What is the relation between Kp and Kc?

The equilibrium constants are related by Kp = Kc · (RT)^(Δn_g), where Δn_g = sum of gaseous product coefficients - sum of gaseous reactant coefficients.

What is Le Chatelier’s Principle?

Le Chatelier’s Principle states that if a dynamic equilibrium is disturbed by changing conditions (temperature, pressure, or concentration), the position of equilibrium moves to counteract the change.

What is a buffer solution and how does it resist pH change?

A buffer solution maintains a nearly constant pH upon addition of small amounts of acid or base. An acidic buffer consists of a weak acid and its salt with a strong base (e.g., CH₃COOH + CH₃COONa), where Henderson-Hasselbalch equation gives pH = pKa + log([Salt]/[Acid]).

What is the common ion effect and its application in qualitative analysis?

The common ion effect is the suppression of dissociation of a weak electrolyte by the addition of a strong electrolyte providing a common ion. For example, NH₄Cl suppresses NH₄OH dissociation in Group III cation precipitation.

ALL UNITS

Unit 06 • Physical Chemistry

Equilibrium

Class 11 Chemistry Chapter 6 complete study module. Dynamic nature of equilibrium, Law of Mass Action, $K_p$ vs $K_c$, Le Chatelier's Principle, acid-base theories (Arrhenius, Brönsted-Lowry, Lewis), pH scale, ionic product of water ($K_w$), ionization of weak electrolytes ($K_a, K_b$), common ion effect, salt hydrolysis, Henderson-Hasselbalch equation for buffer solutions, solubility product ($K_{sp}$), in-text problems (6.16.28), solved NCERT exercises (6.1 to 6.73), revision cheat-sheets, and 3-level tests.

Introduction: Dynamic Equilibrium

Chemical equilibria are fundamental to living systems and the chemical industry. In human physiology, equilibria involving $O_2$ and hemoglobin deliver oxygen from lungs to tissues, while competitive binding by $CO$ accounts for carbon monoxide poisoning.

Dynamic Nature of Equilibrium: When a liquid evaporates in a closed container, molecules with high kinetic energy escape into the vapour phase, while vapour molecules collide with the liquid surface and condense. At equilibrium:

$$\mathbf{\text{Rate of Forward Process} = \text{Rate of Reverse Process}}$$ $$H_2O_{(l)} \rightleftharpoons H_2O_{(g)}$$

The double half arrows ($\rightleftharpoons$) indicate that forward and reverse reactions proceed simultaneously at equal rates. The concentration of reactants and products remains strictly constant with time, but molecular exchange never ceases!

Liquid-Vapour Dynamic Equilibrium & Vapour Pressure (Manometer Setup)

6.1 Equilibrium in Physical Processes

Physical Process	Equilibrium Reaction	Constant Parameter at Given Temperature
Solid ⇌ Liquid	$H_2O_{(s)} \rightleftharpoons H_2O_{(l)}$	Melting point is fixed at constant pressure ($273\text{ K}$ at $1.013\text{ bar}$).
Liquid ⇌ Vapour	$H_2O_{(l)} \rightleftharpoons H_2O_{(g)}$	Vapour pressure ($p_{H_2O}$) is constant at a given temperature.
Solid ⇌ Vapour	$I_{2(s)} \rightleftharpoons I_{2(\text{vap})}$, $\text{Camphor}_{(s)} \rightleftharpoons \text{Camphor}_{(\text{vap})}$	Sublimation pressure / colour intensity of vapour is constant.
Dissolution of Solid in Liquid	$\text{Sugar}_{(s)} \rightleftharpoons \text{Sugar}_{(\text{soln})}$	Solubility is constant in a saturated solution ($\text{Rate}_{\text{dissol}} = \text{Rate}_{\text{cryst}}$).
Dissolution of Gas in Liquid	$CO_{2(g)} \rightleftharpoons CO_{2(aq)}$	$\frac{[CO_{2(aq)}]}{[CO_{2(g)}]} = \text{constant}$ (governed by Henry's Law: $m = k_H \cdot p$).

