Chemical Equilibrium - Wyatt's Notes

Intuition

Chemical equilibrium is like a tug-of-war that never ends — forward and reverse reactions proceed at equal rates: Le Chatelier’s principle predicts how systems respond to stress — they shift to relieve the applied change

Why it matters: Equilibrium principles guide industrial processes, drug design, and understanding natural systems

The key insight: Le Chatelier’s principle predicts how systems respond to stress — they shift to relieve the applied change

Dynamic Equilibrium

Reversible Reactions

Many reactions are reversible: reactants form products, and products can re-form reactants. This is Represented with a double arrow:

\mathrm{A} + \mathrm{B} \rightleftharpoons \mathrm{C} + \mathrm{D}

Closed Systems

Equilibrium can only be established in a closed system — one where no matter can enter or leave.

Dynamic Nature

At equilibrium:

Both forward and reverse reactions continue to occur.
The rates of the forward and reverse reactions are equal.
The concentrations of all species remain constant (not necessarily equal).
Macroscopic properties (colour, pressure, pH) are constant.

Characteristics of Equilibrium

Property	Description
Forward rate	Equals reverse rate
Concentrations	Constant (but not necessarily equal)
Can be approached	From either direction
Dynamic	Both reactions continue
Closed system	Required

The Equilibrium Constant

$K_c$ — Concentration Equilibrium Constant

For the reaction $a\mathrm{A} + b\mathrm{B} \rightleftharpoons c\mathrm{C} + d\mathrm{D}$ :

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

Only gases and aqueous species are included.
Pure solids and pure liquids are NOT included (their activity $= 1$ ).
Square brackets denote equilibrium concentrations in mol/L.
$K_c$ is dimensionless (but concentrations are still used in the calculation).

$K_p$ — Pressure Equilibrium Constant

For gaseous reactions:

K_p = \frac{(p_C)^c(p_D)^d}{(p_A)^a(p_B)^b}

Where $p_X$ is the partial pressure of gas $X$ at equilibrium.

Relationship Between $K_c$ and $K_p$

K_p = K_c(RT)^{\Delta n}

Where $\Delta n = (\mathrm{moles of gaseous products}) - (\mathrm{moles of gaseous reactants})$ .

Example For N $_2$ (g) + 3H $_2$ (g) $\rightleftharpoons$ 2NH $_3$ (g), $\Delta n = 2 - 4 = -2$ .

K_p = K_c(RT)^{-2} = \frac{K_c}{(RT)^2}

The Reaction Quotient ( $Q$ )

The reaction quotient has the same form as $K_c$ but uses current (non-equilibrium) Concentrations:

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

Comparison	Meaning
$Q \lt K$	Reaction proceeds forward (more products)
$Q = K$	System is at equilibrium
$Q \gt K$	Reaction proceeds in reverse (more reactants)

Le Chatelier”s Principle

Statement

If a system at equilibrium is subjected to a change, the system will shift to counteract that change And restore equilibrium.

Effect of Concentration Changes

Change	System Response
Increase [reactant]	Shift to products (right)
Decrease [reactant]	Shift to reactants (left)
Increase [product]	Shift to reactants (left)
Decrease [product]	Shift to products (right)

Effect of Pressure Changes (Gases)

Change	System Response
Increase pressure	Shift to side with fewer moles of gas
Decrease pressure	Shift to side with more moles of gas

Important: Changing pressure by adding an inert gas does NOT shift the equilibrium (partial Pressures of reacting gases are unchanged).

Effect of Temperature Changes

Change	System Response	Effect on $K$
Increase temperature	Shift in endothermic direction	$K$ changes
Decrease temperature	Shift in exothermic direction	$K$ changes

Example For the reaction H $_2$ (g) + I $_2$ (g) $\rightleftharpoons$ 2HI(g), $K_c = 50.5$ at $448\degree\mathrm{C}$ .

