Equilibrium: Class 11 Chemistry · Chapter 6

Chemistry Part 1 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 18 Oct 2026 · Updated 18 Oct 2026

125 min
1 simulation
43 practice questions
35 MCQs

What you will learn in Equilibrium

Explain why physical and chemical equilibria are dynamic, and identify the quantity that stays constant in each kind of physical equilibrium
Write Kc and Kp expressions, including heterogeneous systems, relate Kp to Kc and to rewritten equations, and use K and Q to judge the extent and direction of a reaction
Calculate equilibrium compositions from initial data with ICE tables, and relate K to the standard Gibbs energy change
Predict with Le Chatelier’s principle how concentration, pressure, an inert gas, temperature and a catalyst affect an equilibrium
Classify acids and bases by the Arrhenius, Brønsted–Lowry and Lewis definitions and identify conjugate acid–base pairs
Calculate pH, pOH, Ka, Kb and degree of ionization, and the pH of salt and buffer solutions, including the common ion effect
Relate Ksp to molar solubility and predict precipitation, including the effects of a common ion and of pH

Original companion text aligned to NCERT Chemistry Class 11, Chapter 6, pp. 168–214. Not affiliated with or endorsed by NCERT.

A sealed bottle of soda on a shop shelf looks perfectly still: the liquid level and the gas pressure never change. Inside, carbon dioxide molecules leave the liquid and return to it all the time, at exactly the same rate. Nothing seems to happen because two opposite changes cancel. This chapter builds that picture of equilibrium (साम्यावस्था), first for physical changes and then for reactions.

A pair of half-arrows marks a change that runs both ways at once, as in $C O_{2} (g) \overset{}{⇌} C O_{2} (a q)$ . Once the composition stops changing, the reactants and products present together form the equilibrium mixture.

Reactions differ hugely in how far they go. Hydrogen burnt in oxygen leaves essentially no hydrogen; silver chloride shaken with water hardly dissolves; hydrogen heated with iodine vapour ends with plenty of all three gases. Where a reaction settles depends on temperature, pressure and starting amounts, which chemical plants tune to favour their product. The second half of the chapter applies the same ideas to ions in water: acids, bases, buffers and sparingly soluble salts, the subject of ionic equilibrium.

6.1 Equilibrium in Physical Processes

Changes of state are the easiest place to watch equilibrium, because nothing new is made: molecules only move between solid, liquid and gas. Three pairs matter here: $solid ⇌ liquid$ , $liquid ⇌ gas$ and $solid ⇌ gas$ .

6.1.1 Solid–Liquid Equilibrium

Put ice cubes and water together in a well-insulated flask at 0 °C (273 K) and 1 atm. Hours later the masses of ice and water, and the temperature, are exactly what they were. At the molecular scale the boundary is anything but quiet: water molecules keep locking onto the ice surface while others break free from it. The masses stay fixed only because both exchanges run at the same rate. This balance exists at just one temperature for a given pressure. For a pure substance at 1 atm that temperature is its normal melting point, which is also its normal freezing point. Two features already stand out: the opposing processes run at the same time, and they run at equal rates.

6.1.2 Liquid–Vapour Equilibrium

Picture a dry glass jar with a pressure gauge built into its lid. Place a dish of water inside, seal the jar and keep it at a steady temperature. The gauge climbs quickly at first, then more slowly, and finally stops at a fixed reading, while the water level drops a little (Fig. 6.1).

The explanation is a race between two rates. Evaporation runs at a rate set by the temperature and surface area. Condensation starts at zero and grows as vapour builds up and more molecules strike the water and stick. Once it catches up, the amount of vapour stops changing: $H_{2} O (l) \overset{}{⇌} H_{2} O (g)$ , rate of evaporation = rate of condensation.

3/4Condensation now matches evaporation: the vapour pressure is steady.

Fig. 6.1With the lid on, the vapour pressure levels off when condensation catches up with evaporation.NCERT Fig. 6.1 · p. 169

A sealed jar of water beside a graph of vapour pressure against time; orange arrows show evaporation, blue dashed arrows condensation. Step 1: only evaporation arrows, and the curve rises steeply. Step 2: more vapour and one condensation arrow; the curve bends over. Step 3: three arrows each way and a flat curve labelled steady. Step 4: the lid is lifted, vapour escapes, the water level falls and a dashed curve labelled open jar stays low.

The steady pressure is the equilibrium vapour pressure of water, often just called its vapour pressure, and it rises with temperature. Ethanol or acetone in the same jar gives a higher reading at the same temperature. A liquid with a higher vapour pressure is more volatile and boils at a lower temperature. Leave the dish open instead and the water eventually disappears. Molecules still evaporate as before, but they wander off into the room and very few come back, so condensation never catches up. How long the dish takes to dry depends on the liquid, its amount and the temperature, but no balance is ever reached: an open system cannot reach equilibrium.

Liquid and vapour also coexist at the boiling point: water and steam are in equilibrium at 100 °C under 1 atm (1.013 bar). The temperature at which a liquid and its vapour are in equilibrium at 1 atm is its normal boiling point. Lower the outside pressure and the boiling point falls: in Leh, about 3500 m up, water boils near 88 °C, which is why pressure cookers matter in the hills.

SimulationEvaporation in a closed and an open container

Water particles in a container with a temperature slider and a removable lid. With the lid on, molecules leave the liquid, bounce around and some rejoin it, so the vapour levels off. With the lid off, an escape counter keeps rising.

Fig. 6.2Press Play. With the lid on, vapour piles up until as many molecules return to the water as leave it.

Try this

Keep the lid on and watch the vapour above the water. Does the number of vapour particles keep growing for ever?
Take the lid off. What happens to the escape counter, and why can this container never reach equilibrium?
Put the lid back and raise the temperature. Is there more or less vapour once things settle?

6.1.3 Solid–Vapour Equilibrium

Some solids pass straight into vapour. Iodine crystals sealed in a glass vessel fill it with violet vapour whose colour deepens and then holds steady: from then on $I_{2} (s) \overset{}{⇌} I_{2} (g)$ , with sublimation and deposition at equal rates. Camphor tablets and naphthalene mothballs behave the same way. Ammonium chloride only seems to sublime: its vapour is ammonia and hydrogen chloride, which recombine on cooling.

6.1.4 Equilibrium Involving Dissolution of Solid or Gases in Liquids

Solids in liquids. Stir sugar into a glass of water and a point comes when no more dissolves at that temperature; extra crystals just sit at the bottom. The solution is then saturated. A thick syrup made in hot water, as for gulab jamun, grows crystals on cooling because sugar is less soluble at the lower temperature.

In a saturated solution, $sugar(s) ⇌ sugar(aq)$ is a dynamic equilibrium: sugar dissolves exactly as fast as it crystallises. Radioactive tracers show the two-way traffic. Drop a few crystals of sugar made with radioactive carbon into a saturated solution of ordinary sugar. Soon a detector finds radioactivity in the solution and in the leftover crystals, though the amount dissolved has not changed, which is only possible if both processes keep running.

Gases in liquids. Soda is bottled under a high pressure of carbon dioxide, so a lot of gas dissolves, $C O_{2} (g) \overset{}{⇌} C O_{2} (a q)$ . Henry's law governs this: at a fixed temperature, the amount of a gas that a given quantity of liquid takes up is proportional to the partial pressure of that gas above it. Opening the cap drops the $C O_{2}$ pressure to its tiny value in air, gas bubbles out until the solution matches it, and the drink goes flat, faster when warm because gases are less soluble in warm water.

Table 6.1What stays fixed in some physical equilibriaNCERT Table 6.1 · p. 171

Process	Example	Fixed at a given temperature
Solid ⇌ liquid	$H_{2} O (s) \overset{}{⇌} H_{2} O (l)$	only one temperature (the melting point) at a given pressure
Liquid ⇌ vapour	$H_{2} O (l) \overset{}{⇌} H_{2} O (g)$	vapour pressure, $p_{H_{2} O}$
Solid ⇌ solution	sugar(s) ⇌ sugar(aq)	concentration of the saturated solution (solubility)
Gas ⇌ solution	$C O_{2} (g) \overset{}{⇌} C O_{2} (a q)$	the ratio $[C O_{2} (a q)] / [C O_{2} (g)]$

6.1.5 General Characteristics of Equilibria Involving Physical Processes

Every measurable property, whether mass, pressure, colour or concentration, stops changing.
Yet nothing has stopped: the opposing processes carry on at equal rates, so the state is dynamic but stable.
This needs a closed system held at one temperature.
One quantity, fixed at a given temperature, describes each equilibrium (Table 6.1); its size shows how far the process went before balancing.

6.2 Equilibrium in Chemical Processes – Dynamic Equilibrium

Reactions settle in the same way. Start $A + B \overset{}{⇌} C + D$ with only A and B in a closed vessel. At first only the forward reaction runs, at its fastest. As C and D build up, the reverse reaction starts and speeds up while the forward one slows, until the two rates are equal (Fig. 6.3). From then on every concentration stays constant although both reactions continue: this is chemical equilibrium, and it is dynamic.

3/3The rates are equal, so the concentrations no longer change, though both reactions continue.

Fig. 6.3Concentrations stop changing exactly when the forward and reverse rates become equal.NCERT Fig. 6.2 · p. 172

Two graphs against the same time axis: concentrations of reactants (solid, falling) and products (dashed, rising), and forward (solid) and reverse (dashed) rates. Step 1 shows the first moments: high forward rate, no reverse rate. Step 2 extends the curves as the forward rate falls and the reverse rate rises. Step 3 adds a shaded equilibrium region where both concentrations are flat and the rate lines meet, labelled rates equal.

Each direction needs only its own reactants, so the same equilibrium can be reached from either side. Start with C and D alone and the reverse reaction does the work. At a given temperature, with the same total amount of each kind of atom in the same volume, both routes end at the same composition (Fig. 6.4).

Fig. 6.4From 0.10 M

N_{2} O_{4}

or 0.20 M

N O_{2}

at 350 K: the same equilibrium.NCERT Fig. 6.5 · p. 174

Two concentration–time graphs for dinitrogen tetroxide and nitrogen dioxide at 350 kelvin. Left: starting from 0.10 molar dinitrogen tetroxide only. Right: starting from 0.20 molar nitrogen dioxide only. Dashed guides show both reach the same levels, about 0.091 molar nitrogen dioxide and 0.055 molar dinitrogen tetroxide.

