2 2.8 × 10 –2 28.2%
CH 3 COOH (acetic acid) 1.3 × 10 –3 1.3%
CH 3 CH 2 COOH (propionic acid) 1.1 × 10 –3 1.1%

We might be tempted to conclude from [link] that we can characterize the strength of each acid by the percent ionization of acid moleculesin solution. However, before doing so, we observe the pH of a single acid, nitrous acid, in solution as a function of theconcentration of the acid.

H N O 2 ( aq ) + H 2 O ( l ) H 3 O + ( aq ) + N O 2 - ( aq )

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

% ionization of nitrous acid
c 0 (M) [ H 3 O + ] pH % Ionization
0.50 1.7 -2 1.8 3.3%
0.20 1.0 -2 2.0 5.1%
0.10 7.0 -3 2.2 7.0%
0.050 4.8 -3 2.3 9.7%
0.020 2.9 -3 2.5 14.7%
0.010 2.0 -3 2.7 20.0%
0.005 1.3 -3 2.9 26.7%
0.001 4.9 -4 3.3 49.1%
0.0005 3.0 -4 3.5 60.8%

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

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

K [ H 3 O + ] [ N O 2 - ] [ H N O 2 ] [ H 2 O ]

The concentrations at equilibrium can be calculated from the data in [link] for nitrous acid. [H 3 O + ] is listed and[NO 2 ]=[H 3 O + ].Furthermore, if c 0 is the initial concentration of the acid defined by the number of moles of acid dissolved in solution per liter of solution, thenHA = c 0 – [H 3 O + ].Note that the contribution of [H 2 O(l)] to the value of the function K is simply aconstant. This is because the "concentration" of water in the solution is simply the molar density of water, n H 2 O V 55.5 M , which is not affected by the presence or absence of solute. All ofthe relevant concentrations, along with the function in [link] are calculated and tabulated in [link] .

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Source:  OpenStax, Chemistry of life: bis2a modules 2.0 to 2.3 (including appendix i and ii). OpenStax CNX. Jun 15, 2015 Download for free at https://legacy.cnx.org/content/col11826/1.1
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