# Hydrolysis

Skills to Develop

• Predict the acidity of a salt solution.
• Calculate the pH of a salt solution.
• Calculate the concentrations of various ions in a salt solution.
• Explain hydrolysis reactions.

A salt is formed between the reaction of an acid and a base. Usually, a neutral salt is formed when a strong acid and a strong base are neutralized in the reaction:

$\ce{H+ + OH- \rightleftharpoons H2O} \label{1}$

The bystander ions in an acid-base reaction form a salt solution. Most neutral salts consist of cations and anions listed in the table below. These ions have little tendency to react with water. Thus, salts consisting of these ions are neutral salts. For example: $$\ce{NaCl}$$, $$\ce{KNO3}$$, $$\ce{CaBr2}$$, $$\ce{CsClO4}$$ are neutral salts.

When weak acids and bases react, the relative strength of the conjugated acid-base pair in the salt determines the pH of its solutions. The salt, or its solution, so formed can be acidic, neutral or basic. A salt formed between a strong acid and a weak base is an acid salt, for example $$\ce{NH4Cl}$$. A salt formed between a weak acid and a strong base is a basic salt, for example $$\ce{NaCH3COO}$$. These salts are acidic or basic due to their acidic or basic ions as shown in the Table $$\PageIndex{1}$$.

Table $$\PageIndex{1}$$: Examples of Neutral, Acidic, and Basic Ions
Ions of neutral salts Acidic Ions Basic Ions
Cations Anions Cations Anions Anions
$$\ce{Na+}$$ $$\ce{K+}$$ $$\ce{Cl-}$$ $$\ce{Br-}$$ $$\ce{NH4+}$$ $$\ce{Al^3+}$$ $$\ce{HSO4-}$$ $$\ce{HPO4^2-}$$ $$\ce{F-}$$ $$\ce{C2H3O2-}$$
$$\ce{Rb+}$$ $$\ce{Cs+}$$ $$\ce{I-}$$ $$\ce{ClO4-}$$ $$\ce{Pb^2+}$$ $$\ce{Sn^2+}$$ $$\ce{H2PO4-}$$ $$\ce{PO4^3-}$$ $$\ce{NO2-}$$ $$\ce{HCO3-}$$
$$\ce{Mg^2+}$$ $$\ce{Ca^2+}$$ $$\ce{BrO4-}$$ $$\ce{ClO3-}$$ $$\ce{CN-}$$ $$\ce{CO3^2-}$$
$$\ce{Sr^2+}$$ $$\ce{Ba^2+}$$ $$\ce{NO3-}$$ $$\ce{S^2-}$$ $$\ce{SO4^2-}$$

## Hydrolysis of Acidic Salts

A salt formed between a strong acid and a weak base is an acid salt. Ammonia is a weak base, and its salt with any strong acid gives a solution with a pH lower than 7. For example, let us consider the reaction:

$\ce{HCl + NH4OH \rightleftharpoons NH4+ + Cl- + H2O} \label{2}$

In the solution, the $$\ce{NH4+}$$ ion reacts with water (called hydrolysis) according to the equation:

$\ce{NH4+ + H2O \rightleftharpoons NH3 + H3O+}. \label{3}$

The acidity constant can be derived from $$K_w$$ and $$K_b$$.

\begin{align} K_{\large\textrm a} &= \dfrac{\ce{[H3O+] [NH3]}}{\ce{[NH4+]}} \dfrac{\ce{[OH- ]}}{\ce{[OH- ]}}\\ &= \dfrac{K_{\large\textrm w}}{K_{\large\textrm b}}\\ &= \dfrac{1.00 \times 10^{-14}}{1.75 \times 10^{-5}} = 5.7 \times 10^{-10} \end{align}

Example $$\PageIndex{1}$$

What is the concentration of $$\ce{NH4+}$$, $$\ce{NH3}$$, and $$\ce{H+}$$ in a 0.100 M $$\ce{NH4NO3}$$ solution?