How do we know the reactions keep going? Isotopes give a clean test with the ammonia synthesis, $N_{2} (g) + 3 H_{2} (g) \overset{}{⇌} 2 N H_{3} (g)$ , studied by Fritz Haber. Make two identical equilibrium mixtures, one with ordinary hydrogen and one with deuterium, $D_{2}$ , and then mix them. The total amount of ammonia does not change, yet a mass spectrometer soon finds $N H_{2} D$ , $N H D_{2}$ and $H D$ (Fig. 6.5). H and D atoms can be shuffled between molecules only if ammonia keeps forming and splitting: at equilibrium both reactions are still running.

3/3A mass spectrometer finds NH₂D, NHD₂ and HD: the reactions never stopped.

Fig. 6.5Scrambling of H and D after mixing shows that ammonia keeps forming and breaking down at equilibrium.

A box divided by a wall into a hydrogen mixture (NH3, H2, N2) and a deuterium mixture (ND3, D2, N2). Step 1: the two separate mixtures. Step 2: the wall is removed and the total amount of ammonia stays the same. Step 3: the molecules are relabelled as mixed forms such as NH2D, NHD2 and HD, showing H and D scrambled.

Student activityNCERT Dynamic equilibrium – a student’s activity · p. 173

A model of dynamic equilibrium with two straws

You’ll need

two identical tall, transparent glasses
two drinking straws, one wide and one narrow
water with a drop of food colour
a ruler

Predict first

Glass 1 starts half full and glass 2 empty. You keep moving water from glass 1 to glass 2 with the wide straw and back with the narrow straw. Where will the levels settle?

Equal in both glassesHigher in glass 2Higher in glass 1

Half fill glass 1 with coloured water; leave glass 2 empty. Put the wide straw in glass 1 and the narrow one in glass 2.
Push the wide straw to the bottom of glass 1, close its top with a finger and empty it into glass 2. Then move water from glass 2 to glass 1 with the narrow straw. That is one round.
Measure both levels every 5 rounds until they stop changing, then do 10 more rounds. Repeat with two equal straws.

Your readings (wide straw in glass 1)

Rounds	Level in glass 1 (cm)	Level in glass 2 (cm)
5
10
20
30

Your entries are saved automatically.

Quick check

1 of 3

Which statement describes

N_{2} + 3 H_{2} \overset{}{⇌} 2 N H_{3}

once it has reached equilibrium?

The forward and reverse rates are equal and not zeroBoth reactions have stopped

[N_{2}] = [H_{2}] = [N H_{3}]

Ammonia still forms slowly, so its concentration creeps up

Show answer

Answer (A) The forward and reverse rates are equal and not zero

The rates are equal, so the concentrations are constant, but not equal: K fixes their ratio.

6.3 Law of Chemical Equilibrium and Equilibrium Constant

Three questions follow naturally. How are the concentrations in an equilibrium mixture related? How can we predict them from the starting amounts? And how can we push the composition the way we want? The last one matters every time a factory makes ammonia, hydrogen or quicklime. In 1864 the Norwegian chemists Cato Guldberg and Peter Waage found the answer after studying many reversible reactions: for $A + B \overset{}{⇌} C + D$ the ratio $[C] [D] / ([A] [B])$ at equilibrium has one value at a given temperature, whatever the starting amounts. Concentration was then called 'active mass', hence the name law of mass action.

The coefficients matter. Table 6.2 shows four runs of $N_{2} O_{4} (g) \overset{}{⇌} 2 N O_{2} (g)$ at 350 K, two started from $N_{2} O_{4}$ and two from $N O_{2}$ . The simple ratio $[N O_{2}] / [N_{2} O_{4}]$ drifts by a factor of almost four. Square $[N O_{2}]$ , because its coefficient is 2, and the ratio is the same every time (Fig. 6.6).

Table 6.2Four equilibrium runs of

N_{2} O_{4} \overset{}{⇌} 2 N O_{2}

at 350 K (concentrations in mol L⁻¹)NCERT Tables 6.2, 6.3 · p. 175

Run	Start $[N_{2} O_{4}]$	Start $[N O_{2}]$	Equil. $[N_{2} O_{4}]$	Equil. $[N O_{2}]$	$\frac{[N O_{2}]}{[N_{2} O_{4}]}$	$\frac{[N O_{2}]^{2}}{[N_{2} O_{4}]}$
1	0.030	0	0.0103	0.0394	3.83	0.151
2	0.080	0	0.0409	0.0783	1.91	0.150
3	0	0.20	0.0547	0.0906	1.66	0.150
4	0	0.40	0.1301	0.1397	1.07	0.150

3/3The slope of that line is the equilibrium constant, 0.15 at 350 K.

Fig. 6.6Only

[N O_{2}]^{2} / [N_{2} O_{4}]

is constant: the power is the coefficient of

N O_{2}

in the equation.

Step 1: the equilibrium nitrogen dioxide concentration plotted against dinitrogen tetroxide for four runs falls on a curve. Step 2: a second graph plots the square of the nitrogen dioxide concentration instead, and the points lie on a straight line through the origin. Step 3: a slope triangle on that line marks its gradient, the equilibrium constant, about 0.15.

LawLaw of chemical equilibrium

For a reaction at equilibrium at a fixed temperature, multiply the product concentrations together, each raised to its coefficient in the balanced equation, and divide by the reactant concentrations treated the same way. The result is always the same number, the equilibrium constant $K_{c}$ .

Equilibrium constant

K_{c} = \frac{[C]^{c} [D]^{d}}{[A]^{a} [B]^{b}}

(6.4)

where

$a A + b B ⇌ c C + d D$: the balanced equation
$[]$: equilibrium molar concentration, mol L⁻¹

The brackets are understood to hold equilibrium concentrations, so no 'eq' label is needed, and the subscript c says the constant uses molar concentrations. State symbols are usually left out of the expression. For methane burning at a high temperature, $C H_{4} (g) + 2 O_{2} (g) ⇌ C O_{2} (g) + 2 H_{2} O (g)$ , the expression is $K_{c} = [C O_{2}] [H_{2} O]^{2} / ([C H_{4}] [O_{2}]^{2})$ . Because $K_{c}$ belongs to a particular equation, rewriting the equation changes it. Reversing the equation turns the expression upside down; doubling every coefficient squares it; halving them takes its square root (Table 6.3). Always quote K together with the equation it refers to.

Table 6.3How

K_{c}

changes when the equation is rewrittenNCERT Table 6.4 · p. 177

Change to the equation	New constant	Example: $H_{2} + I_{2} \overset{}{⇌} 2 H I$ , $K_{c} = 50$
Reverse it	$1 / K_{c}$	$2 H I \overset{}{⇌} H_{2} + I_{2}$ : 0.020
Multiply every coefficient by $n$	$K_{c}^{n}$	$2 H_{2} + 2 I_{2} \overset{}{⇌} 4 H I$ : 2500
Halve every coefficient	$K_{c}^{1 / 2}$	$\frac{1}{2} H_{2} + \frac{1}{2} I_{2} ⇌ H I$ : 7.07
Add two equations	$K_{1} \times K_{2}$	see Section 6.11.5

Example 6.1NCERT Problems 6.1, 6.2 · p. 177

Kc from equilibrium concentrations

Methanol is made from 'synthesis gas', $C O (g) + 2 H_{2} (g) ⇌ C H_{3} O H (g)$ . At 500 K an equilibrium mixture has $[C O] = 0.050$ , $[H_{2}] = 0.10$ and $[C H_{3} O H] = 9.0 \times 10^{- 3} M$ . Find $K_{c}$ , and $K_{c}$ for (a) the reverse reaction and (b) $\frac{1}{2} C O + H_{2} ⇌ \frac{1}{2} C H_{3} O H$ .

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

Which is

K_{c}

for

2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)

\frac{[S O_{3}]}{[S O_{2}] [O_{2}]}

\frac{[S O_{2}]^{2} [O_{2}]}{[S O_{3}]^{2}}

\frac{2 [S O_{3}]}{2 [S O_{2}] [O_{2}]}

\frac{[S O_{3}]^{2}}{[S O_{2}]^{2} [O_{2}]}

Show answer

Answer (D) $\frac{[S O_{3}]^{2}}{[S O_{2}]^{2} [O_{2}]}$

Products over reactants, each raised to its coefficient. Multiplying by 2 instead of squaring is a common slip.

6.4 Homogeneous Equilibria

In a homogeneous equilibrium every species is in the same phase: all gases, as in the ammonia synthesis, or all in one solution. Dichromate and chromate ions in water are an example: $C r_{2} O_{7}^{2 -} (a q) + H_{2} O (l) ⇌ 2 C r O_{4}^{2 -} (a q) + 2 H^{+} (a q)$ , orange on the left, yellow on the right.

6.4.1 Equilibrium Constant in Gaseous Systems

For gases it is often easier to measure pressures than concentrations. The ideal gas equation, $p V = n R T$ , gives $p = (n / V) R T = c R T$ , so at a fixed temperature each partial pressure is proportional to the molar concentration of that gas. With $p$ in bar and $c$ in mol L⁻¹, $R = 0.0831 b a r L m o l^{- 1} K^{- 1}$ . The equilibrium law written with partial pressures defines $K_{p}$ .

DerivationHow Kp is related to Kc

Suppose $a$ moles of gas A react with $b$ moles of gas B to give $c$ moles of C and $d$ moles of D. In terms of partial pressures,

K_{p} = \frac{p_{C}^{c} p_{D}^{d}}{p_{A}^{a} p_{B}^{b}}

(6.12)

Now write each partial pressure as concentration × $R T$ , for example $p_{C} = [C] R T$ . The concentrations rebuild $K_{c}$ , while the numerator gains $(R T)^{c + d}$ and the denominator $(R T)^{a + b}$ , so the factors of $R T$ collect into one power:

Kp and Kc

K_{p} = K_{c} (R T)^{Δ n}

(6.15)

where

$Δ n$: (moles of gaseous products) − (moles of gaseous reactants) in the balanced equation

So $K_{p} = K_{c}$ only when $Δ n = 0$ (Fig. 6.7). Pressures must be in bar, the standard pressure, with $1 b a r = 10^{5} P a$ . Solids and liquids do not count in $Δ n$ .

Fig. 6.7Count gas moles:

Δ n

decides the power of

R T

Three cards, each with an equation, circles counting gas molecules before and after, and the result. Carbon monoxide plus steam: two become two, delta n zero, Kp equals Kc. Two sulphur dioxide plus oxygen: three become two, delta n minus one, Kp equals Kc times RT to the minus one. Phosphorus pentachloride: one becomes two, delta n plus one, Kp equals Kc times RT.

