0.21 Acid ionization equilibrium  (Page 5/7)

NH ₃ (aq) + H ₂ O(l) → NH ₄ ⁺ (aq) + OH ^- (aq)

Due to the lone pair of electrons on the highly electronegative N atom, NH ₃ molecules will readily attach a free hydrogen ion forming the ammonium ion NH ₄ ⁺ . When we measure the concentration of OH ^- for various initial concentration of NH ₃ in water, we observe the results in Table 6. We should anticipate that a base ionization equilibrium constant might exist comparable to the acid ionization equilibrium constant, and in Table 6, we have also calculated the value of the function K _b defined as:

${\text{K}}_{\text{b}}=\frac{\left[\text{N}{\text{H}}_{\text{4}}^{+}\right]\left[\text{O}{\text{H}}^{-}\right]}{\left[\text{N}{\text{H}}_{\text{3}}\right]}$

Given that we have dissolved a base in pure water, we might be surprised to discover the presence of positive hydronium ions, H ₃ O ⁺ , in solution, but a measurement of the pH for each of the solutions reveals small amounts. The pH for each solution is also listed in Table 6. The source of these H ₃ O ⁺ ions must be the autoionization of water. Note, however, that in each case in basic solution, the concentration of H ₃ O ⁺ ions is less than that in pure water. Hence, the presence of the base in solution has suppressed the autoionization. Because of this, in each case the pH of a basic solution is greater than 7.

Base ionization is therefore quite analogous to acid ionization observed earlier. We now consider a comparison of the strength of an acid to the strength of a base. To do so, we consider a class of reactions called "neutralization reactions," which occur when we mix an acid solution with a base solution. Since the acid donates protons and the base accepts protons, we might expect, when mixing acid and base, to achieve a solution which is no longer acidic or basic. For example, if we mix together equal volumes of 0.1 M HCl(aq) and 0.1 M NaOH(aq), the following reaction occurs:

HCl(aq) + NaOH(aq) → Na ⁺ (aq) + Cl ^- (aq) + H ₂ O(l)

The resultant solution is simply a salt solution with NaCl dissolved in water. This solution has neither acidic nor basic properties, and the pH is 7; hence the acid and base have neutralized each other. In this case, we have mixed together a strong acid with a strong base. Since both are strong and since we mixed equal molar quantities of each, the neutralization reaction is essentially complete.

We next consider mixing together a weak acid solution with a strong base solution, again with equal molar quantities of acid and base. As an example, we mix 100 mL of 0.1 M acetic acid (CH ₃ COOH) solution with 100 mL of 0.1 M sodium hydroxide. In this discussion, we will abbreviate the acetic acid molecular formula CH ₃ COOH as HA and the acetate ion CH ₃ COO ^- as A ^- . The reaction of HA and NaOH is:

HA(aq) + NaOH(aq) → Na ⁺ (aq) + A ^- (aq) + H ₂ O(l)

A ^- (aq) is the acetate ion in solution, formed when an acetic acid molecule donates the positive hydrogen ion. We have thus created a salt solution again, in this case of sodium acetate in water. Note that the volume of the combined solution is 200 mL, so the concentration of sodium acetate (NaA) in solution is 0.050 M.