If $1.0\mathrm{ mol}$ of H $_2$ and $1.0\mathrm{ mol}$ of I $_2$ are placed in a $1.0\mathrm{ L}$ Flask:

	H $_2$	I $_2$	HI
Initial	1.0	1.0	0
Change	$-x$	$-x$	$+2x$
Equilibrium	$1.0-x$	$1.0-x$	$2x$

K_c = \frac{(2x)^2}{(1.0-x)(1.0-x)} = \frac{4x^2}{(1.0-x)^2} = 50.5

\frac{2x}{1.0-x} = \sqrt{50.5} = 7.11

2x = 7.11 - 7.11x

9.11x = 7.11 \implies x = 0.781

Equilibrium concentrations: [H $_2$ ] = [I $_2$ ] = $0.219\mathrm{ mol/L}$ [HI] = $1.562\mathrm{ mol/L}$ .

Calculating $K$ from Given Data

Determine equilibrium concentrations (using ICE table if needed).
Substitute into the equilibrium expression.
Calculate $K$ .

Industrial Applications

The Haber Process

\mathrm{N}_2(\mathrm{g}) + 3\mathrm{H}_2(\mathrm{g}) \rightleftharpoons 2\mathrm{NH}_3(\mathrm{g}) \quad \Delta H = -92\mathrm{ kJ/mol}

Condition	Le Chatelier Prediction	Industrial Choice
Temperature	Low (exothermic) favours products	$400\mathrm{--}500\degree\mathrm{C}$ (compromise: reasonable rate)
Pressure	High (fewer moles of gas on product side)	$150\mathrm{--}300\mathrm{ atm}$ (compromise: cost/safety)
Catalyst	Does not change position but speeds up rate	Iron catalyst

The Contact Process

2\mathrm{SO}_2(\mathrm{g}) + \mathrm{O}_2(\mathrm{g}) \rightleftharpoons 2\mathrm{SO}_3(\mathrm{g}) \quad \Delta H = -197\mathrm{ kJ/mol}

Condition	Le Chatelier Prediction	Industrial Choice
Temperature	Low (exothermic)	$400\mathrm{--}450\degree\mathrm{C}$
Pressure	High (fewer moles of gas)	$1\mathrm{--}2\mathrm{ atm}$ (cost effective)
Catalyst	V $_2$ O $_5$ catalyst	V $_2$ O $_5$

Acid-Base Equilibrium

The pH Scale

\mathrm{pH} = -\log[\mathrm{H}^+]

pH	Description
$\lt 7$	Acidic
$= 7$	Neutral (at $25\degree\mathrm{C}$ )
$\gt 7$	Basic (alkaline)

Strong and Weak Acids

Property	Strong Acid	Weak Acid
Dissociation	Complete	Partial
$[\mathrm{H}^+]$ vs $c$	$[\mathrm{H}^+] = c$	$[\mathrm{H}^+] \lt c$
Equilibrium	No equilibrium	Equilibrium established
pH	Lower (for same $c$ )	Higher (for same $c$ )
Examples	HCl, HNO $_3$ H $_2$ SO $_4$	CH $_3$ COOH, HF, HCN

Acid Dissociation Constant ( $K_a$ )

For a weak acid HA:

\mathrm{HA} \rightleftharpoons \mathrm{H}^+ + \mathrm{A}^-

K_a = \frac{[\mathrm{H}^+][\mathrm{A}^-]}{[\mathrm{HA}]}

Base Dissociation Constant ( $K_b$ )

For a weak base B:

\mathrm{B} + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{BH}^+ + \mathrm{OH}^-

K_b = \frac{[\mathrm{BH}^+][\mathrm{OH}^-]}{[\mathrm{B}]}

Ionic Product of Water ( $K_w$ )

\mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{H}^+ + \mathrm{OH}^-

K_w = [\mathrm{H}^+][\mathrm{OH}^-] = 1.0 \times 10^{-14}\mathrm{ (at }25\degree\mathrm{C)}

p $K_a$ and p $K_b$

\mathrm{p}K_a = -\log K_a

\mathrm{p}K_b = -\log K_b

Relationship Between $K_a$ and $K_b$

For a conjugate acid-base pair:

K_a \times K_b = K_w = 1.0 \times 10^{-14}

\mathrm{p}K_a + \mathrm{p}K_b = 14

pH of Weak Acids

For a weak acid HA of concentration $c$ :

[\mathrm{H}^+] = \sqrt{K_a \times c}

\mathrm{pH} = \frac{1}{2}(\mathrm{p}K_a - \log c)

(approximation valid when $K_a$ is small: $[\mathrm{HA}]_{\mathrm{eq}} \approx c$ )

Example Calculate the pH of $0.10\mathrm{ M}$ ethanoic acid ( $K_a = 1.8 \times 10^{-5}$ ).