Example 6.4NCERT Problems 6.3, 6.4, 6.9 · p. 179

Equilibrium concentrations by solving a quadratic

For $H_{2} (g) + I_{2} (g) \overset{}{⇌} 2 H I (g)$ , $K_{c} = 50$ at 720 K. A flask starts with $[H_{2}] = 0.20$ , $[I_{2}] = 0.10 M$ and no HI. Find the equilibrium concentrations.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 2

For which reaction is

K_{p} = K_{c}

N_{2} O_{4} (g) \overset{}{⇌} 2 N O_{2} (g)

H_{2} (g) + C l_{2} (g) \overset{}{⇌} 2 H C l (g)

2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)

C a C O_{3} (s) ⇌ C a O (s) + C O_{2} (g)

Show answer

Answer (B) $H_{2} (g) + C l_{2} (g) \overset{}{⇌} 2 H C l (g)$

Only

H_{2} + C l_{2} \overset{}{⇌} 2 H C l

has

Δ n = 2 - 2 = 0

. For the limestone reaction

Δ n = 1

: solids are not counted, but the

C O_{2}

is.

6.5 Heterogeneous Equilibria

When more than one phase takes part, the equilibrium is heterogeneous: water and its vapour in a closed flask, or slaked lime in contact with its saturated solution, $C a (O H)_{2} (s) ⇌ C a^{2 +} (a q) + 2 O H^{-} (a q)$ . Pure solids and liquids simplify the expression. A pure solid has a fixed 'concentration', set by its density and molar mass: a big lump of calcium carbonate and a small crumb hold the same moles per litre of solid. Such constant terms are folded into K, so pure solids and liquids are left out; gases and dissolved species, whose concentrations can change, stay in.

Heating limestone in a closed kiln shows what this means: $C a C O_{3} (s) ⇌ C a O (s) + C O_{2} (g)$ gives

K_{c}^{'} = [C O_{2}] or K_{p} = p_{C O_{2}}

(6.18)

At a given temperature the pressure of $C O_{2}$ over the two solids has one value, however much of each solid is present (Fig. 6.8). It reaches about 1 bar near 900 °C (about 1170 K), which is why lime kilns are run hotter than that. Some solid, however little, must be present for the equilibrium to exist, even though it does not appear in K.

3/3Pump some CO₂ out: more CaCO₃ decomposes until the pressure is back to the same value.

Fig. 6.8At fixed temperature,

p_{C O_{2}}

does not depend on the amount of solid.

Step 1: a heated vessel with a little calcium carbonate and calcium oxide, some carbon dioxide and a pressure gauge. Step 2: a second vessel at the same temperature with much more solid; its gauge reads the same. Step 3: some carbon dioxide is pumped out of the second vessel; more carbonate decomposes, its heap shrinks and the gauge returns to the same reading.

Two more examples. Iron oxide reduced by hydrogen, $F e_{3} O_{4} (s) + 4 H_{2} (g) ⇌ 3 F e (s) + 4 H_{2} O (g)$ , has $K_{c} = [H_{2} O]^{4} / [H_{2}]^{4}$ . Blue copper sulphate pentahydrate can lose two of its waters, forming the trihydrate and water vapour, with $K_{p} = p_{H_{2} O}^{2}$ : the pentahydrate is stable only while the air holds enough water vapour, and in very dry air it slowly loses water.

Units of K. With concentrations in mol L⁻¹ or pressures in bar, K has a unit unless the powers cancel: none for $H_{2} + I_{2} \overset{}{⇌} 2 H I$ , but mol L⁻¹ for $K_{c}$ and bar for $K_{p}$ of $P C l_{5} \overset{}{⇌} P C l_{3} + C l_{2}$ . Thermodynamics divides each concentration by a standard 1 mol L⁻¹ and each pressure by 1 bar, which makes K a pure number; because the standard states differ, $K_{c}$ and $K_{p}$ still differ unless $Δ n = 0$ .

Example 6.6NCERT Problem 6.6 · p. 180

A heterogeneous equilibrium with a solid present

$N H_{4} H S (s) ⇌ N H_{3} (g) + H_{2} S (g)$ has $K_{p} = 0.11$ at 298 K. Some solid is put in a flask already holding ammonia at 0.20 bar. Find the equilibrium partial pressures.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 2

Which is the equilibrium constant for

C (s) + H_{2} O (g) ⇌ C O (g) + H_{2} (g)

\frac{[C O] [H_{2}]}{[C] [H_{2} O]}

[C O] [H_{2}]

\frac{[C O] [H_{2}]}{[H_{2} O]}

\frac{[H_{2} O]}{[C O] [H_{2}]}

Show answer

Answer (C) $\frac{[C O] [H_{2}]}{[H_{2} O]}$

Solid carbon is left out; steam is a gas, so it stays in the denominator.

6.6 Applications of Equilibrium Constants

Before using K, collect what we know about it:

It applies only once the concentrations have stopped changing.
It does not depend on the starting concentrations.
It has one value for each balanced equation at each temperature.
Reversing the equation inverts it; multiplying the equation by a number raises it to that power.

K has three main uses: judging how far a reaction goes, predicting which way a mixture will move, and calculating the equilibrium composition.

6.6.1 Predicting the Extent of a Reaction

Products sit in the numerator of K and reactants in the denominator, so a large K means products dominate the equilibrium mixture and a small K means reactants do. K says nothing about speed: a mixture can take years to approach an equilibrium with a huge K. Rough bands (Fig. 6.9):

$K > 10^{3}$ : products dominate and the reaction is practically complete, as in neutralisation, $H^{+} (a q) + O H^{-} (a q) ⇌ H_{2} O (l)$ , $K = 10^{14}$ at 298 K.
$K < 10^{- 3}$ : reactants dominate and very little product forms, as when silver chloride dissolves in water, $K = 1.8 \times 10^{- 10}$ .
$10^{- 3}$ to $10^{3}$ : both are present in appreciable amounts, as for $H_{2} + I_{2} \overset{}{⇌} 2 H I$ at 720 K ( $K_{c} = 50$ ) or $N_{2} O_{4} \overset{}{⇌} 2 N O_{2}$ at 350 K ( $K_{c} = 0.15$ ).

Fig. 6.9The size of K, on a logarithmic scale, shows how far a reaction goes.NCERT Fig. 6.6 · p. 182

A log K axis from minus 16 to plus 16 with three zones: reactants below minus 3, both between minus 3 and plus 3, products above plus 3. Pins mark silver chloride dissolving (about minus 9.7), acetic acid ionizing (about minus 4.7), hydrogen plus iodine at 720 kelvin (about 1.7) and hydrogen ion plus hydroxide forming water (14).

6.6.2 Predicting the Direction of the Reaction

For a mixture that may not be at equilibrium, work out the reaction quotient $Q$ . It has the same form as K but uses the concentrations ( $Q_{c}$ ) or partial pressures ( $Q_{p}$ ) as they are at that moment:

Reaction quotient

Q_{c} = \frac{[C]^{c} [D]^{d}}{[A]^{a} [B]^{b}} (any moment)

(6.20)

Comparing Q with K shows which way the mixture will move (Fig. 6.10). If $Q < K$ there is too little product, so the net reaction goes forward. If $Q > K$ there is too much, so it goes backward. If $Q = K$ the mixture is already at equilibrium.

1/3Too little product: the net reaction runs forward until Q = K.

Fig. 6.10The mixture always moves so as to bring Q towards K.NCERT Fig. 6.7 · p. 182

Boxes for reactants and products above a log-scale Q axis with K marked in the middle. Step 1: the Q pointer is left of K and a thick arrow points from reactants to products, net forward. Step 2: the pointer sits on K and opposite arrows show no net change. Step 3: the pointer is right of K and the arrow points back to reactants, net backward.

Example 6.7NCERT Problem 6.7 · p. 182

Which way will the mixture go?

For $N_{2} O_{4} (g) \overset{}{⇌} 2 N O_{2} (g)$ , $K_{c} = 4.6 \times 10^{- 3}$ at 298 K. A flask at 298 K contains $[N_{2} O_{4}] = 0.020 M$ and $[N O_{2}] = 0.015 M$ . Is it at equilibrium? If not, which way will it go, and what will you see?

Try it yourself first. Then reveal the solution one step at a time.

6.6.3 Calculating Equilibrium Concentrations

When only the starting amounts are known, an ICE table (Initial, Change, Equilibrium) keeps the bookkeeping straight:

Write the balanced equation.
Under it, list each species' initial concentration, its change in terms of one unknown $x$ (using the coefficients), and its equilibrium concentration.
Substitute into the K expression and solve for $x$ , keeping the root that leaves every concentration positive and possible.
Find each concentration, and check by putting them back into K.

Example 6.8NCERT Problems 6.5, 6.8 · p. 183

Kp and Kc from the total pressure

6.25 g of $P C l_{5}$ (208.2 g mol⁻¹) is heated to 523 K in an empty 1.00 L vessel. At equilibrium, $P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)$ , the total pressure is 2.18 bar. Find $K_{p}$ , $K_{c}$ and the fraction dissociated.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

A reaction has

K_{c} = 4.0

. A mixture of the reactants and products has

Q_{c} = 9.0

. What happens next?

Net reaction goes forwardNothing: it is at equilibriumNet reaction goes backward

K_{c}

rises to 9.0

Show answer

Answer (C) Net reaction goes backward

Q_{c} > K_{c}

: too much product, so net reaction runs backward until

Q_{c} = 4.0

6.7 Relationship between Equilibrium Constant K, Reaction Quotient Q and Gibbs Energy G

Thermodynamics (Unit 5) explains why K has its value. At constant temperature and pressure, a mixture with $Δ G < 0$ moves forward spontaneously, one with $Δ G > 0$ moves backward, and one with $Δ G = 0$ is at equilibrium, with no Gibbs energy left to drive change. $Δ G$ depends on composition through Q:

Δ G = Δ G^{\circ} + R T \ln Q

(6.21)

where

$Δ G^{\circ}$: the standard Gibbs energy change (every species in its standard state)

DerivationLinking K to the standard Gibbs energy

At equilibrium $Δ G = 0$ and $Q = K$ , so Eq. (6.21) becomes $0 = Δ G^{\circ} + R T \ln K$ :

Δ G^{\circ} = - R T \ln K

(6.22)

Taking antilogarithms:

K = e^{- Δ G^{\circ} / R T}

(6.23)

So a negative $Δ G^{\circ}$ gives $K > 1$ , and products are favoured at equilibrium; a positive $Δ G^{\circ}$ gives $K < 1$ , with only a little product. Fig. 6.11 shows why every mixture heads for the same composition: the Gibbs energy of the mixture falls from either end to a minimum, and that minimum is the equilibrium.

3/3At the minimum ΔG = 0 and Q = K: the equilibrium mixture.