SOLUTION

Assume that $$\ce{[NH3]} = x$$, then $$\ce{[H3O+]} = x$$, and you write the concentration below the formula in the reaction:

$$\begin{array}{ccccccc} \ce{NH4+ &+ &H2O &\rightleftharpoons &NH3 &+ &H3O+}\\ 0.100-x &&&&x &&x \end{array}$$

\begin{align} K_{\large\textrm a} &= \textrm{5.7E-10}\\ &= \dfrac{x^2}{0.100-x} \end{align}

Since the concentration has a value much greater than Ka, you may use

\begin{align} x &= (0.100\times\textrm{5.7E(-10)})^{1/2}\\ &= \textrm{7.5E-6} \end{align}

\begin{align} \ce{[NH3]} &= \ce{[H+]} = x = \textrm{7.5E-6 M}\\ \ce{pH} &= -\log\textrm{7.5e-6} = 5.12 \end{align}

$$\ce{[NH4+]} = \textrm{0.100 M}$$

DISCUSSION

Since pH = 5.12, the contribution of $$\ce{[H+]}$$ due to self ionization of water may therefore be neglected.

## Hydrolysis of Basic Salts

A basic salt is formed between a weak acid and a strong base. The basicity is due to the hydrolysis of the conjugate base of the (weak) acid used in the neutralization reaction. For example, sodium acetate formed between the weak acetic acid and the strong base $$\ce{NaOH}$$ is a basic salt. When the salt is dissolved, ionization takes place:

$\ce{NaAc \rightleftharpoons Na+ + Ac-} \label{4}$

In the presence of water, $$\ce{Ac-}$$ undergoes hydrolysis:

$\ce{H2O + Ac- \rightleftharpoons HAc + OH-} \label{5}$

And the equilibrium constant for this reaction is Kb of the conjugate base $$\ce{Ac-}$$ of the acid $$\ce{HAc}$$. Note the following equilibrium constants: Acetic acid ($$K_a=1.75 \times 10^{-5}$$) and Ammonia ($$K_b=1.75 \times 10^{-5}$$)

\begin{align} K_{\large\textrm b} &= \ce{\dfrac{[HAc] [OH- ]}{[Ac- ]}}\\ K_{\large\textrm b} &= \ce{\dfrac{[HAc] [OH- ]}{[Ac- ]} \dfrac{[H+]}{[H+]}}\\ K_{\large\textrm b} &= \ce{\dfrac{[HAc]}{[Ac- ][H+]} [OH- ][H+]}\\ &= \dfrac{K_{\large\textrm w}}{K_{\large\textrm a}}\\ &= \dfrac{\textrm{1.00e-14}}{\textrm{1.75e-5}} = \textrm{5.7e-10} \end{align}

Thus,

$$K_{\large\ce a} K_{\large\ce b} = K_{\large\ce w}$$

or

$$\mathrm{p\mathit K_{\large a} + p\mathit K_{\large b} = 14}$$

for a conjugate acid-base pair. Let us look at a numerical problem of this type.

Example $$\PageIndex{2}$$

Calculate the $$\ce{[Na+]}$$, $$\ce{[Ac- ]}$$, $$\ce{[H+]}$$ and $$\ce{[OH- ]}$$ of a solution of 0.100 M $$\ce{NaAc}$$ (at 298 K). (Ka = 1.8E-5)

SOLUTION

Let x represent $$\ce{[H+]}$$, then

$$\begin{array}{ccccccc} \ce{H2O &+ &Ac- &\rightleftharpoons &HAc &+ &OH-}\\ &&0.100-x &&x &&x \end{array}$$

$$\dfrac{x^2}{0.100-x} = \dfrac{\textrm{1E-14}}{\textrm{1.8E-5}} = \textrm{5.6E-10}$$

Solving for x results in

\begin{align} x &= \sqrt{0.100\times\textrm{5.6E-10}}\\ &= \textrm{7.5E-6} \end{align}

$$\ce{[OH- ]} = \ce{[HAc]} = \textrm{7.5E-6}$$

$$\ce{[Na+]} = \textrm{0.100 F}$$

DISCUSSION

This corresponds to a pH of 8.9 or $$\ce{[H+]} = \textrm{1.3E-9}$$.

