Table of Contents

Chemical Equilibrium

Ionic Equilibrium – Degree of Ionization and Dissociation

Equilibrium Constant – Characteristics and Applications

Le Chatelier’s Principle on Equilibrium

Solubility and Solubility Product

Acid and Base

pH Scale and Acidity

pH and Solutions

Hydrolysis, Salts, and Types

Buffer Solutions

Hydrolysis is a chemical reaction in which a molecule is split into two parts by the addition of a molecule of water.

Salt solutions are capable of undergoing a neutralization reaction, where the cation and anion react with each other to form their conjugate pairs. This reaction can result in an increase in the hydrogen or hydroxide ion concentrations, making the solution acidic or basic or neutral. Thus, salts are strong electrolytes which can ionize fully to form ions and can undergo hydrolysis, depending on the strength of the pairs.

Consider hydrolysis of esters

CH3COOCH3 + H2O → CH3COOH + CH3OH

Methyl acetate + Water → Acetic acid + Methanol

Consider a salt BA ionizing

(\begin{array}{l}B^{+} + A^{-} \rightleftharpoons BA\end{array})

Cation Hydrolysis

\(\begin{array}{l}B^{+} + 2H_{2}O \rightleftharpoons BOH + H_{3}O^{+}\end{array}\)

Anion Hydrolysis

(\begin{array}{l}A^{-} + H_2O \longrightarrow HA + OH^{-}\end{array})

Hydrolysis of Salt

\(\begin{array}{l}Acid + Base\rightleftharpoons Salt + Water\end{array} \)

Factors Determining the Extent of Hydrolysis

Hydrolysis Complete

If the cation/anion and water are stronger than their conjugate pairs, then B⁺ is more acidic than the conjugate hydronium ion and water is more basic than conjugate BOH; and A^- is more basic than the conjugate hydroxide ion and water is more acidic than conjugate HA.

Answer: An ion that undergoes complete hydrolysis.

* $\begin{array}{l}PH_3 + H_3O^{+} \rightarrow PH_{4}^{+} + H_2O\end{array}$ form acid solution.

NH2⁻ and H⁻ form a basic solution.

• $\begin{array}{l}H^{-} + H_2O \rightarrow H_2 + OH^{-}\end{array}$ form basic solution.

No Hydrolysis

If the cation/anion and water are weaker than their conjugate pairs, B⁺ is less acidic than the conjugate hydronium ion and conjugate BOH is more basic than water, A^- is less basic than the conjugate hydroxide ion and water is less acidic than conjugate HA.

Answer: An ion that does not undergo complete hydrolysis.

Cations of strong bases:

• Alkali ions
• Alkaline metal ions

Anions of Strong Acids:

• Chloride
• Nitrate
• Sulphate
• Phosphate
• Chlorate

Na⁺ + H₂O → NaOH + H₃O⁺

H2O + Cl- → HCl + OH-

Limited Hydrolysis

When the cations or anions are not as strong compared to their conjugate pairs, hydrolysis relative to strength will occur and the solution may be either acidic or basic.

Cations of weak bases undergoing limited hydrolysis result in an acidic solution, while anions of weak acids undergoing limited hydrolysis yield a basic solution.

NH⁴⁺, Al³⁺ CH₃COO^-, CN^-. C₂O₄^2-, PO₄^3-

(\begin{array}{l}NH_{4}^{+} + H_2O \leftrightharpoons NH_4OH + H_3O^{+}\end{array})

(\begin{array}{l}CH_3COO^{-} + H_2O \leftrightharpoons CH_3COOH + OH^{-}\end{array})

What are Salts?

Salts are a type of cryptographic primitive used to protect passwords and other sensitive data. They are randomly generated strings of characters that are appended to a user’s password before it is hashed. This makes it more difficult for attackers to guess passwords using brute-force attacks.

Acid and base react to form a salt; salts are classified into four categories.

The salt of a strong acid and strong base is NaCl.

Salt of the strong acid and weak base:

• NH4Cl
• FeCl3
• CuCl2
• AlCl3
• CuSO4

Salt of the weak acid and strong base –

CH3COONa, NaCN, NaHCO3, Na2CO3

The salt of the weak acid and weak base is:

• CH3COONH4
• (NH4)2CO3
• NH4HCO3

pH of Aqueous Solutions of Salt of a Strong Acid and Strong Base

The solution of a strong acid and base will be neutral, with a pH of 7, as anions and cations of these compounds do not undergo hydrolysis with water.

pH of Aqueous Solutions of Salt of a Strong Acid and Weak Base

Examples: NH₄Cl, FeCl₃, CuCl₂, AlCl₃, CuSO₄

Only the cation of the weak base undergoes hydrolysis, resulting in an acidic solution.