Fig. 6.11Whichever side it starts from, the mixture slides downhill in G to the same minimum: equilibrium.

Gibbs energy of the mixture against composition, from pure reactants to pure products, with a minimum a little past the middle. Step 1: a ball near the reactant end with a downhill arrow labelled forward. Step 2: the ball near the product end with a downhill arrow labelled backward. Step 3: the ball rests at the minimum, labelled delta G equals zero, Q equals K.

Example 6.10NCERT Problems 6.10, 6.11 · p. 184

From ΔG° to K and back

(a) ATP hydrolysis has $Δ G^{\circ} = - 30.5 k J m o l^{- 1}$ at 310 K. Find K. (b) $N_{2} O_{4} \overset{}{⇌} 2 N O_{2}$ has $K_{p} = 0.144$ at 298 K. Find $Δ G^{\circ}$ .

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

A reaction has

Δ G^{\circ} = + 10 k J m o l^{- 1}

at 298 K. Which statement is correct?

It cannot happen at all

K > 1

: mostly products at equilibrium

K = 0

K < 1

: little product at equilibrium

Show answer

Answer (D) $K < 1$ : little product at equilibrium

K = e^{- Δ G^{\circ} / R T} \approx 0.018 < 1

: some product still forms.

6.8 Factors Affecting Equilibria

Industry wants the most product for the least energy. Ammonia plants, which supply the fertiliser industry (urea is made from ammonia), face this choice: world output is well over 150 million tonnes a year. If a mixture at equilibrium has a concentration, the pressure or the temperature changed, a net reaction runs until a new equilibrium is reached. Le Chatelier's principle predicts the direction:

PrincipleLe Chatelier’s principle

If a change is made to any of the conditions that fix an equilibrium, the system shifts in the direction that partly counteracts the change, until a new equilibrium is set up. It holds for physical and chemical equilibria alike.

Concentration and pressure changes leave K unchanged but make Q differ from K; a temperature change alters K itself. You can try every factor in Fig. 6.14.

6.8.1 Effect of Concentration Change

Add a reactant or a product, and net reaction runs in the direction that uses up the added substance.
Remove a reactant or a product, and net reaction runs in the direction that replaces it.

Add nitrogen to an ammonia mixture at equilibrium. $Q_{c} = [N H_{3}]^{2} / ([N_{2}] [H_{2}]^{3})$ falls below $K_{c}$ , so the forward reaction runs, using up some $N_{2}$ and $H_{2}$ and making more $N H_{3}$ (Fig. 6.12). At the new equilibrium $[N_{2}]$ is lower than just after the addition but still higher than it was originally: the change is only partly undone.

4/4N₂ settles above its old level but below the peak; there is now more NH₃.

Fig. 6.12Adding

N_{2}

makes

Q < K

; the forward reaction then uses up part of the added nitrogen.NCERT Fig. 6.8 · p. 185

Concentrations of nitrogen (solid), hydrogen (dashed) and ammonia (dotted) against time. Step 1: all three are flat. Step 2: nitrogen jumps up when it is added. Step 3: nitrogen and hydrogen curve down and ammonia curves up. Step 4: a guide marks the old nitrogen level; the new level is above it but below the peak, and there is more ammonia.

Removing a product works just as well and is used on a large scale. Ammonia is liquefied and drawn off from the reactor gases, and in a lime kiln the $C O_{2}$ is swept away so that $C a C O_{3} \overset{}{\to} C a O + C O_{2}$ goes to completion. Continuous removal keeps Q below K, so the forward reaction never stops.

ExperimentNCERT Effect of concentration – an experiment · p. 186

Pushing the iron(III)–thiocyanate equilibrium

You’ll need

dilute iron(III) chloride or nitrate
dilute potassium thiocyanate
dilute oxalic acid
four test tubes and a dropper
white paper

Predict first

Once the red colour has formed, you add a few more drops of thiocyanate solution. The red colour will

deepenfadestay exactly the same

Add 2 drops of thiocyanate to 1 mL of iron(III) solution in 20 mL of water. A red colour forms: $F e^{3 +} + S C N^{-} ⇌ [F e (S C N)]^{2 +}$ (yellow, colourless, deep red). Share it among four tubes; tube 1 is the reference.
Tube 2: add 3 drops of thiocyanate. Tube 3: add 3 drops of iron(III) solution. Tube 4: add oxalic acid drop by drop. Compare each with tube 1 against white paper.

Your observations

Tube	Your prediction	Colour compared with tube 1
2: more SCN⁻
3: more Fe³⁺
4: oxalic acid

Your entries are saved automatically.

6.8.2 Effect of Pressure Change

Compressing a gas mixture raises every partial pressure. If the reaction changes the number of gas molecules, Le Chatelier's principle predicts a shift towards the side with fewer moles of gas, which lowers the pressure part of the way back (Fig. 6.13). If $Δ n = 0$ , pressure does not affect the composition. Solids and liquids hardly change volume, so only the gases count.

3/3Some NO₂ molecules pair up: fewer molecules, so the pressure falls part of the way back.

Fig. 6.13Halving the volume of

2 N O_{2} \overset{}{⇌} N_{2} O_{4}

shifts it towards

N_{2} O_{4}

, the side with fewer gas molecules.

A cylinder with a piston holding single nitrogen dioxide molecules and paired dinitrogen tetroxide molecules. Step 1: eight singles and two pairs fill the cylinder. Step 2: the piston halves the volume and the same molecules are crowded together. Step 3: now four singles and four pairs, ten molecules instead of twelve, as some nitrogen dioxide pairs up.

Q says the same. For $2 N O_{2} (g) \overset{}{⇌} N_{2} O_{4} (g)$ , $Q_{c} = [N_{2} O_{4}] / [N O_{2}]^{2}$ . Halving the volume doubles every concentration: the numerator doubles but the denominator quadruples, so $Q_{c} = K_{c} / 2$ and the forward reaction runs until $Q = K$ . Likewise, high pressure pushes the water-gas reaction $C (s) + H_{2} O (g) ⇌ C O (g) + H_{2} (g)$ to the left.

6.8.3 Effect of Inert Gas Addition

Adding argon, or any non-reacting gas, at constant volume raises the total pressure but leaves the partial pressures and concentrations of the reacting gases unchanged, so Q still equals K and nothing shifts. (At constant total pressure the vessel must expand, lowering every partial pressure, and the shift is towards more gas moles.)

6.8.4 Effect of Temperature Change

Temperature is different: it changes K itself. Treat heat as a reactant or product. For an exothermic reaction ( $Δ H < 0$ ) heat is a product, so warming shifts the equilibrium left and K falls; for an endothermic reaction K rises with temperature. The ammonia synthesis has $Δ H = - 92.4 k J m o l^{- 1}$ , so low temperatures favour ammonia, but there the reaction is far too slow.

DemonstrationNCERT Effect of temperature – an experiment · p. 187

Brown gas in hot and cold water

Predict first

Two identical sealed tubes of brown

N O_{2}

are put in ice water and in hot water. For

2 N O_{2} (g) \overset{}{⇌} N_{2} O_{4} (g)

Δ H = - 57.2 k J m o l^{- 1}

. What happens in the ice water?

The brown colour deepensThe brown colour fadesThere is no visible change

In a fume cupboard your teacher seals brown $N O_{2}$ (from copper and concentrated nitric acid) in two identical tubes and lets them settle at room temperature.
One tube goes into ice with salt, the other into water at about 90 °C. Compare the colours after a few minutes, then swap the tubes.

6.8.5 Effect of a Catalyst

A catalyst provides a new pathway with a lower activation energy. It lowers the barrier for the forward and the reverse reaction by exactly the same amount, so both speed up by the same factor. Equilibrium is reached sooner, but its composition, and K, are unchanged. A catalyst appears neither in the balanced equation nor in the K expression.

The ammonia synthesis shows the balance in practice. Low temperature gives a good yield but a slow reaction; high temperature is fast but gives little ammonia. Haber's iron catalyst gives a useful rate at about 450–500 °C, where the yield is still acceptable, and a high pressure, about 200 atm, helps because 4 moles of gas give 2. In the contact process, $2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)$ has a huge K but is far too slow without vanadium(V) oxide (or platinum) as catalyst. A catalyst cannot rescue a reaction with a tiny K: it would only reach a poor equilibrium faster.

Fig. 6.14Press Play to watch Q approach K, then add or remove substances, change the volume or temperature, add argon or a catalyst, and see how the mixture responds.

Quick check

1 of 2

For

N_{2} O_{4} (g) \overset{}{⇌} 2 N O_{2} (g)

Δ H = + 57.2 k J m o l^{- 1}

, which change increases the amount of

N O_{2}

at equilibrium?

Compressing the mixtureAdding a catalystRaising the temperatureAdding argon at constant volume

Show answer

Answer (C) Raising the temperature

The forward reaction is endothermic, so heating raises K. Compression favours

N_{2} O_{4}

; a catalyst or argon at constant volume changes nothing.

6.9 Ionic Equilibrium in Solution

The iron(III)–thiocyanate equilibrium was already a reaction between ions, and many equilibria in water involve ions and the molecules that form them. A sugar solution does not conduct electricity, but a salt solution does, and more salt conducts better.

Michael Faraday sorted substances by this test. Electrolytes conduct when dissolved in water; non-electrolytes do not. Strong electrolytes such as sodium chloride are present almost entirely as ions, while weak ones such as acetic acid stay mostly as molecules, less than 5 % ionized (Fig. 6.15). In a weak electrolyte ions and molecules are in equilibrium, $C H_{3} C O O H (a q) ⇌ H^{+} (a q) + C H_{3} C O O^{-} (a q)$ : an ionic equilibrium. How far a substance ionizes depends on the strength of the bond that must break and on how strongly water solvates the ions. Strictly, dissociation separates ions already present in a solid (NaCl) and ionization splits a neutral molecule into ions (HCl); here the two words are used interchangeably.

3/3In the weak electrolyte, molecules and ions are in dynamic equilibrium.

Fig. 6.15A strong electrolyte is fully ionized; a weak electrolyte sets up an equilibrium between molecules and a few ions.

Step 1: a beaker of sodium chloride solution containing only positive and negative ions, labelled conducts well. Step 2: a beaker of acetic acid solution with eight neutral molecules and just one pair of ions, labelled conducts weakly. Step 3: opposite arrows link the molecules and ions in the acetic acid, labelled molecules in equilibrium with ions.

