# 8.5 Hydrolysis of salt solutions  (Page 4/5)

 Page 4 / 5

## 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, to predict the pH, we must know both the K a of the weak acid and the K b of the weak base. If K a > K b , the solution is acidic, and if K b > K a , the solution is basic.

## Determining the acidic or basic nature of salts

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

(a) KBr

(b) NaHCO 3

(c) NH 4 Cl

(d) Na 2 HPO 4

(e) NH 4 F

## 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 K a of ${\text{HCO}}_{3}{}^{\text{−}}$ is 4.7 $×$ 10 −11 ,and its K b is $\frac{1.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-14}}{4.3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-7}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}2.3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-8}.$

Since K b >> K a , 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 K a of ${\text{HPO}}_{4}{}^{\text{2−}}$ is 4.2 $×$ 10 −13 ,

and its K b is $\frac{1.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-14}}{6.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-8}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}1.6\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-7}.$ Because K b >> K a , 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 K a and the K b of the two ions. K a of ${\text{NH}}_{4}{}^{\text{+}}$ is 5.6 $×$ 10 −10 , which seems very small, yet the K b of F is 1.4 $×$ 10 −11 , so the solution is acidic, since K a > K b .

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

(a) K 2 CO 3

(b) CaCl 2

(c) KH 2 PO 4

(d) (NH 4 ) 2 CO 3

(e) AlBr 3

(a) basic; (b) neutral; (c) acidic; (d) basic; (e) acidic

## The ionization of hydrated metal ions

If we measure the pH of the solutions of a variety of metal ions we will find that these ions act as weak acids when in solution. The aluminum ion is an example. When aluminum nitrate dissolves in water, the aluminum ion reacts with water to give a hydrated aluminum ion, $\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{6}{}^{3+},$ dissolved in bulk water. What this means is that the aluminum ion has the strongest interactions with the six closest water molecules (the so-called first solvation shell), even though it does interact with the other water molecules surrounding this $\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{6}{}^{3+}$ cluster as well:

$\text{Al}{\left({\text{NO}}_{3}\right)}_{3}\left(s\right)+6{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{6}{}^{3+}\left(aq\right)+3{\text{NO}}_{3}{}^{\text{−}}\left(aq\right)$

We frequently see the formula of this ion simply as “Al 3+ ( aq )”, without explicitly noting the six water molecules that are the closest ones to the aluminum ion and just describing the ion as being solvated in water (hydrated). This is similar to the simplification of the formula of the hydronium ion, H 3 O + to H + . However, in this case, the hydrated aluminum ion is a weak acid ( [link] ) and donates a proton to a water molecule. Thus, the hydration becomes important and we may use formulas that show the extent of hydration:

$\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{6}{}^{3+}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}⇌\phantom{\rule{0.2em}{0ex}}{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{5}\left(\text{O}{\text{H}\right)}^{2+}\left(aq\right)\phantom{\rule{4em}{0ex}}{K}_{\text{a}}=1.4\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}$

As with other polyprotic acids, the hydrated aluminum ion ionizes in stages, as shown by:

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