[\mathrm{H}^+] = \sqrt{1.8 \times 10^{-5} \times 0.10} = \sqrt{1.8 \times 10^{-6}} = 1.34 \times 10^{-3}\mathrm{ mol/L}

\mathrm{pH} = -\log(1.34 \times 10^{-3}) = 2.87

Buffers

Definition

A buffer solution resists changes in pH when small amounts of acid or base are added. It Consists of:

A weak acid and its conjugate base, OR
A weak base and its conjugate acid.

Henderson-Hasselbalch Equation

\mathrm{pH} = \mathrm{p}K_a + \log\frac{[\mathrm{A}^-]}{[\mathrm{HA}]}

Buffer Capacity

The buffer is most effective when:

[\mathrm{HA}] = [\mathrm{A}^-] \implies \mathrm{pH} = \mathrm{p}K_a

A buffer works best within $\pm 1$ pH unit of its $\mathrm{p}K_a$ .

How Buffers Work

Adding acid (H $^+$ ): the conjugate base (A $^-$ ) reacts with the added H $^+$ to form more HA, minimising pH change.
Adding base (OH $^-$ ): the weak acid (HA) reacts with the added OH $^-$ to form A $^-$ and H $_2$ O, minimising pH change.

Example A buffer contains $0.20\mathrm{ M}$ CH $_3$ COOH ( $\mathrm{p}K_a = 4.76$ ) and $0.30\mathrm{ M}$ CH $_3$ COONa. Calculate the pH.

\mathrm{pH} = 4.76 + \log\!\left(\frac{0.30}{0.20}\right) = 4.76 + \log(1.5) = 4.76 + 0.18 = 4.94

Neutralisation and Indicators

Strong Acid + Strong Base

\mathrm{pH} = 7 \mathrm{ at equivalence point}

Weak Acid + Strong Base

\mathrm{pH} \gt 7 \mathrm{ at equivalence point}

Strong Acid + Weak Base

\mathrm{pH} \lt 7 \mathrm{ at equivalence point}

Indicators

An acid-base indicator is a weak acid that has different colours in its protonated and deprotonated Forms.

Indicator	pH Range	Colour Change
Methyl orange	3.1—4.4	Red $\to$ yellow
Bromothymol blue	6.0—7.6	Yellow $\to$ blue
Phenolphthalein	8.3—10.0	Colourless $\to$ pink

Solubility Product ( $K_{sp}$ )

Definition

For a sparingly soluble salt $\mathrm{M}_a\mathrm{X}_b$ :

\mathrm{M}_a\mathrm{X}_b(s) \rightleftharpoons a\mathrm{M}^{b+}(aq) + b\mathrm{X}^{a-}(aq)

K_`\{sp}` = [\mathrm{M}^{b+}]^a[\mathrm{X}^{a-}]^b

Common $K_{sp}$ Values

Salt	$K_{sp}$
AgCl	$1.8 \times 10^{-10}$
AgBr	$5.0 \times 10^{-13}$
AgI	$8.3 \times 10^{-17}$
BaSO $_4$	$1.1 \times 10^{-10}$
CaCO $_3$	$3.4 \times 10^{-9}$
PbCl $_2$	$1.7 \times 10^{-5}$

Predicting Precipitation

Compare the ion product ( $Q$ ) with $K_{sp}$ :

Condition	Result
$Q \lt K_{sp}$	Unsaturated (no precipitate)
$Q = K_{sp}$	Saturated (at equilibrium)
$Q \gt K_{sp}$	Supersaturated (precipitate forms)

Common Ion Effect

The solubility of a salt decreases when a common ion is present.