6.10 Acids, Bases and Salts

Acids, bases and salts are everywhere. The stomach secretes hydrochloric acid, vinegar is dilute acetic acid, lemons and amla contain citric and ascorbic acids, and tamarind contains tartaric acid. Acid comes from the Latin acidus, sour. Acids turn blue litmus red and release hydrogen with active metals; bases turn red litmus blue and feel soapy, like washing soda. (Never taste laboratory chemicals.) Acids and bases react to give salts such as sodium chloride, potassium nitrate and barium sulphate.

Why does salt dissolve so readily? The electrostatic force holding $N a^{+}$ and $C l^{-}$ together is divided by the dielectric constant of the medium, about 80 for water, so in water it is roughly 80 times weaker. Each ion also gathers a shell of oriented water molecules, oxygen ends towards $N a^{+}$ and hydrogen ends towards $C l^{-}$ . This hydration releases energy and keeps the ions apart and free to move.

6.10.1 Arrhenius Concept of Acids and Bases

Svante Arrhenius defined an acid as a substance that produces hydrogen ions, $H^{+} (a q)$ , in water, and a base as one that produces hydroxide ions, $O H^{-} (a q)$ : $H X (a q) \overset{}{\to} H^{+} (a q) + X^{-} (a q)$ and $M O H (a q) ⟶ M^{+} (a q) + O H^{-} (a q)$ .

A bare proton (radius about $10^{- 15} m$ ) has such an intense field that it never floats free in water: it bonds to a lone pair of a water molecule, giving the pyramidal hydronium ion, $H X + H_{2} O \overset{}{\to} H_{3} O^{+} + X^{-}$ . Hydronium is further hydrated (as $H_{5} O_{2}^{+}$ , $H_{9} O_{4}^{+}$ and so on), and so is hydroxide. Here $H^{+} (a q)$ and $H_{3} O^{+} (a q)$ both mean the hydrated proton. The Arrhenius definitions work only in water, and they cannot explain why ammonia, which has no $O H$ group, is a base.

6.10.2 The Brönsted–Lowry Acids and Bases

In 1923 Johannes Brønsted (Denmark) and Thomas Lowry (England) independently widened the idea: an acid is a proton donor and a base is a proton acceptor. In $N H_{3} (a q) + H_{2} O (l) ⇌ N H_{4}^{+} (a q) + O H^{-} (a q)$ (Fig. 6.16) water gives a proton to ammonia, so water is the acid and ammonia the base; in the reverse reaction $N H_{4}^{+}$ is the acid and $O H^{-}$ the base.

3/3NH₃/NH₄⁺ and H₂O/OH⁻ are conjugate pairs: each differs by one proton.

Fig. 6.16A Brønsted acid–base reaction is a proton transfer; each acid and the base it becomes form a conjugate pair.

The equation ammonia plus water in equilibrium with ammonium plus hydroxide, with base, acid, acid and base written under the species. Step 1: a proton sits on the water molecule. Step 2: a curved arrow carries the proton to ammonia. Step 3: brackets join ammonia with ammonium and water with hydroxide, each labelled conjugate pair.

Definition

Conjugate acid–base pair

An acid and a base that differ by exactly one proton. The acid has one $H^{+}$ more than its conjugate base.

Example $N H_{4}^{+}$ and $N H_{3}$ ; $H_{2} O$ and $O H^{-}$ .

With hydrogen chloride, water plays the other role: in $H C l + H_{2} O \overset{}{\to} H_{3} O^{+} + C l^{-}$ it takes the proton, so here it behaves as a base. $C l^{-}$ is the conjugate base of HCl and $H_{3} O^{+}$ the conjugate acid of water. A species that can give or take a proton, like water, is amphiprotic. The stronger an acid, the weaker its conjugate base, and the other way round.

Example 6.12NCERT Problems 6.12–6.14 · p. 191

Finding conjugate acids and bases

Give (a) the conjugate base of $H C O O H$ , $H_{2} P O_{4}^{-}$ and $N H_{4}^{+}$ ; (b) the conjugate acid of $C H_{3} N H_{2}$ and $C O_{3}^{2 -}$ ; (c) both partners of the amphiprotic ion $H P O_{4}^{2 -}$ .

Try it yourself first. Then reveal the solution one step at a time.

6.10.3 Lewis Acids and Bases

G. N. Lewis (1923) shifted the focus to electron pairs: a Lewis acid accepts an electron pair and a Lewis base donates one. Brønsted bases are Lewis bases too, since they take a proton by offering a lone pair. The gain is that species with no proton can be acids: boron trifluoride, with only six electrons around boron, takes ammonia's lone pair, $B F_{3} + N H_{3} \overset{}{\to} F_{3} B - N H_{3}$ .

Other Lewis acids include electron-deficient molecules such as $A l C l_{3}$ , metal ions such as $F e^{3 +}$ , $C u^{2 +}$ and $M g^{2 +}$ , and $C O_{2}$ , which accepts a pair from $O H^{-}$ to form $H C O_{3}^{-}$ . Molecules and ions with lone pairs, such as $H_{2} O$ , $N H_{3}$ , $O H^{-}$ and $C l^{-}$ , are Lewis bases.

Example 6.15NCERT Problem 6.15 · p. 192

Classifying Lewis acids and bases

Classify each species as a Lewis acid or base and say how it acts: (a) $A g^{+}$ (b) $C N^{-}$ (c) $C O_{2}$ (d) $H_{2} O$ .

Try it yourself first. Then reveal the solution one step at a time.

6.11 Ionization of Acids and Bases

Strong acids (perchloric, hydrochloric, hydrobromic, hydroiodic, nitric, and sulphuric in its first step) and strong bases (the hydroxides of lithium, sodium, potassium, caesium and barium) are almost completely ionized in water. In Brønsted terms, a strong acid is a good proton donor and a strong base a good proton acceptor.

For a weak acid HA, the reaction $H A + H_{2} O \overset{}{⇌} H_{3} O^{+} + A^{-}$ is a contest between two acids, HA and $H_{3} O^{+}$ , for the job of giving away a proton. The equilibrium lies on the side of the weaker acid and weaker base: the stronger acid hands its proton to the stronger base. For HCl, HBr, HI, $H N O_{3}$ and $H C l O_{4}$ it lies far to the right, so $C l^{-}$ , $B r^{-}$ , $I^{-}$ , $N O_{3}^{-}$ and $C l O_{4}^{-}$ are much weaker bases than water. Weak acids such as HF, $H N O_{2}$ and acetic acid stay mostly as molecules, and their conjugate bases are relatively strong. $N H_{2}^{-}$ , $O^{2 -}$ and $H^{-}$ are such strong bases that they pull protons off water.

Indicators are weak acids whose acid form, HIn, and conjugate base, $I n^{-}$ , have different colours: $H I n (a q) + H_{2} O (l) ⇌ H_{3} O^{+} (a q) + I n^{-} (a q)$ . Phenolphthalein is colourless as HIn and pink as $I n^{-}$ ; bromothymol blue is yellow in acid and blue in alkali. Which form dominates depends on $[H^{+}]$ , which is why indicators can signal the end point of a titration.

6.11.1 The Ionization Constant of Water and its Ionic Product

Because water is amphiprotic, one water molecule can hand a proton to another: $H_{2} O (l) + H_{2} O (l) ⇌ H_{3} O^{+} (a q) + O H^{-} (a q)$ . Its equilibrium constant, $K = [H_{3} O^{+}] [O H^{-}] / [H_{2} O]$ , contains the concentration of pure water. That is constant, about 1000 g L⁻¹ ÷ 18.0 g mol⁻¹ = 55.6 mol L⁻¹, so it is absorbed into a new constant, the ionic product of water:

Ionic product of water

K_{w} = [H^{+}] [O H^{-}]

(6.27)

In pure water at 298 K, $[H^{+}] = [O H^{-}] = 1.0 \times 10^{- 7} M$ by measurement, so

At 298 K

K_{w} = (1.0 \times 10^{- 7})^{2} = 1.0 \times 10^{- 14}

(6.28)

Only about 2 molecules in every $10^{9}$ are ionized at any moment ( $10^{- 7} / 55.6 = 1.8 \times 10^{- 9}$ ), so the equilibrium lies far towards un-ionized water. Like any equilibrium constant, $K_{w}$ changes with temperature; it rises when water is warmed. Any aqueous solution is acidic if $[H_{3} O^{+}] > [O H^{-}]$ , neutral if they are equal and basic if $[H_{3} O^{+}] < [O H^{-}]$ .

6.11.2 The pH Scale

Hydrogen ion concentrations span many powers of ten, so they are reported on a logarithmic scale. Strictly, pH is the negative base-10 logarithm of the activity of hydrogen ions, $a_{H^{+}}$ , a pure number. In dilute solutions (below about 0.01 M) the activity equals the molarity divided by 1 mol L⁻¹, so

Definition of pH

p H = - \log a_{H^{+}} = - \log [H^{+}]

So 0.001 M nitric acid has pH 3, and a solution with $[O H^{-}] = 10^{- 3} M$ has $[H^{+}] = 10^{- 11} M$ and pH 11. Pure water at 298 K has pH 7. At 298 K, acidic solutions have pH < 7, neutral ones pH = 7 and basic ones pH > 7. Taking negative logarithms of $K_{w}$ gives a second useful relation:

At 298 K

p H + p O H = p K_{w} = 14

(6.29)

Because the scale is logarithmic, one pH unit is a factor of 10 in $[H^{+}]$ , and two units a factor of 100 (Fig. 6.17). $K_{w}$ does change with temperature, but over the usual working range the effect on pH is small and usually ignored.

3/3One pH unit means a factor of 10 in [H⁺]; two units mean 100.

Fig. 6.17Typical pH values; each step of one unit is a tenfold change in

[H^{+}]

.NCERT Table 6.5 · p. 195

A pH bar from 0 to 14, coloured from vermillion through green to blue. Step 1: pins mark lemon juice near 2.3, curd near 4.5 and milk near 6.7. Step 2: pins below add pure water at 7.0, blood at 7.4, baking soda solution near 8.3 and lime water near 12.4. Step 3: brackets over pH 9 to 10 and 10 to 11 each stand for a tenfold change in hydrogen ion concentration.

Universal indicator paper gives a quick estimate. Papers carrying four strips of different dyes, each changing colour over a different range, can read pH 1–14 to about ±0.5. A pH meter measures an electrical potential that depends on pH and reads to 0.01 or better; pen-sized meters are now cheap enough for farmers testing soil.

Example 6.16NCERT Problems 6.16, 6.17 · p. 194

pH from [H⁺], and a very dilute strong acid

(a) A tamarind drink has $[H^{+}] = 2.5 \times 10^{- 3} M$ . Find its pH. (b) Find the pH of $2.0 \times 10^{- 8} M$ nitric acid.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 2

What is the pH of 0.0050 M barium hydroxide,

B a (O H)_{2}

, at 298 K?

