Chemistry » Acid-Base Equilibria » Hydrolysis of Salt Solutions

# Salts of Weak Bases and Strong Acids

## Salts of Weak Bases and Strong Acids

When we neutralize a weak base with a strong acid, the product is a salt containing the conjugate acid of the weak base. This conjugate acid is a weak acid. For example, ammonium chloride, NH4Cl, is a salt formed by the reaction of the weak base ammonia with the strong acid HCl:

$${\text{NH}}_{3}\left(aq\right)+\text{HCl}\left(aq\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{NH}}_{4}\text{Cl}\left(aq\right)$$

A solution of this salt contains ammonium ions and chloride ions. The chloride ion has no effect on the acidity of the solution since HCl is a strong acid. Chloride is a very weak base and will not accept a proton to a measurable extent. However, the ammonium ion, the conjugate acid of ammonia, reacts with water and increases the hydronium ion concentration:

$${\text{NH}}_{4}{}^{\text{+}}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}⇌\phantom{\rule{0.2em}{0ex}}{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{NH}}_{3}\left(aq\right)$$

The equilibrium equation for this reaction is simply the ionization constant. Ka, for the acid $${\text{NH}}_{4}{}^{\text{+}}:$$

$$\cfrac{\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]\left[{\text{NH}}_{3}\right]}{\left[{\text{NH}}_{4}{}^{\text{+}}\right]}\phantom{\rule{0.2em}{0ex}}={K}_{\text{a}}$$

We will not find a value of Ka for the ammonium ion in this appendix. However, it is not difficult to determine Ka for $${\text{NH}}_{4}{}^{\text{+}}$$ from the value of the ionization constant of water, Kw, and Kb, the ionization constant of its conjugate base, NH3, using the following relationship:

$${K}_{\text{w}}={K}_{\text{a}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{K}_{\text{b}}$$

This relation holds for any base and its conjugate acid or for any acid and its conjugate base.

## Example

### The pH of a Solution of a Salt of a Weak Base and a Strong Acid

Aniline is an amine that is used to manufacture dyes. It is isolated as aniline hydrochloride, $$\left[{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}{}^{\text{+}}\right]\text{Cl},$$ a salt prepared by the reaction of the weak base aniline and hydrochloric acid. What is the pH of a 0.233 M solution of aniline hydrochloride?

$${\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}{}^{\text{+}}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}⇌\phantom{\rule{0.2em}{0ex}}{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{2}\left(aq\right)$$

### Solution

The new step in this example is to determine Ka for the $${\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}{}^{\text{+}}$$ ion. The $${\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}{}^{\text{+}}$$ ion is the conjugate acid of a weak base. The value of Ka for this acid is not listed in this appendix, but we can determine it from the value of Kb for aniline, C6H5NH2, which is given as 4.3 $$×$$ 10−10 (see this table and this appendix):

$${K}_{\text{a}}\phantom{\rule{0.2em}{0ex}}\left(\text{for}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}{}^{\text{+}}\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{K}_{\text{b}}\phantom{\rule{0.2em}{0ex}}\left(\text{for}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{2}\right)={K}_{\text{w}}=1.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-14}$$

$${K}_{\text{a}}\phantom{\rule{0.2em}{0ex}}\left(\text{for}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}{}^{\text{+}}\right)\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\cfrac{{K}_{\text{w}}}{{K}_{\text{b}}\phantom{\rule{0.2em}{0ex}}\left(\text{for}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{2}\right)}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\cfrac{1.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-14}}{4.3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-10}}\phantom{\rule{0.2em}{0ex}}=2.3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}$$

Now we have the ionization constant and the initial concentration of the weak acid, the information necessary to determine the equilibrium concentration of H3O+, and the pH:

With these steps we find [H3O+] = 2.3 $$×$$ 10−3M and pH = 2.64