# Salt Hydrolysis

Acid-base neutralization reactions are defined as an acid + base reacting to give water and a salt written generally as

$\mathrm{\textcolor{Red}{HA}}(aq) + \mathrm{\textcolor{green}{BOH}}(aq) \rightleftharpoons \mathrm{H_2O}(l) + \mathrm{\textcolor{Peach}{B}\textcolor{TealBlue}{A}}(aq)$

where AB is the salt written in molecular form.

In the ionic form, the salt would be written as

$\mathrm{\textcolor{Peach}{B}\textcolor{TealBlue}{A}} (aq) \longrightarrow \textcolor{Peach}{\mathrm{B^+}}(aq) + \textcolor{TealBlue}{\mathrm{A^-}}(aq)$

## Acidic Salt

Consider the reaction of the conjugate acid, B+, with water.

$\textcolor{Peach}{\mathrm{B^+}}(aq) + \mathrm{H_2O}(l) \rightleftharpoons \mathrm{H_3O^+}(aq) + \textcolor{green}{\mathrm{B}}(aq)$ This reaction is equivalent to writing the following equation

$\textcolor{Peach}{\mathrm{HB^+}}(aq) + \mathrm{H_2O}(l) \rightleftharpoons \mathrm{H_3O^+}(aq) + \textcolor{green}{\mathrm{B}}(aq)$

The proton is now explicitly shown to illustrate the proton being donated to water. B+ and HB+ are the same (think of it as a NH4+ cation). The positive charge on B is due to the extra proton that is bound to it.

If the conjugate acid, HB+, is strong enough to react with water, it will produce H3O+ thereby lowering the pH of solution.

A salt that, when added to water, lowers the pH of the solution, is known as an acidic salt.

## Basic salt

Consider the reaction of the conjugate base, A, with water

$\textcolor{TealBlue}{\mathrm{A}^-}(aq) + \mathrm{H_2O}(l) \rightleftharpoons \mathrm{OH^-} + \textcolor{Red}{\mathrm{HA}}$

If the conjugate base, A, is strong enough to react with water, it will produce OH, thereby raising the pH of the solution.

A salt that, when added to water, raises the pH of the solution, is known as a basic salt.

## Strong acid + strong base

A reaction between a strong acid and strong base gives a neutral solution.

$\mathrm{H\textcolor{TealBlue}{Cl}}(aq) + \mathrm{\textcolor{Peach}{Na}OH}(aq) \rightleftharpoons \mathrm{H_2O}(l) + \mathrm{\textcolor{Peach}{Na}\textcolor{TealBlue}{Cl}}(aq)$

Group 1A and 2A metals do not react with water to produce hydroxide. Na+, a group 1A metal, therefore will not undergo acid ionization and no OH is produced and the pH does not change.

$\mathrm{\textcolor{Peach}{Na^+}}(aq) + \mathrm{H_2O}(l) \longrightarrow \mathrm{no~reaction}$

The anion, Cl, does not undergo base ionization as it is too weak to react with water (its Kb << 1). The chloride anion is a conjugate base to a strong acid, and therefore, will never become HCl (strong acids dissociate completely). To illustrate the reaction that will not take place, we write the following

$\mathrm{\textcolor{TealBlue}{Cl^-}}(aq) + \mathrm{H_2O}(l) \longrightarrow \mathrm{no~reaction}$

Because neither component of the salt, NaCl, react with water, NaCl is a neutral salt.

Summary

• Group 1A and 2A metals do not react with water to change the pH (neutral cations)
• Conjugate bases of strong acids do not react with water to change the pH (neutral anions)
• Neutral salts do not affect the pH of a solution

## Strong acid + weak base

A reaction between a strong acid and weak base gives a slightly acidic solution.

$\mathrm{H\textcolor{TealBlue}{Cl}}(aq) + \mathrm{NH_3}(aq) \rightleftharpoons \mathrm{\textcolor{Peach}{NH_4}\textcolor{TealBlue}{Cl}}(aq)$

The above reaction can be written out in two steps to illustrate the explicit production of both water and salt from this neutralization reaction.

\begin{align} \mathrm{NH_3}(aq) + \mathrm{H_2O}(l) \rightleftharpoons \mathrm{\textcolor{Peach}{NH_4}OH}(aq) \end{align} \begin{align} \mathrm{HCl}(aq) + \mathrm{\textcolor{Peach}{NH_4}OH}(aq) \rightleftharpoons \mathrm{H_2O}(l) + \mathrm{\textcolor{Peach}{NH_4}\textcolor{TealBlue}{Cl}}(aq) \end{align}

We have already discussed that Cl will not react with water.
However, NH4+ will since it is a conjugate acid to a weak base. The Ka of ammonium ion is less than 1. Since the following reaction will occur

$\mathrm{\textcolor{Peach}{NH_4^+}}(aq) + \mathrm{H_2O}(l) \rightleftharpoons \mathrm{H_3O^+}(aq) + \mathrm{\textcolor{green}{NH_3}}(aq)$

NH4+ will cause the pH of the solution to decrease giving a slightly acidic solution.

Summary

• Conjugate acids to weak bases will react with water to lower the pH of solution
• Salts that lower the pH of solution are called acidic salts
• Cations that lower the pH of solution are conjugates to a weak base or small, high-charge metals (such as Al3+, Fe3+, Cr3+ and Sc3+) and are collectively called acidic cations.

## Weak acid + strong base

A reaction between a weak and strong base gives a slightly basic solution.