Example The $K_{sp}$ of AgCl is $1.8 \times 10^{-10}$ . Calculate the solubility of AgCl in:

(a) Pure water:

S^2 = 1.8 \times 10^{-10} \implies s = 1.34 \times 10^{-5}\mathrm{ mol/L}

(b) $0.10\mathrm{ M}$ NaCl:

[\mathrm{Ag}^+][\mathrm{Cl}^-] = s \times 0.10 = 1.8 \times 10^{-10}

S = \frac{1.8 \times 10^{-10}}{0.10} = 1.8 \times 10^{-9}\mathrm{ mol/L}

The solubility is much lower due to the common ion effect.

IB Exam-Style Questions

Question 1 (Paper 1 style)

For the equilibrium $2\mathrm{SO}_2(\mathrm{g}) + \mathrm{O}_2(\mathrm{g}) \rightleftharpoons 2\mathrm{SO}_3(\mathrm{g})$ What happens when the pressure is increased?

The product side has 2 moles of gas, the reactant side has 3. The equilibrium shifts to the right (fewer moles of gas), increasing [SO $_3$ ]. Note that $K_p$ is a constant at a given temperature and Does not change; only the equilibrium position shifts.

Question 2 (Paper 2 style)

For the reaction $\mathrm{PCl}_5(\mathrm{g}) \rightleftharpoons \mathrm{PCl}_3(\mathrm{g}) + \mathrm{Cl}_2(\mathrm{g})$ $1.0\mathrm{ mol}$ of PCl $_5$ is placed in a $2.0\mathrm{ L}$ flask at $250\degree\mathrm{C}$ . At Equilibrium, $0.40\mathrm{ mol}$ of Cl $_2$ is present.

(a) Calculate $K_c$ .

	PCl $_5$	PCl $_3$	Cl $_2$
Initial	0.50	0	0
Change	$-0.20$	$+0.20$	$+0.20$
Equilibrium	0.30	0.20	0.20

K_c = \frac{[\mathrm{PCl}_3][\mathrm{Cl}_2]}{[\mathrm{PCl}_5]} = \frac{(0.20)(0.20)}{0.30} = \frac{0.04}{0.30} = 0.133

(b) Calculate the percentage dissociation of PCl $_5$ .

\%\mathrm{ dissociation} = \frac{0.20}{0.50} \times 100\% = 40\%

Question 3 (Paper 2 style)

Calculate the pH of $0.050\mathrm{ M}$ NH $_3$ ( $K_b = 1.8 \times 10^{-5}$ ).

[\mathrm{OH}^-] = \sqrt{K_b \times c} = \sqrt{1.8 \times 10^{-5} \times 0.050} = \sqrt{9.0 \times 10^{-7}} = 9.49 \times 10^{-4}

\mathrm{pOH} = -\log(9.49 \times 10^{-4}) = 3.02

\mathrm{pH} = 14 - 3.02 = 10.98

Question 4 (Paper 1 style)

Which salt is most soluble?

A. AgCl ( $K_{sp} = 1.8 \times 10^{-10}$ ) B. AgBr ( $K_{sp} = 5.0 \times 10^{-13}$ ) C. AgI ( $K_{sp} = 8.3 \times 10^{-17}$ ) D. BaSO $_4$ ( $K_{sp} = 1.1 \times 10^{-10}$ )

Answer: D — BaSO $_4$ has the highest $K_{sp}$ But this comparison is only valid for salts with The same stoichiometry (1:1). For a fair comparison of solubility, convert to molar solubility. All Four are 1:1 salts, so the highest $K_{sp}$ gives the highest solubility: BaSO $_4$ .

flowchart TD
    A[1_Equilibrium] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

