Equilibrium • Topic 2 of 3

Ionic Equilibrium: Acids & Bases

When electrolytes dissolve in water they dissociate into ions, and a dynamic equilibrium is set up between the un-ionised molecules and the ions. Strong electrolytes (HCl, NaOH, $H_2SO_4$) ionise almost completely; weak electrolytes ($CH_3COOH$, $NH_3$, $H_2CO_3$) ionise only partially, giving a true equilibrium governed by an ionisation constant.

Three theories of acids and bases

Arrhenius: an acid gives $H^+$ and a base gives $OH^-$ in water. Simple but limited to aqueous solutions.
Brønsted–Lowry: an acid is a proton donor, a base a proton acceptor. Every acid has a conjugate base (acid minus $H^+$) and every base a conjugate acid. For example $HCl/Cl^-$ and $H_2O/H_3O^+$ are conjugate pairs.
Lewis: an acid is an electron-pair acceptor (e.g. $BF_3$, $H^+$) and a base an electron-pair donor (e.g. $NH_3$, $OH^-$). The most general definition.

Ionisation constants $K_a$ and $K_b$

For a weak acid $HA\rightleftharpoons H^+ + A^-$, $K_a=\dfrac{[H^+][A^-]}{[HA]}$. For a weak base $BOH\rightleftharpoons B^+ + OH^-$, $K_b=\dfrac{[B^+][OH^-]}{[BOH]}$. A larger $K_a$ means a stronger acid. If the initial concentration is $c$ and the degree of ionisation is $\alpha$, then (Ostwald's dilution law) $K_a=\dfrac{c\alpha^2}{1-\alpha}\approx c\alpha^2$ for small $\alpha$, so $\alpha=\sqrt{K_a/c}$ — dissociation increases on dilution.

Ionic product of water and the pH scale

Water self-ionises: $2H_2O\rightleftharpoons H_3O^+ + OH^-$, with $K_w=[H^+][OH^-]=1.0\times10^{-14}$ at $298\ \text{K}$. In pure water $[H^+]=[OH^-]=10^{-7}\ \text{M}$. The pH is defined as

$$pH=-\log[H^+]$$

so $pH+pOH=14$ at $298\ \text{K}$. Acidic: $pH<7$; neutral: $pH=7$; basic: $pH>7$. For a strong acid (fully ionised) $[H^+]$ equals its concentration; for a weak acid $[H^+]=\sqrt{K_a c}$, giving $pH=\tfrac12(pK_a-\log c)$.

Salt hydrolysis

The ions of a salt can react with water and shift the pH from 7.

Strong acid + strong base (NaCl): no hydrolysis → neutral.
Weak acid + strong base ($CH_3COONa$): anion hydrolyses → basic ($pH>7$).
Strong acid + weak base ($NH_4Cl$): cation hydrolyses → acidic ($pH<7$).

Solved Examples

Worked Example

Identify the conjugate base of (i) $HNO_3$ and (ii) $H_2PO_4^-$, and the conjugate acid of $NH_3$.

Solution

Conjugate base = species after losing one $H^+$.
$HNO_3$ loses $H^+$ → $NO_3^-$.
$H_2PO_4^-$ loses $H^+$ → $HPO_4^{2-}$.
Conjugate acid = species after gaining one $H^+$: $NH_3+H^+\rightarrow NH_4^+$.

Answer: $NO_3^-$; $HPO_4^{2-}$; conjugate acid of $NH_3$ is $NH_4^+$.

Worked Example

Calculate the pH of a $0.001\ \text{M}$ HCl solution.

Solution

HCl is a strong acid, fully ionised, so $[H^+]=0.001=10^{-3}\ \text{M}$.
Apply $pH=-\log[H^+]=-\log(10^{-3})$.
Compute: $pH=3$.

Answer: $pH=3$.

Worked Example

The $K_a$ of acetic acid is $1.8\times10^{-5}$. Find $[H^+]$ and pH of a $0.1\ \text{M}$ solution.

Solution

For a weak acid $[H^+]=\sqrt{K_a\,c}=\sqrt{1.8\times10^{-5}\times 0.1}$.
$=\sqrt{1.8\times10^{-6}}=1.34\times10^{-3}\ \text{M}$.
$pH=-\log(1.34\times10^{-3})=3-\log 1.34\approx 3-0.127$.

Answer: $[H^+]\approx1.34\times10^{-3}\ \text{M}$, $pH\approx 2.87$.

Worked Example

Calculate the degree of ionisation $\alpha$ of a $0.02\ \text{M}$ weak acid with $K_a=2\times10^{-5}$.

Solution

Use Ostwald's law for small $\alpha$: $\alpha=\sqrt{K_a/c}$.
$\alpha=\sqrt{\dfrac{2\times10^{-5}}{0.02}}=\sqrt{1\times10^{-3}}$.
$\alpha=3.16\times10^{-2}$, i.e. about $3.16\%$.

Answer: $\alpha\approx0.0316$ (about $3.2\%$).

Worked Example

The pH of a solution is $9$. Find $[H^+]$ and $[OH^-]$ at $298\ \text{K}$.

Solution

$[H^+]=10^{-pH}=10^{-9}\ \text{M}$.
Use $K_w=[H^+][OH^-]=10^{-14}$.
$[OH^-]=\dfrac{10^{-14}}{10^{-9}}=10^{-5}\ \text{M}$.

Answer: $[H^+]=10^{-9}\ \text{M}$, $[OH^-]=10^{-5}\ \text{M}$ (basic).

Worked Example

Predict whether aqueous solutions of (i) $NaCl$, (ii) $CH_3COONa$ and (iii) $NH_4Cl$ are acidic, basic or neutral.

Solution

$NaCl$: strong acid + strong base → no hydrolysis.
$CH_3COONa$: salt of weak acid + strong base; acetate ion hydrolyses to give $OH^-$.
$NH_4Cl$: salt of strong acid + weak base; ammonium ion hydrolyses to give $H^+$.

Answer: (i) neutral, (ii) basic ($pH>7$), (iii) acidic ($pH<7$).

Key Points

Brønsted–Lowry: acids donate protons, bases accept them; every acid has a conjugate base differing by one $H^+$. Lewis: acid = electron-pair acceptor, base = electron-pair donor.
Weak acids/bases ionise partially: $K_a=\dfrac{[H^+][A^-]}{[HA]}$; Ostwald's law gives $\alpha=\sqrt{K_a/c}$, so ionisation rises on dilution.
Water self-ionises with $K_w=[H^+][OH^-]=10^{-14}$ at $298\ \text{K}$; $pH=-\log[H^+]$ and $pH+pOH=14$.
Strong acid: $[H^+]=c$; weak acid: $[H^+]=\sqrt{K_a c}$ so $pH=\tfrac12(pK_a-\log c)$.
Salt hydrolysis: salt of weak acid + strong base is basic; salt of strong acid + weak base is acidic; strong acid + strong base is neutral.

Quick Check — 4 MCQs

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Q1.According to Brønsted–Lowry theory, a base is a:

Explanation: A Brønsted–Lowry base accepts a proton ($H^+$).

Q2.The conjugate base of $H_2O$ is:

Explanation: Removing one $H^+$ from $H_2O$ gives $OH^-$.

Q3.The pH of $0.01\ \text{M}$ HCl is:

Explanation: $[H^+]=10^{-2}$, so $pH=-\log(10^{-2})=2$.

Q4.An aqueous solution of $NH_4Cl$ is:

Explanation: It is the salt of a strong acid and weak base; the $NH_4^+$ ion hydrolyses to give an acidic solution.

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