Note that $$\dfrac{K_{\large\ce w}}{K_{\large\ce a}} = K_{\large\ce b}$$ of $$\ce{Ac-}$$, so that Kb rather than Ka may be given as data in this question.

## Salts of Weak Acids and Weak Bases

A salt formed between a weak acid and a weak base can be neutral, acidic, or basic depending on the relative strengths of the acid and base.

• If Ka(cation) > Kb(anion) the solution of the salt is acidic.
• If Ka(cation) = Kb(anion) the solution of the salt is neutral.
• If Ka(cation) < Kb(anion) the solution of the salt is basic.

Example $$\PageIndex{3}$$

Arrange the three salts according to their acidity. $$\ce{NH4CH3COO}$$ (ammonium acetate), $$\ce{NH4CN}$$ (ammonium cyanide), and $$\ce{NH4HC2O4}$$ (ammonium oxalate).

• $$K_{\large\ce a}(\textrm{acetic acid}) = \textrm{1.85E-5}$$,
• $$K_{\large\ce a}(\textrm{hydrogen cyanide}) = \textrm{6.2E-10}$$,
• $$K_{\large\ce a}(\textrm{oxalic acid}) = \textrm{5.6E-2}$$,
• $$K_{\large\ce b}(\ce{NH3}) = \textrm{1.8E-5}$$.

SOLUTION

ammonium oxalate -- acidic, $$K_{\large\ce a}(\ce o) > K_{\large\ce b}(\ce{NH3})$$
ammonium acetate -- neutral, $$K_{\large\ce a} = K_{\large\ce b}$$
ammonium cyanide -- basic, $$K_{\large\ce a}(\ce c) < K_{\large\ce b}(\ce{NH3})$$

## Questions

1. The reaction of an acid and a base always produces a salt as the by-product, true or false? (t/f)
2. Is a solution of sodium acetate acidic, neutral or basic?
3. Are solutions of ammonium chloride acidic, basic or neutral?
4. Calculate the pH of a 0.100 M $$\ce{KCN}$$ solution.
$$K_{\large\ce a}(\ce{HCN}) = \textrm{6.2e-10}$$, $$K_{\large\ce b}(\ce{CN-}) = \textrm{1.6E-5}$$.
5. The symbol $$K_{\large\ce b}(\ce{HS-})$$ is the equilibrium constant for the reaction:
1. $$\ce{HS- + OH- \rightleftharpoons S^2- + H2O}$$
2. $$\ce{HS- + H2O \rightleftharpoons H2S + OH-}$$
3. $$\ce{HS- + H2O \rightleftharpoons H3O+ + S^2-}$$
4. $$\ce{HS- + H3O+ \rightleftharpoons H2S + H2O}$$
6. What symbol would you use for the equilibrium constant of

$$\ce{HS- \rightleftharpoons H+ + S^2-}$$

## Solutions

Consider...
Water is the real product, while the salt is formed from the spectator ions.

Consider...
Acetic acid is a weak acid that forms a salt with a strong base, $$\ce{NaOH}$$. The salt solution turns bromothymol-blue blue.

Consider...
Ammonium hydroxide does not have the same strength as a base as $$\ce{HCl}$$ has as an acid. Ammonium chloride solutions turn bromothymol-blue yellow.

Consider...

$$\begin{array}{ccccccccccc} \ce{KCN &\rightarrow &K+ &+ &CN- &&&&&&}\\ \ce{&&& &CN- &+ &H2O &\rightleftharpoons &HCN &+ &OH-}\\ &&& &(0.100-x) &&&&x &&x \end{array}$$

\begin{align} x &= (0.100\times\textrm{1.5E-5})^{1/2}\\ &= \textrm{1.2E-3}\\ \ce{pOH} &= 2.9\\ \ce{pH} &= 11.1 \end{align}

This is the ionization of $$\ce{HS-}$$; Ka for $$\ce{HS-}$$, or $$K_{\large\ce a_{\Large 2}}$$ for $$\ce{H2S}$$.