Summary

Concept	Formula
$K_c$	$K_c = \dfrac{[\mathrm{C}]^c[\mathrm{D}]^d}{[\mathrm{A}]^a[\mathrm{B}]^b}$
$K_p$	$K_p = K_c(RT)^{\Delta n}$
pH	$\mathrm{pH} = -\log[\mathrm{H}^+]$
$K_w$	$K_w = [\mathrm{H}^+][\mathrm{OH}^-] = 10^{-14}$
Henderson-Hasselbalch	$\mathrm{pH} = \mathrm{p}K_a + \log\dfrac{[\mathrm{A}^-]}{[\mathrm{HA}]}$
$K_{sp}$	$K_{sp} = [\mathrm{M}^{b+}]^a[\mathrm{X}^{a-}]^b$

Example For the reaction N $_2$ O $_4$ (g) $\rightleftharpoons$ 2NO $_2$ (g), $\Delta H^\circ = +57\mathrm{ kJ/mol}$ and $K = 0.115$ at $298\mathrm{ K}$ . Find $K$ at $350\mathrm{ K}$ .

\ln\!\left(\frac{K_2}{0.115}\right) = -\frac{57000}{8.314}\left(\frac{1}{350} - \frac{1}{298}\right)

= -6856 \times (-0.000500) = 3.428

\frac{K_2}{0.115} = e^{3.428} = 30.8

K_2 = 3.54

As expected for an endothermic reaction, $K$ increases with temperature.

Quantitative Le Chatelier Calculations

When a change is made to a system at equilibrium, a new equilibrium is established. The new Concentrations can be found by setting up a new ICE table.

Example For the reaction PCl $_5$ (g) $\rightleftharpoons$ PCl $_3$ (g) + Cl $_2$ (g), $K_c = 0.0211$ at $500\mathrm{ K}$ .

At equilibrium in a $1\mathrm{ L}$ flask: [PCl $_5$ ] $= 0.200\mathrm{ M}$ [PCl $_3$ ] $= [Cl_2]$ $= 0.0650\mathrm{ M}$ .

If $0.100\mathrm{ mol}$ of Cl $_2$ is added, find the new equilibrium concentrations.

After adding Cl $_2$ : [PCl $_5$ ] $= 0.200$ [PCl $_3$ ] $= 0.0650$ [Cl $_2$ ] $= 0.165$ .

Q_c = \frac{(0.0650)(0.165)}{0.200} = 0.0536 \gt K_c = 0.0211

The reaction shifts left. Let $x$ be the moles of Cl $_2$ that react:

	PCl $_5$	PCl $_3$	Cl $_2$
Initial	0.200	0.0650	0.165
Change	$+x$	$-x$	$-x$
Equilibrium	$0.200+x$	$0.0650-x$	$0.165-x$

\frac{(0.0650-x)(0.165-x)}{0.200+x} = 0.0211

(0.0650-x)(0.165-x) = 0.0211(0.200+x)

0.01073 - 0.2300x + x^2 = 0.004220 + 0.0211x

X^2 - 0.2511x + 0.00651 = 0

X = \frac{0.2511 \pm \sqrt{0.06305 - 0.02604}}{2} = \frac{0.2511 \pm 0.1923}{2}

$x = 0.2217$ or $x = 0.0294$ .

Since $x \le 0.0650$ : $x = 0.0294$ .

New equilibrium: [PCl $_5$ ] $= 0.229\mathrm{ M}$ [PCl $_3$ ] $= 0.0356\mathrm{ M}$ [Cl $_2$ ] $= 0.136\mathrm{ M}$ .

Acid-Base Extended: Polyprotic Acids

Polyprotic acids can donate more than one proton.

Carbonic acid (H $_2$ CO $_3$ ):

\mathrm{H}_2\mathrm{CO}_3 \rightleftharpoons \mathrm{H}^+ + \mathrm{HCO}_3^- \quad K_{a1} = 4.3 \times 10^{-7}

\mathrm{HCO}_3^- \rightleftharpoons \mathrm{H}^+ + \mathrm{CO}_3^{2-} \quad K_{a2} = 4.8 \times 10^{-11}

Note: $K_{a1} \gg K_{a2}$ So the first dissociation dominates.

pH of Salt Solutions

Salt of strong acid + strong base: neutral (pH = 7).
Salt of strong acid + weak base: acidic (pH $\lt$ 7).
Salt of weak acid + strong base: basic (pH $\gt$ 7).
Salt of weak acid + weak base: depends on relative $K_a$ and $K_b$ .