Say, h be the degree of hydrolysis.

NH4+ + Cl- ⇆ NH4Cl

NH4+ + H2O ⇆ NH4OH + H+

At the Start, Concentration (mole/L): C = 0

C$_{(1-h)}$ C$_h$ C$_h$

(\begin{array}{l}K_h = \frac{[NH_4OH][H^+]}{[NH_4^+]} \times \frac{[CH_2]}{1-h} = \frac{[CH_2]^2}{1-h} = [CH_2]^2 \end{array})

1 - h = 1

$$h = \sqrt{\frac{KH}{c}} \quad \quad [H^{+}] = c \cdot h = \sqrt{cKH}$$

For the equilibrium,

For the equilibrium

(\begin{array}{l}NH_4OH \leftrightharpoons NH_{4}^{+} + OH^{-}\ \text{Equilibrium constant of the base} = K_b = \frac{[NH_{4}^{+}][OH^{-}]}{[NH_4OH]}\end{array} )

For the equilibrium, it is essential to maintain a balance.

(\begin{array}{l}H_2O \rightleftharpoons H^+ + OH^- \ \text{Ionic product of water} = K_w = [H^+][OH^-] = 10^{-14}\end{array})

(\begin{array}{l}[H^{+}][OH^{-}] = \frac{[NH_{4}OH][H^{+}]}{[NH_{4}^{+}]} \times \frac{[NH_{4}^{+}][OH^{-}]}{[NH_{4}OH]}\end{array})

(\begin{array}{l}Kb= \frac{Kw}{Kh}\end{array})

(\begin{array}{l}[H^{+}]= ch=\sqrt{cKh} = \sqrt{c \frac {Kw}{Kb}} = \frac{1}{2}[\log Kb - \log Kw - \log c]\end{array} )

pH Aqueous Solutions of Salt of Strong Acid and Weak Base

(\begin{array}{l} pH = \frac{1}{2}\left[pK_w - pK_b - \log{c}\right] \end{array})

pH of Aqueous Solutions of Salt of Weak Acid and Strong Base

Examples:

• CH3COONa
• NaCN
• NaHCO3
• Na2CO3

(\begin{array}{l}CH_3COO^{-} + Na^{+} \rightleftharpoons CH_3COONa\end{array})

\(\begin{array}{l}CH_3COOH + OH^{-} \rightleftharpoons CH_3COO^{-} + H_2O\end{array} \)

At the Start, Concentration (mole/L) : C = 0

C(1-h)ChCh at equilibrium

(\begin{array}{l} \text{Hydrolysis constant} = Kh = \frac{[CH_3COOH][OH^{-}]}{[CH_3COO^{-}]} = \frac{[CH_3COOH]\cdot[OH^-]}{[CH_3COO^-]} = \frac{[CH_3COOH]\cdot[H^+]}{[CH_3COO^-]} = \frac{[CH_3COOH]^2}{[CH_3COO^-]\cdot[H^+]} = \frac{[CH_3COOH]^2}{[CH_3COO^-]\cdot[CH_3COOH]} = \frac{[CH_3COOH]^2}{[CH_3COO^-][CH_3COOH]} = \frac{[CH_3COOH]^2}{[CH_3COO^-]\cdot[CH_3COOH]} = \frac{[CH_3COOH]^2}{[CH_3COOH]^2} = 1 \end{array})

1 - h = 1

(\begin{array}{l}h = \sqrt{\frac{K}{c}h}; [OH^{-}] = \sqrt{cK}\end{array})

#For the Equilibrium

$$CH_3COOH \leftrightharpoons CH_3COO^- + H^+ \ \text{Equilibrium constant of acid}, Ka=\frac{[CH_3COO^-][H^+]}{CH_3COOH}$$

For the equilibrium,

For the equilibrium,

(\begin{array}{l}H_{2}O^{+} + OH^{-}\ \text{Ionic product of water} = Kw = [H^{+}][OH^{-}] = 10^{-14}\end{array})

(\frac{[CH_{3}COOH][OH^{-}]}{[CH_{3}COO^{-}]} \times \frac{[CH_{3}COO^{-}][H^{+}]}{[CH_{3}COOH]} = [H^{+}][OH^{-}])

Kh = Kw/Ka

Kw = Kh * Ka

So, $$[H^-] = ch = \sqrt{cKh} = \sqrt {c\frac{Kw}{Ka}}$$

$$\begin{array}{l}=-\log \sqrt{c\frac{K_w}{K_a}}\ = -\frac{1}{2}[\log K_w-\log K_a +\log c]\ pOH = \frac{1}{2}[pK_w – pK_a – \log c]\end{array} $$