Show answer

Answer 12

Each formula unit gives 2

O H^{-}

[O H^{-}] = 0.010 M

, pOH = 2.00, pH = 12.00.

6.11.3 Ionization Constants of Weak Acids

Let a weak acid HX have initial concentration $c$ , and let a fraction $α$ , its degree of ionization, of the molecules ionize:

Table 6.4ICE table for

H X (a q) + H_{2} O (l) ⇌ H_{3} O^{+} (a q) + X^{-} (a q)

NCERT p. 195

	$[H X]$	$[H_{3} O^{+}]$	$[X^{-}]$
Initial	$c$	0	0
Change	$- c α$	$+ c α$	$+ c α$
Equilibrium	$c (1 - α)$	$c α$	$c α$

Acid ionization constant

K_{a} = \frac{[H^{+}] [X^{-}]}{[H X]} = \frac{c α^{2}}{1 - α}

(6.30)

$K_{a}$ is the acid ionization constant. At a given temperature, the larger $K_{a}$ , the stronger the acid. Taking 1 M as the standard concentration makes $K_{a}$ a pure number. As with pH, $p K_{a} = - \log K_{a}$ (6.31), so a smaller $p K_{a}$ means a stronger acid. Table 6.5 lists some values.

When $α$ is small, $1 - α \approx 1$ , so $K_{a} \approx c α^{2}$ , $α \approx \sqrt{K_{a} / c}$ and $[H^{+}] \approx \sqrt{K_{a} c}$ . Diluting a weak acid therefore raises its degree of ionization, although $[H^{+}]$ falls. The approximation is safe if $α < 0.05$ . Percent ionization (6.32) is $100 α$ : ionized ÷ initial concentration × 100.

Table 6.5Ionization constants of some weak acids and bases at 298 KNCERT Tables 6.6, 6.7 · p. 195

Substance	Where you meet it	$K_{a}$ or $K_{b}$	$p K$
Formic acid	ant stings	$1.8 \times 10^{- 4}$	3.74
Lactic acid	curd, tired muscles	$1.4 \times 10^{- 4}$	3.86
Benzoic acid	food preservative	$6.3 \times 10^{- 5}$	4.20
Acetic acid	vinegar	$1.8 \times 10^{- 5}$	4.74
Carbonic acid (first step)	soda water	$4.3 \times 10^{- 7}$	6.37
Hypochlorous acid	bleach, pool water	$3.0 \times 10^{- 8}$	7.52
Hydrocyanic acid	bitter almond kernels	$6.2 \times 10^{- 10}$	9.21
Methylamine (base)	fishy smell	$4.4 \times 10^{- 4}$	3.36
Ammonia (base)	household cleaners	$1.8 \times 10^{- 5}$	4.74
Aniline (base)	dye industry	$4.3 \times 10^{- 10}$	9.37

A general route to the pH of a weak electrolyte: identify the acids and bases present; write each proton-transfer equilibrium; take the one with the largest K as the main reaction (water's own ionization usually adds a negligible amount); set up its ICE table in terms of $α$ or $x$ ; solve, checking any approximation; then find the concentrations and the pH.

Example 6.18NCERT Problems 6.18–6.20 · p. 196

pH and degree of ionization of a weak acid

(a) Find the pH and percent ionization of 0.050 M formic acid ( $K_{a} = 1.8 \times 10^{- 4}$ ). (b) 0.10 M of an unknown monobasic acid has pH 3.20. Find $K_{a}$ .

Try it yourself first. Then reveal the solution one step at a time.

6.11.4 Ionization of Weak Bases

A weak base B ionizes partly by taking a proton from water, $B (a q) + H_{2} O (l) ⇌ B H^{+} (a q) + O H^{-} (a q)$ . (For a hydroxide MOH, $M O H (a q) ⇌ M^{+} (a q) + O H^{-} (a q)$ .) Water's constant concentration is absorbed into the base ionization constant:

Base ionization constant

K_{b} = \frac{[B H^{+}] [O H^{-}]}{[B]} = \frac{c α^{2}}{1 - α}

(6.33)

Values are included in Table 6.5, and $p K_{b} = - \log K_{b}$ (6.34). Many organic bases are amines, ammonia with one or more hydrogens replaced by carbon groups. Methylamine, nicotine, codeine and quinine are all weak bases with small $K_{b}$ .

Example 6.21NCERT Problems 6.21, 6.22 · p. 198

Kb of a base from the pH of its solution

0.010 M methylamine, $C H_{3} N H_{2}$ , has pH 11.28. Find $K_{b}$ .

Try it yourself first. Then reveal the solution one step at a time.

6.11.5 Relation between Ka and Kb

DerivationKa × Kb = Kw for a conjugate pair

Take the pair $N H_{4}^{+}$ / $N H_{3}$ . As an acid, $N H_{4}^{+} + H_{2} O \overset{}{⇌} H_{3} O^{+} + N H_{3}$ has $K_{a}$ ; as a base, $N H_{3} + H_{2} O \overset{}{⇌} N H_{4}^{+} + O H^{-}$ has $K_{b}$ . Adding the two equations gives $2 H_{2} O \overset{}{⇌} H_{3} O^{+} + O H^{-}$ , whose constant is $K_{w}$ . When equations are added, their constants multiply:

K_{net} = K_{1} \times K_{2}

(6.35)

K_{a} \times K_{b} = K_{w}

(6.36)

The same follows algebraically: multiply the top and bottom of the $K_{b}$ expression by $[H^{+}]$ , and it splits into $[H^{+}] [O H^{-}]$ divided by $[H^{+}] [B] / [B H^{+}]$ , that is, $K_{w} / K_{a}$ . Taking negative logarithms:

p K_{a} + p K_{b} = p K_{w} = 14

(6.37)

For ammonia, $K_{b} = 1.8 \times 10^{- 5}$ , so the ammonium ion has $K_{a} = 1.0 \times 10^{- 14} / 1.8 \times 10^{- 5} = 5.6 \times 10^{- 10}$ . A strong acid has a very weak conjugate base, and the other way round.

Example 6.23NCERT Problem 6.23 · p. 199

A weak base and its conjugate acid

Find the degree of ionization and the pH of 0.10 M pyridine ( $K_{b} = 1.7 \times 10^{- 9}$ ), and $K_{a}$ for the pyridinium ion.

Try it yourself first. Then reveal the solution one step at a time.

6.11.6 Di- and Polybasic Acids and Di- and Polyacidic Bases

Some acid molecules can give up two or three protons, for example oxalic, sulphuric, carbonic and phosphoric acid: they are polybasic (polyprotic). They ionize in steps, each with its own constant. For $H_{2} X$ , $H_{2} X \overset{}{⇌} H^{+} + H X^{-}$ has $K_{a 1} = [H^{+}] [H X^{-}] / [H_{2} X]$ , and $H X^{-} \overset{}{⇌} H^{+} + X^{2 -}$ has $K_{a 2} = [H^{+}] [X^{2 -}] / [H X^{-}]$ . A tribasic acid such as $H_{3} P O_{4}$ has three.

Successive constants fall steeply, because pulling a positive proton away from an ion that is already negative is harder. Phosphoric acid has $K_{a 1} = 7.5 \times 10^{- 3}$ , $K_{a 2} = 6.2 \times 10^{- 8}$ and $K_{a 3} = 4.8 \times 10^{- 13}$ ; carbonic acid has $K_{a 1} = 4.3 \times 10^{- 7}$ and $K_{a 2} = 4.7 \times 10^{- 11}$ . So nearly all the $H_{3} O^{+}$ comes from the first step. Bases that accept two or more protons, such as ethane-1,2-diamine or carbonate, likewise have one $K_{b}$ per step.

6.11.7 Factors Affecting Acid Strength

Why are some acids stronger than others? The full answer is complicated, but broadly an H–A bond gives up $H^{+}$ more easily when it is weaker and when it is more polar. Which factor wins depends on where A sits in the periodic table:

Compare

Two trends in acid strength

Down a group: bond strength wins

A gets larger, the H–A bond longer and weaker.
$H F ≪ H C l < H B r < H I$
$H_{2} S$ is a stronger acid than $H_{2} O$ .

Across a period: polarity wins

Atoms are similar in size; electronegativity of A rises.
$C H_{4} < N H_{3} < H_{2} O < H F$
A more polar H–A bond releases $H^{+}$ more easily.

6.11.8 Common Ion Effect in the Ionization of Acids and Bases

Add sodium acetate to acetic acid. The extra acetate pushes $C H_{3} C O O H \overset{}{⇌} H^{+} + C H_{3} C O O^{-}$ to the left, so $[H^{+}]$ falls and the acid ionizes even less. Adding $H^{+}$ from a strong acid has the same effect. This is a case of Le Chatelier's principle.

Definition

Common ion effect

The shift of an equilibrium caused by adding a substance that supplies an ion already taking part in it. It suppresses the ionization of a weak electrolyte and lowers the solubility of a sparingly soluble salt.

ExampleSodium acetate lowers the ionization of acetic acid.

Example 6.24NCERT Problem 6.24 · p. 201

Common ion and partial neutralisation

With $K_{a} = 1.8 \times 10^{- 5}$ , find the pH of (a) 0.10 M acetic acid; (b) 0.10 M acetic acid + 0.10 M sodium acetate; (c) 40 mL of 0.10 M acetic acid + 20 mL of 0.10 M NaOH.

Try it yourself first. Then reveal the solution one step at a time.

6.11.9 Hydrolysis of Salts and the pH of their Solutions

Salts are fully ionized in water, but some of their ions then react with water. Cations of strong bases ( $N a^{+}$ , $K^{+}$ , $C a^{2 +}$ , $B a^{2 +}$ ) and anions of strong acids ( $C l^{-}$ , $B r^{-}$ , $N O_{3}^{-}$ , $C l O_{4}^{-}$ ) are only hydrated, so NaCl gives a neutral solution. Ions from a weak acid or base react with water, called hydrolysis, and shift the pH:

Weak acid + strong base, such as sodium acetate: $C H_{3} C O O^{-} + H_{2} O ⇌ C H_{3} C O O H + O H^{-}$ . Hydroxide builds up, so the solution is basic (pH > 7).
Strong acid + weak base, such as ammonium chloride: $N H_{4}^{+} + H_{2} O \overset{}{⇌} N H_{3} + H_{3} O^{+}$ . The solution is acidic (pH < 7). (Older books write the base as $N H_{4} O H$ .)
Weak acid + weak base, such as ammonium acetate: both ions hydrolyse, and the pH depends on which is the stronger.