Example Calculate the pH of $0.10\mathrm{ M}$ sodium ethanoate (CH $_3$ COONa). $K_a$ (CH $_3$ COOH) $= 1.8 \times 10^{-5}$ .

K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}

[\mathrm{OH}^-] = \sqrt{K_b \times c} = \sqrt{5.56 \times 10^{-10} \times 0.10} = 7.46 \times 10^{-6}

\mathrm{pOH} = 5.13, \quad \mathrm{pH} = 14 - 5.13 = 8.87

The solution is basic, as expected for the salt of a weak acid and strong base.

Additional IB Exam-Style Questions

Question 5 (Paper 2 style)

For the equilibrium: $2\mathrm{SO}_2(\mathrm{g}) + \mathrm{O}_2(\mathrm{g}) \rightleftharpoons 2\mathrm{SO}_3(\mathrm{g})$ $\Delta H = -198\mathrm{ kJ/mol}$ .

(a) Explain the effect of increasing temperature on the equilibrium yield of SO $_3$ .

Since the forward reaction is exothermic ( $\Delta H \lt 0$ ), increasing temperature shifts the Equilibrium to the left (endothermic direction) by Le Chatelier’s principle. This decreases the Yield of SO $_3$ and decreases $K$ (since $\ln K \propto -\Delta H / T$ Increasing $T$ for an Exothermic reaction reduces $K$ ).

(b) Explain the effect of increasing pressure on the equilibrium yield of SO $_3$ .

There are 3 moles of gas on the left and 2 on the right. Increasing pressure shifts the equilibrium To the right (fewer moles), increasing the yield of SO $_3$ .

(c) Explain why a catalyst does not change the equilibrium yield.

A catalyst increases the rate of both forward and reverse reactions equally, so the equilibrium Position is unchanged. It only helps the system reach equilibrium faster.

Question 6 (Paper 1 style)

What is the pH of a $0.010\mathrm{ M}$ solution of Ba(OH) $_2$ ?

Ba(OH) $_2$ is a strong base that dissociates completely: Ba(OH) $_2$ $\to$ Ba $^{2+}$ + 2OH $^-$ .

[\mathrm{OH}^-] = 2 \times 0.010 = 0.020\mathrm{ M}

\mathrm{pOH} = -\log(0.020) = 1.70

\mathrm{pH} = 14 - 1.70 = 12.30

Question 7 (Paper 2 style)

The solubility of PbI $_2$ at $25\degree\mathrm{C}$ is $1.4 \times 10^{-3}\mathrm{ mol/L}$ .

(a) Calculate $K_{sp}$ for PbI $_2$ .

\mathrm{PbI}_2(s) \rightleftharpoons \mathrm{Pb}^{2+}(aq) + 2\mathrm{I}^-(aq)

[\mathrm{Pb}^{2+}] = 1.4 \times 10^{-3}\mathrm{ M}, \quad [\mathrm{I}^-] = 2(1.4 \times 10^{-3}) = 2.8 \times 10^{-3}\mathrm{ M}

K_`\{sp}` = [\mathrm{Pb}^{2+}][\mathrm{I}^-]^2 = (1.4 \times 10^{-3})(2.8 \times 10^{-3})^2 = (1.4 \times 10^{-3})(7.84 \times 10^{-6})

= 1.1 \times 10^{-8}

(b) Will a precipitate form when $50\mathrm{ mL}$ of $0.010\mathrm{ M}$ Pb(NO $_3$ ) $_2$ is mixed With $50\mathrm{ mL}$ of $0.020\mathrm{ M}$ KI?

After mixing (volumes double, concentrations halve):

[\mathrm{Pb}^{2+}] = 0.005\mathrm{ M}, \quad [\mathrm{I}^-] = 0.010\mathrm{ M}

Q = [\mathrm{Pb}^{2+}][\mathrm{I}^-]^2 = (0.005)(0.010)^2 = 5.0 \times 10^{-7}

Since $Q = 5.0 \times 10^{-7} \gt K_{sp} = 1.1 \times 10^{-8}$ A precipitate of PbI $_2$ will form.