(\begin{array}{l}pH = 14 + \frac{1}{2}[pKw - pKa - \log c] \= 14 - \frac{1}{2}[-pKw + pKa + \log c]\end{array})

(\begin{array}{l}pH = \frac{1}{2}\left[28 - pK_w + pK_a + \log c\right]\= \frac{1}{2}\left[2pK_w - pK_w + pK_a + \log c\right]\= \frac{1}{2}\left[pK_w + pK_a + \log c\right]\end{array})

The pH of an aqueous solution of salt, a weak acid, and a strong base

$$\begin{array}{l} pH = \frac{1}{2}\left[pK_w + pK_b + \log c\right] \end{array}$$

pH of Aqueous Solutions of Salt of Weak Acid and Weak Base

Examples:

• CH3COONH4
• (NH4)2CO3
• NH4HCO3

(\begin{array}{l}CH_3COO^{-} + NH_4^{+} \rightleftharpoons CH_3COO NH_4\end{array})

(\begin{array}{l}CH_3COO^{-} + NH_{4}^{+} + H_2O \rightarrow CH_3COOH + NH_4OH\end{array})

At Start, Concentration (mole/L): C_C = 0

C(1-h)² Ch²

(\begin{array}{l}\text{Hydrolysis constant} = Kh = \frac{[CH_3COOH][NH_4OH]}{[CH_3COO^{-}][NH_{4}^{+}]} = \frac{[CH_3COOH][NH_4OH]}{[CH_3COO^{-}][NH_{4}^{+}]} = \frac{h^{2}}{(1-h)^{2}} ;\end{array} )

(\begin{array}{l}\frac{h^{2}}{1-h} = Kh; h = \sqrt{Kh}\cdot \frac{1}{1-h}\end{array} )

For the equilibrium:

$$\begin{array}{l}CH_3COOH \leftrightarrow CH_3COO^{-} + H^{+}\end{array}$$

(\begin{array}{l}Ka = \frac{[H^{+}][CH_3COO^{-}]}{[CH_3COOH]}\ Ka\times \frac{[CH_3COOH]}{[CH_3COO^{-}]} = [H^{+}]\end{array})

(\begin{array}{l}[H^{+}] = Ka\sqrt{kh}\end{array})

For the equilibrium:

(\begin{array}{l}NH_4OH \leftrightarrow NH_{4}^{+} + OH^{-}\ \text{Equilibrium constant of the base} = K_b = \frac{[NH_{4}^{+}][OH^{-}]}{NH_4OH}\end{array})

For the equilibrium:

(\begin{array}{l}H_2O \rightleftharpoons H^+ + OH^- \quad \text{Ionic product of water} = Kw = [H^+][OH^-] = 10^{-14}\end{array})

(\begin{array}{l}\frac{[CH_3COOH][NH_4OH]}{[CH_3COO^{-}][NH_{4}^{+}]} \times \frac{[CH_{3}COO^{-}][H^{+}]}{[CH_{3}COOH]} \times \frac{[NH_{4}^{+}][OH^{-}]}{NH_4OH} = [H^{+}][OH^{-}]\end{array} )

(\begin{array}{l}Kh = \frac{Kw}{KaKb}\end{array})

(\begin{array}{l}[H^{+}] = \sqrt{\frac{KwKa}{Kb}}\end{array})

(\begin{array}{l}pH = \frac{1}{2}[pKw+pKa-pKb]\= \frac{1}{2}[-\log Kw-\log Ka + \log Kb]\= -\log[\sqrt{\frac{KwKa}{Kb}}]\ = -\log[H^{+}]\end{array} )

pH of Aqueous Solutions of Salt of Weak Acid and Strong Base

(\begin{array}{l} pH = \frac{1}{2}[pK_w + pK_b - pK_a] \end{array})

e) pH of Polyprotic Acids and Their Salts

1) pH of Polyprotic Acids:

Polyprotic acids are acids that can ionize to give two or more hydrogen ions.

Examples: Sulphuric, phosphoric, carbonic, oxalic acid are some.

These acids ionize in steps, but the ionization may cease after the first ionization unless it is necessary.

Due to the production of common ions in earlier steps.

Orthophosphoric acid can ionize in three steps to yield three hydrogen ions as follows.

