Part 1 · Chapter 7

Ionic Equilibrium

Equilibrium in water — acids and bases, the pH scale, buffers that resist change, and the fine balance of dissolving salts

Fundamentals of Chemistry Prof. Mithun Mondal Reading time ≈ 46 min
i What you'll learn
  • The Arrhenius, Brønsted–Lowry and Lewis pictures of acids and bases, and conjugate pairs.
  • The ionic product of water, the pH scale, and the link \(\text{pH}+\text{pOH}=14\).
  • The strength constants \(K_a,\ K_b\), Ostwald's dilution law, and \(K_a\,K_b=K_w\).
  • The common ion effect and how a buffer resists pH change.
  • The Henderson–Hasselbalch equation and the pH of salt solutions.
  • The solubility product \(K_{sp}\) and how to predict precipitation.
Section 7-1

Theories of Acids & Bases

Three theories, each broader than the last, define what an acid and a base are. Arrhenius: an acid releases \(\ce{H+}\) in water, a base releases \(\ce{OH-}\). Brønsted–Lowry: an acid is a proton donor, a base a proton acceptor — freeing the idea from water. Lewis: an acid is an electron-pair acceptor, a base an electron-pair donor — the most general of all.

TheoryAcidBase
Arrheniusgives \(\ce{H+}\)gives \(\ce{OH-}\)
Brønsted–Lowryproton donorproton acceptor
Lewiselectron-pair acceptorelectron-pair donor
Why three theories? Lewis explains why \(\ce{BF3}\) and metal cations act as acids though they have no proton to give. Each theory is a wider lens — use the narrowest one that fits the problem.
Section 7-2

Conjugate Acid–Base Pairs

In the Brønsted view, every acid becomes its conjugate base after donating a proton, and every base becomes its conjugate acid after accepting one. The two differ by exactly one proton.

A conjugate pair in action
\[ \underset{\text{acid}}{\ce{HCl}} + \underset{\text{base}}{\ce{H2O}} \rightleftharpoons \underset{\text{conj. acid}}{\ce{H3O+}} + \underset{\text{conj. base}}{\ce{Cl-}} \]
A strong acid has a weak conjugate base, and vice versa — strength is reciprocal across a pair.
Section 7-3

Ionization of Water & the pH Scale

Water itself ionizes slightly — it is amphoteric, acting as both acid and base. The product of the ion concentrations is the ionic product of water, constant at a given temperature.

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Water's ionic product and pH
\(K_w=[\ce{H+}][\ce{OH-}]=10^{-14}\ \text{at }298\ \text{K};\quad \text{pH}=-\log[\ce{H+}]\)

Taking logs gives \(\text{pH}+\text{pOH}=14\). Neutral water has \([\ce{H+}]=[\ce{OH-}]=10^{-7}\), so \(\text{pH}=7\); acidic solutions have \(\text{pH}<7\), basic ones \(\text{pH}>7\).

0 acidic 7 neutral 14 basic pH
The pH scale — each unit is a tenfold change in \([\ce{H+}]\)
Section 7-4

Ka, Kb and Acid Strength

Strong acids and bases ionize almost completely; weak ones only partly, setting up an equilibrium described by an ionization constant. A larger \(K_a\) (or smaller \(\text{p}K_a\)) means a stronger acid.

Ionization constants
\[ K_a=\frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]},\qquad K_b=\frac{[\ce{BH+}][\ce{OH-}]}{[\ce{B}]},\qquad K_a\,K_b=K_w \]
For any conjugate pair the product \(K_a K_b\) equals \(K_w\) — the stronger the acid, the weaker its conjugate base.
Section 7-5

Ostwald's Dilution Law

For a weak electrolyte of concentration \(c\) and degree of ionization \(\alpha\), the ionization constant follows directly from the equilibrium expression.

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Ostwald's dilution law
\(K_a=\dfrac{c\alpha^2}{1-\alpha}\approx c\alpha^2\quad(\alpha\ll1),\qquad \alpha=\sqrt{\dfrac{K_a}{c}}\)

So \([\ce{H+}]=\sqrt{K_a\,c}\) and \(\text{pH}=\tfrac{1}{2}(\text{p}K_a-\log c)\). As a solution is diluted, \(\alpha\) rises — dilution promotes ionization.

Section 7-6

The Common Ion Effect

Adding an ion already present in an equilibrium pushes that equilibrium back — a direct consequence of Le Chatelier. Adding \(\ce{CH3COONa}\) to acetic acid floods the solution with \(\ce{CH3COO-}\), suppressing the acid's ionization and raising its pH.