For the first two types, the hydrolysing ion is simply a weak base or acid whose constant is $K_{w} / K_{a}$ or $K_{w} / K_{b}$ , so its pH is found as in Section 6.11.4. For a salt of a weak acid and a weak base, a fuller treatment shows that the degree of hydrolysis does not depend on concentration and

p H = 7 + \frac{1}{2} (p K_{a} - p K_{b})

(6.38)

where

$p K_{a}$: of the weak acid
$p K_{b}$: of the weak base

The solution is basic if $p K_{a} > p K_{b}$ , that is, if the acid is the weaker partner, and acidic if $p K_{a} < p K_{b}$ .

Example 6.25NCERT Problem 6.25 · p. 202

pH of a salt of a weak acid and a weak base

Find the pH of ammonium formate solution. Formic acid has $p K_{a} = 3.74$ and ammonia has $p K_{b} = 4.75$ .

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

Which salt gives a basic solution in water?

K N O_{3}

N H_{4} C l

N a_{2} C O_{3}

N a C l

Show answer

Answer (C) $N a_{2} C O_{3}$

Carbonate, from a weak acid, takes protons from water and releases

O H^{-}

N H_{4} C l

is acidic; the others are neutral.

6.12 Buffer Solutions

Blood stays between pH 7.35 and 7.45; a drift of a few tenths is a medical emergency. Many reactions, medicines, eye drops and shampoos also need a fixed pH. Solutions that resist a change in pH when they are diluted or when a little acid or alkali is added are buffer solutions. A mixture of acetic acid and sodium acetate buffers near pH 4.75, and one of ammonium chloride and ammonia near pH 9.25. The trick is to hold a reserve of both a weak acid and its conjugate base. Added $H^{+}$ is mopped up by the base and added $O H^{-}$ by the acid (Fig. 6.18), so $[H^{+}]$ hardly changes; Fig. 6.19 lets you compare a buffer with plain water. You will study buffer capacity in higher classes.

2/3Acetate ions take up the added H⁺: pH only falls to 4.65.

Fig. 6.180.010 mol of acid or alkali moves a litre of 0.10 M acetate buffer by under 0.1 pH; plain water would change by 5 units.

Bars for acetic acid and acetate beside a pH meter. Step 1: equal bars, pH 4.74. Step 2: hydrogen ions are added; acetate turns into acetic acid, the bars change and the pH reads 4.65, while plain water would drop from 7 to 2. Step 3: hydroxide is added instead; acetic acid turns into acetate and the pH reads 4.83, while water would rise to 12.

6.12.1 Designing Buffer Solution

DerivationThe Henderson–Hasselbalch equation

For an acidic buffer, start from the weak acid's equilibrium, $H A + H_{2} O \overset{}{⇌} H_{3} O^{+} + A^{-}$ , and rearrange $K_{a}$ for $[H_{3} O^{+}]$ :

[H_{3} O^{+}] = K_{a} \frac{[H A]}{[A^{-}]}

Take $- \log$ of both sides:

Henderson–Hasselbalch

p H = p K_{a} + \log \frac{[A^{-}]}{[H A]}

(6.39)

The weak acid barely ionizes, so $[H A]$ is the concentration of acid taken; nearly all the $A^{-}$ comes from the fully ionized salt. So

p H = p K_{a} + \log \frac{[salt]}{[acid]}

(6.40)

With equal concentrations of acid and salt, $\log 1 = 0$ and $p H = p K_{a}$ . So to make a buffer, choose an acid whose $p K_{a}$ is close to the pH you want, then fine-tune the ratio: acetic acid ( $p K_{a}$ 4.76) with sodium acetate for pH near 4.8.

A basic buffer uses a weak base with its conjugate acid, such as $N H_{3}$ with $N H_{4} C l$ : $p O H = p K_{b} + \log ([B H^{+}] / [B])$ (6.41). Using $p H = 14 - p O H$ and $p K_{a} + p K_{b} = 14$ gives $p H = p K_{a} + \log ([B] / [B H^{+}])$ (6.42), with $p K_{a}$ of $B H^{+}$ : 9.25 for ammonium, so equal amounts give pH 9.25. Diluting a buffer changes both concentrations by the same factor, so the ratio, and the pH, stay the same.

ExampleNCERT p. 203

Designing an acetate buffer

How many grams of sodium acetate, $C H_{3} C O O N a$ (82.0 g mol⁻¹), must be dissolved in 500 mL of 0.10 M acetic acid ( $p K_{a} = 4.76$ ) to make a buffer of pH 5.00?

Try it yourself first. Then reveal the solution one step at a time.

Fig. 6.19Choose water, a strong acid, a weak acid, a weak base or a buffer, set the concentration, then add acid or alkali and watch the pH.

Quick check

1 of 3

Find the pH of a buffer that is 0.20 M in

N H_{3}

and 0.10 M in

N H_{4} C l

(

p K_{b} = 4.75

Show answer

Answer 9.55

p O H = 4.75 + \log (0.10 / 0.20) = 4.45

, so pH = 9.55. Or:

p H = 9.25 + \log (0.20 / 0.10) = 9.55

6.13 Solubility Equilibria of Sparingly Soluble Salts

Ionic solids differ enormously in solubility. Anhydrous calcium chloride draws water vapour from the air and is used as a drying agent, while silver chloride and barium sulphate are often called insoluble. Solubility is a contest between the lattice enthalpy, holding the ions together, and the solvation enthalpy, released (always negative) as solvent molecules surround the ions. In a non-polar solvent little energy is released, so salts stay undissolved. Roughly, a salt dissolves well only if solvation pays for breaking the lattice. Solubility also depends on temperature. By their solubility at room temperature, salts are often grouped as soluble (above 0.1 M), slightly soluble (0.01–0.1 M) and sparingly soluble (below 0.01 M). Here we look at the equilibrium between a sparingly soluble salt and its saturated solution.

6.13.1 Solubility Product Constant

Shake silver chloride with water and a tiny amount dissolves, after which $A g C l (s) ⇌ A g^{+} (a q) + C l^{-} (a q)$ is at equilibrium. The solid's concentration is constant, so it is absorbed into K:

Solubility product

K_{s p} = [A g^{+}] [C l^{-}]

(6.43)

$K_{s p}$ is the solubility product constant; for AgCl it is $1.8 \times 10^{- 10}$ at 298 K. If $S$ mol dissolve per litre (the molar solubility), $S^{2} = 1.8 \times 10^{- 10}$ and $S = 1.34 \times 10^{- 5} m o l L^{- 1}$ . Barium sulphate, the 'barium meal' for stomach X-rays, has $K_{s p} = 1.1 \times 10^{- 10}$ : little enough dissolves to keep toxic $B a^{2 +}$ harmless.

In general, a salt $M_{x} X_{y}$ that gives $x$ cations $M^{p +}$ and $y$ anions $X^{q -}$ (with $x p = y q$ ) has $[M^{p +}] = x S$ and $[X^{q -}] = y S$ :

K_{s p} = (x S)^{x} (y S)^{y} = x^{x} y^{y} S^{x + y}

(6.44)

S = {(\frac{K_{s p}}{x^{x} y^{y}})}^{1 / (x + y)}

(6.45)

Table 6.6Solubility products of some salts at 298 KNCERT Table 6.9 · p. 205

Salt	$K_{s p}$	$K_{s p}$ in terms of $S$
$A g C l$	$1.8 \times 10^{- 10}$	$S^{2}$
$A g B r$	$5.0 \times 10^{- 13}$	$S^{2}$
$B a S O_{4}$	$1.1 \times 10^{- 10}$	$S^{2}$
$C a F_{2}$	$3.9 \times 10^{- 11}$	$4 S^{3}$
$M g (O H)_{2}$	$1.8 \times 10^{- 11}$	$4 S^{3}$
$A g_{2} C r O_{4}$	$1.1 \times 10^{- 12}$	$4 S^{3}$
$P b I_{2}$	$7.1 \times 10^{- 9}$	$4 S^{3}$
$F e (O H)_{3}$	$2.8 \times 10^{- 39}$	$27 S^{4}$
$C a_{3} (P O_{4})_{2}$	$2.0 \times 10^{- 33}$	$108 S^{5}$

When the solution is not at equilibrium, the same product of ion concentrations is the ionic product $Q_{s p}$ . If $Q_{s p} < K_{s p}$ , more solid can dissolve. If $Q_{s p} = K_{s p}$ , the solution is saturated. If $Q_{s p} > K_{s p}$ , solid precipitates until $Q_{s p}$ has fallen to $K_{s p}$ (Fig. 6.20).

2/3 $Q_{s p} = K_{s p}$ : dissolving and precipitating balance.

Fig. 6.20

Q_{s p} < K_{s p}

: solid dissolves;

Q_{s p} = K_{s p}

: saturated;

Q_{s p} > K_{s p}

: solid precipitates.

A beaker with solid silver chloride at the bottom and silver and chloride ions in solution. Step 1: few ions and two arrows up from the solid, labelled unsaturated. Step 2: more ions and one arrow each way, labelled saturated. Step 3: extra chloride ions and small new flakes of solid, labelled silver chloride comes out.

Example 6.26NCERT Problem 6.26 · p. 205

Molar solubility of a salt with five ions per formula

Calcium phosphate, $C a_{3} (P O_{4})_{2}$ , the mineral in bones and teeth, has $K_{s p} = 2.0 \times 10^{- 33}$ . Find its molar solubility in pure water, assuming the ions do not react with water.

Try it yourself first. Then reveal the solution one step at a time.

Example 6.27NCERT Problem 6.27 · p. 205

A smaller Ksp does not always mean less soluble

Which is more soluble in water: $M g (O H)_{2}$ ( $K_{s p} = 1.8 \times 10^{- 11}$ ) or AgCl ( $K_{s p} = 1.8 \times 10^{- 10}$ )?

Try it yourself first. Then reveal the solution one step at a time.

6.13.2 Common Ion Effect on Solubility of Ionic Salts

Raise the concentration of either ion and $Q_{s p}$ exceeds $K_{s p}$ , so some salt precipitates until $Q_{s p} = K_{s p}$ again; lower one ion and more salt dissolves. Soluble salts behave the same way, with activities in place of molarities. Bubbling HCl gas through saturated brine precipitates very pure sodium chloride. In gravimetric analysis an excess of the precipitating ion drives precipitation to completion, for example chloride as AgCl, sulphate as $B a S O_{4}$ and iron(III) as its hydrated oxide.

Example 6.28NCERT Problem 6.28 · p. 206

Solubility with a common ion

Find the molar solubility of AgCl ( $K_{s p} = 1.8 \times 10^{- 10}$ ) in 0.010 M NaCl, and compare it with pure water.

