0.14 Acid-base equilibrium  (Page 5/7)

Due to the lone pair of electrons on the highly electronegative N atom, $N{H}_3$ molecules will readily attach a free hydrogen ion forming the ammonium ion $N{H}_4^{+}$ . When we measure the concentration of $O{H}^{-}$ for various initial concentration of $N{H}_3$ in water, we observe the results in [link] . We should anticipate that a base ionization equilibrium constant might exist comparable to the acidionization equilibrium constant, and in [link] , we have also calculated the value of the function $K_b$ defined as:

Given that we have dissolved a base in pure water, we might be surprised to discover the presence of positivehydrogen ions, $H_3{O}^{+}$ , in solution, but a measurement of the pH for each of the solutionsreveals small amounts. The pH for each solution is also listed in [link] . 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 ofthis, 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 comparisonof the strength of an acid to the strength of a base. To do so, we consider a class of reactions called "neutralizationreactions" which occur when we mix an acid solution with a base solution. Since the acid donates protons and the base acceptsprotons, we might expect, when mixing acid and base, to achieve a solution which is no longer acidic or basic. For example, if we mixtogether equal volumes of 0.1M $H\mathrm{Cl}\left(\mathrm{aq}\right)$ and 0.1M $\mathrm{Na}OH\left(\mathrm{aq}\right)$ , the following reaction occurs:

The resultant solution is simply a salt solution with $\mathrm{Na}\mathrm{Cl}$ dissolved in water. This solution has neither acidic nor basic properties, and the pH is 7; hence the acid and base haveneutralized each other. In this case, we have mixed together a strong acid with a strong base. Since both are strong and since wemixed 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 molarquantities of acid and base. As an example, we mix 100ml of 0.1M acetic acid( $HA$ ) solution with 100ml of 0.1M sodium hydroxide. In this discussion,we will abbreviate the acetic acid molecular formula $C{H}_{3}COOH$ as $HA$ and the acetate ion $C{H}_{3}CO{O}^{-}$ as $A^{-}$ . The reaction of $HA$ and $\mathrm{Na}OH$ is:

$A^{-}\left(\mathrm{aq}\right)$ is the acetate ion in solution, formed when an acetic acid molecule donates the positive hydrogen ion. We have thus created a saltsolution again, in this case of sodium acetate in water. Note thatthe volume of the combined solution is 200ml, so the concentration of sodium acetate( $\mathrm{Na}A$ ) in solution is 0.050M.