0.21 Acid ionization equilibrium  (Page 3/7)

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 Reaction (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 Reaction (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:

$\text{K}=\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[\text{N}{\text{O}}_{\text{2}}^{-}\right]}{\left[\text{HN}{\text{O}}_{\text{2}}\right]\left[{\text{H}}_{\text{2}}\text{O}\right]}$

The concentrations at equilibrium can be calculated from the data in Table 3 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, then [HA] = c ₀ -[H ₃ O ⁺ ]. Note that the contribution of [H ₂ O(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 = 55.5 M, 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:

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 10-15%. 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, K _a , defined as:

${\text{K}}_{\text{a}}=\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[\text{N}{\text{O}}_{\text{2}}^{-}\right]}{\left[\text{HN}{\text{O}}_{\text{2}}\right]}$

From an average of the data in Table 4, we can calculate that, at 25 ºC for nitrous acid, K _a = 5∙10 ^-4 . Acid ionization constants for the other weak acids in Table 2 are listed in Table 5.