Question 8 (Paper 1 style)

Which indicator would be most suitable for the titration of a weak acid (CH $_3$ COOH) with a strong Base (NaOH)?

A. Methyl orange (pH range 3.1—4.4) B. Bromothymol blue (pH range 6.0—7.6) C. Phenolphthalein (pH Range 8.3—10.0)

Answer: C. A weak acid-strong base titration has an equivalence point with pH $\gt$ 7, so Phenolphthalein is appropriate.

Practice Problems

Question 1: ICE Table and $K_c$ Calculation

For the reaction $\mathrm{N}_2\mathrm{O}_4(g) \rightleftharpoons 2\mathrm{NO}_2(g)$ $1.00\mathrm{ mol}$ of $\mathrm{N}_2\mathrm{O}_4$ is placed in a $2.00\mathrm{ L}$ flask at $350\mathrm{ K}$ . At equilibrium, the concentration of $\mathrm{NO}_2$ is $0.200\mathrm{ mol/L}$ . Calculate $K_c$ .

Answer

Initial $[\mathrm{N}_2\mathrm{O}_4] = 1.00 / 2.00 = 0.500\mathrm{ mol/L}$

	$\mathrm{N}_2\mathrm{O}_4$	$\mathrm{NO}_2$
Initial	$0.500$	$0$
Change	$-0.100$	$+0.200$
Equilibrium	$0.400$	$0.200$

Since $[\mathrm{NO}_2]_{\mathrm{eq}} = 0.200\mathrm{ mol/L}$ and it increases by $2x$ , $x = 0.100$ .

$K_c = \frac{[\mathrm{NO}_2]^2}{[\mathrm{N}_2\mathrm{O}_4]} = \frac{(0.200)^2}{0.400} = \frac{0.0400}{0.400} = 0.100$

Question 2: Le Chatelier's Principle

For the exothermic reaction $\mathrm{N}_2(g) + 3\mathrm{H}_2(g) \rightleftharpoons 2\mathrm{NH}_3(g)$ Predict and explain the Effect of each change on the equilibrium yield of $\mathrm{NH}_3$ :

(a) Increasing pressure

(b) Increasing temperature

Answer

(a) There are 4 moles of gas on the left and 2 on the right. Increasing pressure shifts the Equilibrium to the right (fewer moles of gas), increasing the yield of $\mathrm{NH}_3$ .

(b) The forward reaction is exothermic ( $\Delta H \lt 0$ ). Increasing temperature shifts the Equilibrium to the left (endothermic direction), decreasing the yield of $\mathrm{NH}_3$ .

(c) A catalyst increases the rate of both forward and reverse reactions equally. It does not Change the equilibrium position or the yield of $\mathrm{NH}_3$ . It only helps the system reach Equilibrium faster.

Question 3: Buffer pH Calculation

A buffer solution is prepared by mixing $100\mathrm{ mL}$ of $0.20\mathrm{ M}$ $\mathrm{CH}_3\mathrm{COOH}$ ( $\mathrm{p}K_a = 4.76$ ) with $50\mathrm{ mL}$ of $0.20\mathrm{ M}$ $\mathrm{NaOH}$ . Calculate the pH of the resulting buffer.

Answer

The $\mathrm{NaOH}$ reacts with $\mathrm{CH}_3\mathrm{COOH}$ :

$n(\mathrm{CH}_3\mathrm{COOH})_{\mathrm{initial}} = 0.100 \times 0.20 = 0.0200\mathrm{ mol}$

$n(\mathrm{NaOH}) = 0.050 \times 0.20 = 0.0100\mathrm{ mol}$

After reaction:

$n(\mathrm{CH}_3\mathrm{COOH})_{\mathrm{remaining}} = 0.0200 - 0.0100 = 0.0100\mathrm{ mol}$

$n(\mathrm{CH}_3\mathrm{COO}^-)_{\mathrm{formed}} = 0.0100\mathrm{ mol}$

Total volume = $150\mathrm{ mL} = 0.150\mathrm{ L}$ :

