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Equilibrium in a Solution of a Salt of a Weak Acid and a Weak Base

Equilibrium in a Solution of a Salt of a Weak Acid and a Weak Base

In a solution of a salt formed by the reaction of a weak acid and a weak base, the salt’s cation will be a weak acid (the conjugate acid of the weak base reactant) and its anion will be a weak base (the conjugate base of the weak acid reactant). To predict the pH of the salt solution, we must know both the Ka of the acidic cation and the Kb of the basic anion. If Ka > Kb, the solution is acidic, and if Kb > Ka, the solution is basic.

Example

Determining the Acidic or Basic Nature of Salts

Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:

(a) KBr

(b) NaHCO3

(c) NH4Cl

(d) Na2HPO4

(e) NH4F

Solution

Consider each of the ions separately in terms of its effect on the pH of the solution, as shown here:

(a) The K+ cation and the Br anion are both spectators, since they are the cation of a strong base (KOH) and the anion of a strong acid (HBr), respectively. The solution is neutral.

(b) The Na+ cation is a spectator, and will not affect the pH of the solution; while the \({\text{HCO}}_{3}{}^{\text{−}}\) anion is amphiprotic. The Ka of \({\text{HCO}}_{3}{}^{\text{−}}\) is 4.7 \(×\) 10−11,and its Kb is \(\cfrac{1.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-14}}{4.3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-7}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}2.3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-8}.\)

Since Kb >> Ka, the solution is basic.

(c) The \({\text{NH}}_{4}{}^{\text{+}}\) ion is acidic and the Cl ion is a spectator. The solution will be acidic.

(d) The Na+ cation is a spectator, and will not affect the pH of the solution, while the \({\text{HPO}}_{4}{}^{\text{2−}}\) anion is amphiprotic. The Ka of \({\text{HPO}}_{4}{}^{\text{2−}}\) is 4.2 \(×\) 10−13,

and its Kb is \(\cfrac{1.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-14}}{6.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-8}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}1.6\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-7}.\) Because Kb >> Ka, the solution is basic.

(e) The \({\text{NH}}_{4}{}^{\text{+}}\) ion is listed as being acidic, and the F ion is listed as a base, so we must directly compare the Ka and the Kb of the two ions. Ka of \({\text{NH}}_{4}{}^{\text{+}}\) is 5.6 \(×\) 10−10, which seems very small, yet the Kb of F is 1.4 \(×\) 10−11, so the solution is acidic, since Ka > Kb.

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