(\begin{array}{l}H_3PO_4 \rightleftharpoons H^+ + H_2PO_{4}^{-} ;;;;K_{a_1} = \frac{[H^+][H_2PO_4]^{-}}{[H_3PO_4]}\end{array})

(\begin{array}{l}H_2PO_{4}^{-} \leftrightharpoons H^{+} + HPO_{4}^{2-} ;;;;K_{a_2} = \frac{[H^{+}][HPO_{4}^{2-}]}{[H_2PO_{4}^{-}]}\end{array})

(\begin{array}{l}H_2PO_{4}^{-} \leftrightarrow H^{+} + PO_{4}^{-};;;;K_a = \frac{[H^{+}][PO_{4}^{3-}]}{[H_2PO_{4}^{2-}]}\end{array} )

For all acids, Ka1 > Ka2 > Ka3…

So, the pH of acids is calculated based on the first ionization constant only.

For weak acids, $$K_{a_1} = \frac{[H^+][H_2PO_4^{3-}]}{[H_3PO_4]} = \frac{Ch-Ch}{C(1-h)} = \frac{ch^2}{(1-h)}, \quad h \ll 1, \quad (1-\alpha) = 1$$

(\therefore K_{a_1} = \sqrt{\frac{[H^+] \cdot Ka_1}{c}} ;;;; [H^+] ; of ; ion = cH )

and

(\begin{array}{l}pH = -\frac{1}{2}\log Ka1 - \frac{1}{2}\log C;;;;; pH = \frac{1}{2}(pKal-\log C)\end{array} )

2) pH of Salts of Polyprotic Acids: K3PO4, Na2CO3, FeCl3, (NH4)C2O4

Salts will be completely ionized

(\begin{array}{l}K_{h1} = \frac{[OH^{-}][HPO_{4}^{2-}]}{[PO_{4}^{3-}]}\end{array})

Phosphate hydrolyzes as follows:

(\begin{array}{l}PO_4^{3-} + H_2O \rightleftharpoons [HPO_4^{2-}] + [OH^{-}]\end{array})

(\begin{array}{l}K_{h2} = \frac{[OH^{-}][H_{2}PO_{4}^{-}]}{[HPO_{4}^{2-}]}\end{array})

(\begin{array}{l}\ce{[HPO4]^{2-} + H2O \rightleftharpoons H2PO4^{-} + OH^{-}}\end{array} )

(\begin{array}{l}K_{h3} = \frac{[OH^{-}][H_{3}PO_{4}]}{[H_{2}PO_{4}^{-}]}\end{array})

$\begin{array}{l}[H_{2}PO_{4}^{-}] + [H_{2}O] \rightleftharpoons [H_{3}PO_{4}]+[OH]\end{array}$

[PO43-] + H2O –> [HPO42-] + [OH–]

At equilibrium,

moles $\div$ L $\times$ $(1-h) \times$ C$_h \times$ C$_h$

(\begin{array}{l}Kh1 = \frac{Ch^{2}}{1-h}, \quad h < 1, \quad (1-h) = 1\end{array})

(\begin{array}{l}h = \sqrt{\frac{Ch^{2}}{C}} ; [H^{+}] = \frac{Kw}{OH^{[-]}} = \frac{Kw}{\sqrt{CCh^{2}}}\end{array})

(\begin{array}{l}Ka3 = \frac{[H^{+}][PO_{4}^{3-}]}{[HPO_{4}^{2-}]}\end{array} )

(\begin{array}{l}K_h \times K_a = \frac{[OH^{-}][HPO_4^{2-}]}{[PO_4^{3-}]} \times \frac{[H^{+}][PO_4^{3-}]}{[OH^{-}]} = [H^{+}][OH^{-}] = K_w\end{array})

Hence $$Kh1 = \frac{Kw}{Ka3}$$

(\begin{array}{l}[H^{+}] = \sqrt{\frac{K_wK_{a3}}{C}} \\ pH = -\frac{1}{2}(\log K_w + \log K_{a3} - \log C)\end{array})

( pH = -\frac{1}{2}(pK_w + pK_{a3} + \log C) )

The pH of Salts of Polyprotic Acids

(\begin{array}{l}K_{3PO_{4}} = -\frac{1}{2}(pK_{w} + pK_{a3} + \log C)\end{array})

Hydrolysis of Amphoteric Anion

Examples: NaHCO3, NaHS

Amphoteric anions can ionize to give hydrogen ions and hydrolyze to give hydroxide ions.

⇒ Ionization \begin{array}{l} HCO_3^- + H_2O \rightleftharpoons CO_3^{2-} + H_3O^+ \ \text{with}\ Ka1 \end{array}

⇒ Hydrolysis $$HCO_{3}^{-} + H_2O \rightleftharpoons H_2CO_3 + OH^{-} \text{with}\ Ka2$$

\(\begin{array}{l} pH_{(HCO_3^-)} = \frac{pK_a1 + pK_a2}{2} \end{array}\)