$[\mathrm{CH}_3\mathrm{COOH}] = \frac{0.0100}{0.150} = 0.0667\mathrm{ M}$

$[\mathrm{CH}_3\mathrm{COO}^-] = \frac{0.0100}{0.150} = 0.0667\mathrm{ M}$

$\mathrm{pH} = \mathrm{p}K_a + \log\frac{[\mathrm{A}^-]}{[\mathrm{HA}]} = 4.76 + \log\frac{0.0667}{0.0667} = 4.76 + \log(1) = 4.76$

Question 4: Solubility Product and Common Ion Effect

The $K_{sp}$ of $\mathrm{PbCl}_2$ is $1.7 \times 10^{-5}$ at $25\degree\mathrm{C}$ .

(a) Calculate the molar solubility of $\mathrm{PbCl}_2$ in pure water.

(b) Calculate the molar solubility of $\mathrm{PbCl}_2$ in $0.10\mathrm{ M}$ $\mathrm{NaCl}$ Solution.

Answer

(a) Let $s$ = molar solubility of $\mathrm{PbCl}_2$ :

$\mathrm{PbCl}_2(s) \rightleftharpoons \mathrm{Pb}^{2+}(aq) + 2\mathrm{Cl}^-(aq)$

$K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{Cl}^-]^2 = s \times (2s)^2 = 4s^3$

$s^3 = \frac{1.7 \times 10^{-5}}{4} = 4.25 \times 10^{-6}$

$s = 1.62 \times 10^{-2}\mathrm{ mol/L}$

(b) In $0.10\mathrm{ M}$ $\mathrm{NaCl}$ , $[\mathrm{Cl}^-]_{\mathrm{initial}} = 0.10\mathrm{ M}$ :

$K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{Cl}^-]^2 = s \times (0.10 + 2s)^2 \approx s \times (0.10)^2$

$s = \frac{1.7 \times 10^{-5}}{0.010} = 1.7 \times 10^{-3}\mathrm{ mol/L}$

The solubility decreases significantly due to the common ion effect.

Question 5: Weak Acid pH

Calculate the pH of a $0.050\mathrm{ M}$ solution of $\mathrm{HF}$ . ( $K_a = 6.8 \times 10^{-4}$ )

Answer

$[\mathrm{H}^+] = \sqrt{K_a \times c} = \sqrt{6.8 \times 10^{-4} \times 0.050} = \sqrt{3.4 \times 10^{-5}} = 5.83 \times 10^{-3}\mathrm{ mol/L}$

$\mathrm{pH} = -\log(5.83 \times 10^{-3}) = 2.23$

A-Level Equilibria: Chemical Equilibrium

Common Pitfalls

Misapplying Le Chatelier’s principle. It predicts the direction of change, not the extent.
Confusing $K_c$ and $K_p$ — $K_c$ uses concentrations; $K_p$ uses partial pressures, and they only apply to their respective phases.
Confusing the terms ‘molar’ and ‘molecular’. Molar refers to per mole ( $\text{mol}^{-1}$ ), while molecular refers to individual molecules.
Assuming that a strong acid always has a lower pH than a weak acid without considering concentration.
Forgetting to balance equations before performing calculations. Always check that atoms and charges balance on both sides.
Misidentifying the limiting reagent. Compare mole ratios rather than comparing masses.

Cross-References

Topic	Site	Link
[Chemical Equilibrium]	A-Level	View
[Chemical Equilibrium]	IB	View
[Chemical Equilibrium]	DSE	View

Worked Examples

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.

Chemical Equilibrium - Wyatt's Notes

Intuition

Dynamic Equilibrium

Reversible Reactions

Closed Systems

Dynamic Nature

Characteristics of Equilibrium

The Equilibrium Constant

KcK_cKc​ — Concentration Equilibrium Constant

KpK_pKp​ — Pressure Equilibrium Constant

Relationship Between KcK_cKc​ and KpK_pKp​

See Also

$K_c$ — Concentration Equilibrium Constant

$K_p$ — Pressure Equilibrium Constant

Relationship Between $K_c$ and $K_p$