The same idea, two uses. The common ion effect both makes buffers work and lowers the solubility of a sparingly soluble salt when a shared ion is added. One principle underlies the next several sections.
Section 7-7

Buffer Solutions

A buffer resists changes in pH when small amounts of acid or base are added. It is a mixture of a weak acid and its conjugate base (acidic buffer) or a weak base and its conjugate acid (basic buffer). Added \(\ce{H+}\) is mopped up by the conjugate base; added \(\ce{OH-}\) by the weak acid.

buffer HA / A⁻ add H⁺ add OH⁻ pH barely changes
A buffer neutralizes added acid or base, holding pH nearly steady
⚗️
Henderson–Hasselbalch equation
\(\text{pH}=\text{p}K_a+\log\dfrac{[\text{salt}]}{[\text{acid}]},\qquad \text{pOH}=\text{p}K_b+\log\dfrac{[\text{salt}]}{[\text{base}]}\)

A buffer works best when \([\text{salt}]=[\text{acid}]\), so \(\text{pH}=\text{p}K_a\) — choose the weak acid whose \(\text{p}K_a\) is closest to the target pH.

Section 7-8

Salt Hydrolysis

A dissolved salt can react with water, shifting the pH away from 7. Whether it does — and which way — depends on the strength of the parent acid and base.

Salt of…ExampleResulting solution
Strong acid + strong base\(\ce{NaCl}\)Neutral (no hydrolysis)
Weak acid + strong base\(\ce{CH3COONa}\)Basic \((\text{pH}>7)\)
Strong acid + weak base\(\ce{NH4Cl}\)Acidic \((\text{pH}<7)\)
Weak acid + weak base\(\ce{CH3COONH4}\)Depends on \(K_a\) vs \(K_b\)
Follow the weaker parent. The ion from the weaker parent hydrolyses and dictates the pH: a salt of a weak acid turns basic; a salt of a weak base turns acidic.
Section 7-9

The Solubility Product

A sparingly soluble salt sets up an equilibrium between solid and its dissolved ions. The constant for this is the solubility product \(K_{sp}\) — the product of ion concentrations in a saturated solution, each raised to its stoichiometric power.

Solubility product and solubility
\[ \ce{A_xB_y(s) <=> x A^{y+} + y B^{x-}}:\qquad K_{sp}=[\ce{A^{y+}}]^x[\ce{B^{x-}}]^y \]
For \(\ce{AgCl}\), \(K_{sp}=s^2\); for \(\ce{CaF2}\), \(K_{sp}=4s^3\), where \(s\) is the molar solubility.
solid saturated add common ion → solubility falls
Equilibrium between solid and ions; a common ion drives it back
Section 7-10

Predicting Precipitation

To know whether mixing two solutions yields a precipitate, compute the ionic product \(Q_{sp}\) (the same form as \(K_{sp}\) but with actual concentrations) and compare it with \(K_{sp}\).

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The precipitation test
\(Q_{sp}K_{sp}\): precipitate forms

This is the \(Q\) versus \(K\) logic of Chapter 6 applied to dissolving — the system precipitates until \(Q_{sp}\) falls back to \(K_{sp}\).

Worked Examples

Putting It to Work

1 pH of a strong acid

Problem. Find the pH of a \(0.01\ \text{M}\) solution of \(\ce{HCl}\).

Solution. A strong acid ionizes fully, so \([\ce{H+}]=0.01\):

Working
\[ \text{pH}=-\log(10^{-2})=2 \]
2 pH of a weak acid

Problem. Find the pH of \(0.1\ \text{M}\) acetic acid \((K_a=1.8\times10^{-5})\).

Solution. Use \([\ce{H+}]=\sqrt{K_a c}\):

Working
\[ [\ce{H+}]=\sqrt{1.8\times10^{-5}\times0.1}=1.34\times10^{-3}\ \Rightarrow\ \text{pH}\approx2.87 \]
3 Buffer pH

Problem. A buffer is \(0.2\ \text{M}\) in acetic acid and \(0.3\ \text{M}\) in sodium acetate \((\text{p}K_a=4.74)\). Find its pH.

Solution. Apply Henderson–Hasselbalch:

Working
\[ \text{pH}=4.74+\log\frac{0.3}{0.2}=4.74+0.18=4.92 \]
4 Hydrolysis of a salt

Problem. Will a solution of \(\ce{NH4Cl}\) be acidic, basic or neutral? Explain.