Try it yourself first. Then reveal the solution one step at a time.

pH matters too. Salts of weak acids, such as phosphates, carbonates and fluorides, dissolve more in acid. $H^{+}$ turns some of the anion $X^{-}$ into HX, lowering $[X^{-}]$ , so more salt dissolves to keep $[M^{+}] [X^{-}] = K_{s p}$ . For a 1 : 1 salt MX, the fraction of the dissolved anion left as $X^{-}$ is $f = K_{a} / (K_{a} + [H^{+}])$ , so $K_{s p} = S \cdot f S$ and

Solubility of a salt of a weak acid

S = \sqrt{K_{s p} \frac{K_{a} + [H^{+}]}{K_{a}}}

(6.46)

A lower pH (larger $[H^{+}]$ ) gives a larger S. This is also why acidic drinks attack tooth enamel, which is a calcium phosphate mineral.

Recap

Equilibrium: key points and summary

Equilibrium is dynamic: in a closed system opposing processes run at equal rates, so properties stay constant (Section 6.1, Section 6.2).
Each physical equilibrium has one fixed quantity at a given temperature: melting point, vapour pressure, solubility or Henry's-law ratio (Table 6.1).
$K_{c}$ is products over reactants, each concentration raised to its coefficient; it depends only on temperature and on how the equation is written (Section 6.3).
$K_{p} = K_{c} (R T)^{Δ n}$ ; pure solids and liquids are left out of K (Section 6.4.1, Section 6.5).
Large K: mostly products; $Q < K$ : net forward; ICE tables give compositions; $Δ G^{\circ} = - R T \ln K$ (Section 6.6, Section 6.7).
Le Chatelier: added substances are used up, pressure favours fewer gas moles, heat favours the endothermic side, catalysts change only the rate (Section 6.8).
Arrhenius, Brønsted–Lowry (conjugate pairs) and Lewis (electron pairs) acids and bases (Section 6.10).
$K_{w} = 10^{- 14}$ at 298 K; pH + pOH = 14 (Section 6.11.1, Section 6.11.2).
$K_{a} = c α^{2} / (1 - α)$ , $K_{a} K_{b} = K_{w}$ ; common ions suppress ionization; salts hydrolyse (Section 6.11).
Buffers follow $p H = p K_{a} + \log ([salt] / [acid])$ ; $K_{s p}$ gives molar solubility, and $Q_{s p} > K_{s p}$ means precipitation (Section 6.12, Section 6.13).

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Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbook	Practice question here
NCERT Exercises 6.1	Practice Q1
NCERT Exercises 6.2, 6.6	Practice Q2
NCERT Exercises 6.3	Practice Q3
NCERT Exercises 6.4	Practice Q4
NCERT Exercises 6.5, 6.10	Practice Q5
NCERT Exercises 6.7, 6.13	Practice Q6
NCERT Exercises 6.8, 6.33	Practice Q7
NCERT Exercises 6.9	Practice Q8
NCERT Exercises 6.11	Practice Q9
NCERT Exercises 6.12, 6.21	Practice Q10
NCERT Exercises 6.14, 6.15	Practice Q11
NCERT Exercises 6.16, 6.22	Practice Q12
NCERT Exercises 6.17	Practice Q13
NCERT Exercises 6.18	Practice Q14
NCERT Exercises 6.19, 6.34	Practice Q15
NCERT Exercises 6.20	Practice Q16
NCERT Exercises 6.23	Practice Q17
NCERT Exercises 6.24	Practice Q18
NCERT Exercises 6.25, 6.26	Practice Q19
NCERT Exercises 6.27	Practice Q20
NCERT Exercises 6.28	Practice Q21
NCERT Exercises 6.29	Practice Q22
NCERT Exercises 6.30	Practice Q23
NCERT Exercises 6.31	Practice Q24
NCERT Exercises 6.32	Practice Q25
NCERT Exercises 6.35, 6.37, 6.38, 6.39	Practice Q26
NCERT Exercises 6.36, 6.40	Practice Q27
NCERT Exercises 6.41, 6.42, 6.55, 6.56	Practice Q28
NCERT Exercises 6.43, 6.44	Practice Q29
NCERT Exercises 6.45	Practice Q30
NCERT Exercises 6.46, 6.53, 6.59	Practice Q31
NCERT Exercises 6.47, 6.50, 6.60	Practice Q32
NCERT Exercises 6.48, 6.49, 6.57, 6.58	Practice Q33
NCERT Exercises 6.51, 6.52, 6.54	Practice Q34
NCERT Exercises 6.61, 6.62, 6.63, 6.64	Practice Q35
NCERT Exercises 6.65	Practice Q36
NCERT Exercises 6.66	Practice Q37
NCERT Exercises 6.67, 6.68	Practice Q38
NCERT Exercises 6.69, 6.71, 6.73	Practice Q39
NCERT Exercises 6.70	Practice Q40
NCERT Exercises 6.72	Practice Q41

Quick check: MCQs on Equilibrium

6.1.5 General Characteristics of Equilibria Involving Physical Processes3 questions
6.2 Equilibrium in Chemical Processes – Dynamic Equilibrium3 questions
6.3 Law of Chemical Equilibrium and Equilibrium Constant3 questions
6.4.1 Equilibrium Constant in Gaseous Systems2 questions
6.5 Heterogeneous Equilibria2 questions
6.6.3 Calculating Equilibrium Concentrations3 questions
6.7 Relationship between Equilibrium Constant K, Reaction Quotient Q and Gibbs Energy G3 questions
6.8.5 Effect of a Catalyst2 questions
6.10.3 Lewis Acids and Bases3 questions
6.11.2 The pH Scale2 questions
6.11.9 Hydrolysis of Salts and the pH of their Solutions3 questions
6.12.1 Designing Buffer Solution3 questions
6.13.2 Common Ion Effect on Solubility of Ionic Salts3 questions

Glossary · 15

Key terms (English–Hindi)

buffer solutionबफर विलयनbuffer vilayan: A weak acid with its conjugate base, or a weak base with its conjugate acid, that resists pH change.First seen in the chapter →

common ion effectसम-आयन प्रभावsam-aayan prabhav: Suppression of an ionization or solubility by adding an ion already in the equilibrium.
conjugate acid–base pairसंयुग्मी अम्ल-क्षारक युग्मsanyugmi amla-ksharak yugm: An acid and a base that differ by one proton.

degree of ionizationआयनन की मात्राaayanan ki matra: The fraction $α$ of a weak electrolyte that is ionized.First seen in the chapter →
dynamic equilibriumगतिक साम्यgatik saamya: A state of a closed system in which opposing processes run at equal rates, so no measurable property changes.First seen in the chapter →

electrolyteविद्युत्-अपघट्यvidyut-apaghatya: A substance whose aqueous solution conducts because it contains ions.First seen in the chapter →
equilibrium constantसाम्य स्थिरांकsaamya sthirank: The value, fixed at a given temperature, of products over reactants with each concentration (or pressure) raised to its coefficient.First seen in the chapter →

heterogeneous equilibriumविषमांगी साम्यvishamangi saamya: An equilibrium involving more than one phase.First seen in the chapter →
hydrolysis of saltsलवणों का जल-अपघटनlavanon ka jal-apghatan: Reaction of a salt’s ions with water that makes the solution acidic or basic.First seen in the chapter →

ionic product of waterजल का आयनिक गुणनफलjal ka aayanik gunanphal: $K_{w} = [H^{+}] [O H^{-}]$ , $10^{- 14}$ at 298 K.First seen in the chapter →

Le Chatelier’s principleला-शातलिए सिद्धांतla-shatliye siddhant: An equilibrium disturbed by a change of concentration, pressure or temperature shifts so as to partly undo the change.First seen in the chapter →
Lewis acidलूइस अम्लlewis amla: An electron-pair acceptor; a Lewis base is a donor.First seen in the chapter →

pH: The negative base-10 logarithm of the hydrogen ion activity, about $- \log [H^{+}]$ .First seen in the chapter →

reaction quotientअभिक्रिया भागफलabhikriya bhagphal: The K expression evaluated at any moment; comparing Q with K gives the direction of net reaction.First seen in the chapter →

solubility product constantविलेयता गुणनफलvileyata gunanphal: The equilibrium constant $K_{s p}$ between a sparingly soluble salt and its ions.First seen in the chapter →

6.1 Equilibrium in Physical Processes

6.1.1 Solid–Liquid Equilibrium

6.1.2 Liquid–Vapour Equilibrium

6.1.3 Solid–Vapour Equilibrium

6.1.4 Equilibrium Involving Dissolution of Solid or Gases in Liquids

6.1.5 General Characteristics of Equilibria Involving Physical Processes

6.2 Equilibrium in Chemical Processes – Dynamic Equilibrium

6.3 Law of Chemical Equilibrium and Equilibrium Constant

6.4 Homogeneous Equilibria

6.4.1 Equilibrium Constant in Gaseous Systems

6.5 Heterogeneous Equilibria

6.6 Applications of Equilibrium Constants

6.6.1 Predicting the Extent of a Reaction

6.6.2 Predicting the Direction of the Reaction

6.6.3 Calculating Equilibrium Concentrations

6.7 Relationship between Equilibrium Constant K, Reaction Quotient Q and Gibbs Energy G

6.8 Factors Affecting Equilibria

6.8.1 Effect of Concentration Change

6.8.2 Effect of Pressure Change

6.8.3 Effect of Inert Gas Addition

6.8.4 Effect of Temperature Change

6.8.5 Effect of a Catalyst

6.9 Ionic Equilibrium in Solution

6.10 Acids, Bases and Salts

6.10.1 Arrhenius Concept of Acids and Bases

6.10.2 The Brönsted–Lowry Acids and Bases

6.10.3 Lewis Acids and Bases

6.11 Ionization of Acids and Bases

6.11.1 The Ionization Constant of Water and its Ionic Product

6.11.2 The pH Scale

6.11.3 Ionization Constants of Weak Acids

6.11.4 Ionization of Weak Bases

6.11.5 Relation between Ka and Kb

6.11.6 Di- and Polybasic Acids and Di- and Polyacidic Bases

6.11.7 Factors Affecting Acid Strength

6.11.8 Common Ion Effect in the Ionization of Acids and Bases

6.11.9 Hydrolysis of Salts and the pH of their Solutions

6.12 Buffer Solutions

6.12.1 Designing Buffer Solution

6.13 Solubility Equilibria of Sparingly Soluble Salts

6.13.1 Solubility Product Constant

6.13.2 Common Ion Effect on Solubility of Ionic Salts

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