$\mathrm{H\textcolor{TealBlue}{F}}(aq) + \mathrm{\textcolor{Peach}{K}OH}(aq) \rightleftharpoons \mathrm{H_2O}(l) + \mathrm{\textcolor{Peach}{K}\textcolor{TealBlue}{F}}(aq)$

We have already discussed that Group 1A and 2A metals do not react with water to change the pH of solution. Therefore, K+ does not affect the pH of solution. However, F can react with water as it is a conjugate base to a weak acid, thereby raising the pH of the solution.

$\mathrm{\textcolor{TealBlue}{F^-}}(aq) + \mathrm{H_2O}(l) \rightleftharpoons \mathrm{\textcolor{red}{HF}} + \mathrm{OH^-}(aq)$

Summary

• Conjugate bases to weak acids will react with water to raise the pH of solution
• Salts that raise the pH of solution are called basic salts
• Anions that are conjugates to weak acids (e.g. CN, NO2, CH3COO) are called basic anions

## Weak acid + weak base

In this scenario, one must examine the Ka and Kb values of the acid and base being mixed. Consider the following reaction

$\mathrm{\textcolor{red}{HF}}(aq) + \mathrm{\textcolor{green}{NH_3}}(aq) \rightleftharpoons \mathrm{\textcolor{Peach}{NH_4}\textcolor{TealBlue}{F}}(aq)$

Both components can react with water to change the pH of solution.

• F is a conjugate base to the weak acid, HF. F can react with water to produce OH (raise pH). * NH4+ is a conjugate acid to a weak base, NH3. NH4+ can react with water to produce H3O+ (lower pH).

Since both components can raise/lower the pH of solution, which one will win? You have to compare the Ka of the acid to the Kb of the base.

• Ka(HF) = 3.5 × 10–4; therefore, Kb(F) = 2.86 × 10–11
• Kb(NH3) = 1.76 × 10–5; therefore, Ka(NH4+) = 5.68 × 10–10

Since the Ka of NH4+ is greater than the Kb of F, the acid (NH4+) will react slightly more with water to produce slightly more H3O+ giving a slightly acidic solution!

Summary

• If Ka of conjugate acid is greater than the Kb of the conjugate base, the solution will be slight acidic * If Ka of conjugate acid is less than the Kb of the conjugate base, the solution will be slight basic

## Practice Problems

### Question 1

Ammonium chloride, NH4Cl, is dissolved in pure water. Is the resulting solution acidic, basic, or neutral?

Solution

Acidic

NH4+ is an acidic cation that is a conjugate acid to a weak base, NH3. It is able to react with water to produce H3O+.

$\mathrm{NH_4^+}(aq) + \mathrm{H_2O}(l) \rightleftharpoons \mathrm{H_3O^+}(aq) + \mathrm{NH_3}(aq)$

### Question 2

Sodium cyanide, NaCN, is dissolved in pure water. Is the resulting solution acidic, basic, or neutral?

Solution

Basic

CN is a basic anion that is a conjugate base to a weak acid, HCN. It is able to react with water to produce OH.

$\mathrm{CN^-}(aq) + \mathrm{H_2O}(l) \rightleftharpoons \mathrm{OH^-}(aq) + \mathrm{HCN}(aq)$

### Question 3

Potassium nitrate, KNO3, is dissolved in pure water. Is the resulting solution acidic, basic, or neutral?

Solution

Acidic

NO3 is an basic anion that is a conjugate base to a strong acid, HNO3 It will not react with water to produce OH.

### Question 4

What is the pH and pOH (to two decimal places) of a 50.0 mL 0.300 M KF aqueous solution (at 25 °C)? pKb(F) = 10.83

Solution

Salt dissociates and delivers a weak base (F).

$\mathrm{NaF}(s) \longrightarrow \mathrm{Na^+}(aq) + \mathrm{F^-}(aq)$ This base is a conjugate to a weak acid (HF) and can therefore react with water to produce OH.

$\mathrm{F^-}(aq) + \mathrm{H_2O}(l) \rightleftharpoons \mathrm{OH^-}(aq) + \mathrm{HF}(aq)$ Make ICE table. Proceeds right.

F(aq) + H2O(l) $$\rightleftharpoons$$ OH(aq) + HF(aq)
I 0.300 0 0
C -x +x +x
E 0.300 – x x x

\begin{align*} K_{\mathrm{b}} &= 10^{-\mathrm{p}K_{\mathrm{b}}} \\ &= 10^{-10.83} \\ &= 1.48\times 10^{-11}\\[3ex] \dfrac{[\mathrm{OH^-}][\mathrm{HF}]}{[\mathrm{F^-}]} = K_{\mathrm{b}}\\[1.5ex] \dfrac{(x)(x)}{0.300-x} &= 1.48\times 10^{-11}\\[1.5ex] \dfrac{x^2}{0.300} &= 1.48\times 10^{-11}\\[1.5ex] x &= 2.11\times 10^{-6}\\[3ex] \% x &= \dfrac{2.11\times 10^{-6}}{0.300}\times 100\% = 7\times 10^{-4}\% \\[3ex] \mathrm{pOH} &= -\log(2.11\times 10^{-6}) \\ &= 5.68 \\[3ex] \mathrm{pH} &= \mathrm{p}K_{\mathrm{w}} - \mathrm{pOH} \\ &= 14 - 5.68 \\ &= 8.32 \end{align*}

VIDEO