Solution. \(\ce{NH4Cl}\) is the salt of a strong acid \((\ce{HCl})\) and a weak base \((\ce{NH3})\). The \(\ce{NH4+}\) ion hydrolyses:

Working
\[ \ce{NH4+ + H2O <=> NH3 + H3O+}\ \Rightarrow\ \text{acidic, pH}<7 \]
5 Solubility from Ksp

Problem. The \(K_{sp}\) of \(\ce{AgCl}\) is \(1.8\times10^{-10}\). Find its molar solubility in pure water.

Solution. For \(\ce{AgCl <=> Ag+ + Cl-}\), \(K_{sp}=s^2\):

Working
\[ s=\sqrt{1.8\times10^{-10}}=1.34\times10^{-5}\ \text{mol L}^{-1} \]
6 Common ion effect on solubility

Problem. Find the solubility of \(\ce{AgCl}\) \((K_{sp}=1.8\times10^{-10})\) in \(0.1\ \text{M}\ \ce{NaCl}\).

Solution. Now \([\ce{Cl-}]\approx0.1\) from the salt, so \(K_{sp}=s\times0.1\):

Working
\[ s=\frac{1.8\times10^{-10}}{0.1}=1.8\times10^{-9}\ \text{mol L}^{-1} \]

The common ion has cut the solubility roughly ten-thousand-fold.

Review

Chapter Summary

✓ Theories

Arrhenius, Brønsted–Lowry (proton transfer) and Lewis (electron pairs); conjugate pairs differ by one proton.

✓ pH

\(K_w=10^{-14}\); \(\text{pH}=-\log[\ce{H+}]\); \(\text{pH}+\text{pOH}=14\).

✓ Strength

\(K_a,K_b\) with \(K_aK_b=K_w\); Ostwald: \(\alpha=\sqrt{K_a/c}\).

✓ Buffers

Common ion effect; \(\text{pH}=\text{p}K_a+\log\tfrac{[\text{salt}]}{[\text{acid}]}\).

✓ Hydrolysis

The ion of the weaker parent hydrolyses and sets the pH.

✓ Solubility

\(K_{sp}\) from saturated ions; precipitate when \(Q_{sp}>K_{sp}\).

Practice

Problems

Decide first whether the species is strong or weak, then choose the right relation. Take \(K_w=10^{-14}\) at \(298\ \text{K}\) throughout. Difficulty rises down the list.

  1. Find the pH of \(0.001\ \text{M}\ \ce{NaOH}\).
  2. Identify the conjugate base of \(\ce{H2SO4}\) and the conjugate acid of \(\ce{NH3}\).
  3. Calculate the \([\ce{OH-}]\) in a solution of pH \(4\).
  4. The degree of ionization of a \(0.1\ \text{M}\) weak acid is \(1.3\%\). Find its \(K_a\).
  5. Find the pH of \(0.2\ \text{M}\) ammonia \((K_b=1.8\times10^{-5})\).
  6. A buffer contains equal concentrations of \(\ce{NH4Cl}\) and \(\ce{NH3}\ (\text{p}K_b=4.74)\). Find its pH.
  7. State whether solutions of \(\ce{Na2CO3},\ \ce{KCl}\) and \(\ce{NH4NO3}\) are acidic, basic or neutral.
  8. The \(K_{sp}\) of \(\ce{CaF2}\) is \(3.2\times10^{-11}\). Find its molar solubility.
  9. How does the solubility of \(\ce{Mg(OH)2}\) change on adding \(\ce{NaOH}\)? Explain.
  10. Will a precipitate of \(\ce{AgCl}\) form when \(10^{-3}\ \text{M}\ \ce{Ag+}\) is mixed with an equal volume of \(10^{-3}\ \text{M}\ \ce{Cl-}\)? \((K_{sp}=1.8\times10^{-10})\)
  11. What ratio of \([\text{salt}]/[\text{acid}]\) gives a buffer of pH \(5.0\) from an acid of \(\text{p}K_a=4.74\)?
  12. A \(0.1\ \text{M}\) solution of a weak acid has pH \(3\). Find \(K_a\) and the degree of ionization.
Tip: the whole chapter is one equilibrium idea wearing different clothes. Strong species give \([\ce{H+}]\) directly; weak ones need \([\ce{H+}]=\sqrt{K_a c}\); buffers use Henderson–Hasselbalch; salts hydrolyse toward the weaker parent; and dissolving obeys \(Q_{sp}\) versus \(K_{sp}\). Identify which case you are in before reaching for a formula.