General Characteristics of Physical Equilibria

Equilibrium is achievable only in a closed system at a specified constant temperature.
Both opposing forward and reverse processes occur at identical rates (dynamic stability).
All measurable macroscopic properties (pressure, density, concentration, colour intensity) remain constant.
The magnitude of the equilibrium constant indicates the extent to which the process proceeds before balance is reached.

6.2 & 6.3 Law of Chemical Equilibrium and Equilibrium Constant ($K_c$)

In 1864, Cato Guldberg and Peter Waage formulated the Law of Mass Action: at a given temperature, the rate of a chemical reaction is directly proportional to the product of active masses (molar concentrations) of the reacting species.

For a general reversible reaction: $aA + bB \rightleftharpoons cC + dD$

$$\mathbf{K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}}$$ Where $[A], [B], [C], [D]$ are equilibrium molar concentrations in $\text{mol L}^{-1}$.

Attainment of Chemical Equilibrium: Reaction Rates and Concentration Profiles

Rules for Manipulating Equilibrium Constants

Chemical Equation Modification	New Equilibrium Constant	Mathematical Example
Reverse reaction	$K' = \frac{1}{K}$	$2HI \rightleftharpoons H_2 + I_2 \implies K' = \frac{1}{K_c}$
Multiply by coefficient $n$	$K'' = (K)^n$	$2A + 2B \rightleftharpoons 2C \implies K'' = K_c^2$
Divide by factor $n$ ($1/n$)	$K''' = (K)^{1/n} = \sqrt[n]{K}$	$\frac{1}{2}N_2 + \frac{3}{2}H_2 \rightleftharpoons NH_3 \implies K''' = \sqrt{K_c}$
Add two reactions	$K_{\text{net}} = K_1 \times K_2$	Step 1 + Step 2 $\implies$ product of individual constants

6.1

The following concentrations were obtained for the formation of NH₃ from N₂ and H₂ at equilibrium at 500 K: [N₂] = 1.5 × 10⁻² M, [H₂] = 3.0 × 10⁻² M, and [NH₃] = 1.2 × 10⁻² M. Calculate the equilibrium constant Kc.

Reaction: $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$
$$K_c = \frac{[NH_3]^2}{[N_2][H_2]^3} = \frac{(1.2 \times 10^{-2})^2}{(1.5 \times 10^{-2})(3.0 \times 10^{-2})^3}$$ $$= \frac{1.44 \times 10^{-4}}{(1.5 \times 10^{-2})(27 \times 10^{-6})} = \frac{1.44 \times 10^{-4}}{4.05 \times 10^{-7}} = \mathbf{3.55 \times 10^2\text{ L}^2\text{mol}^{-2}}$$

Kc = 3.55 × 10² L² mol⁻².

6.2

At equilibrium, the concentrations of N₂ = 3.0 × 10⁻³ M, O₂ = 4.2 × 10⁻³ M and NO = 2.8 × 10⁻³ M in a sealed vessel at 800 K. What will be Kc for: N₂(g) + O₂(g) ⇌ 2NO(g)?

$$K_c = \frac{[NO]^2}{[N_2][O_2]} = \frac{(2.8 \times 10^{-3})^2}{(3.0 \times 10^{-3})(4.2 \times 10^{-3})} = \frac{7.84 \times 10^{-6}}{1.26 \times 10^{-5}} = \mathbf{0.622}$$

Kc = 0.622 (Dimensionless because Δn = 0).

6.4 Gaseous Systems: Relation Between $K_p$ and $K_c$

For gas-phase reactions, using partial pressures ($p_i$) expressed in bar is standard practice. Using the ideal gas law $p = \frac{n}{V}RT = cRT$:

$$\mathbf{K_p = K_c (RT)^{\Delta n_g}}$$ Where $\Delta n_g = \sum n_{\text{gaseous products}} - \sum n_{\text{gaseous reactants}}$, $R = 0.0831\text{ bar L mol}^{-1}\text{K}^{-1}$ (or $0.0821\text{ atm L mol}^{-1}\text{K}^{-1}$).

If $\mathbf{\Delta n_g = 0} \implies \mathbf{K_p = K_c}$ (e.g., $H_2 + I_2 \rightleftharpoons 2HI$). Both are dimensionless.
If $\mathbf{\Delta n_g > 0} \implies \mathbf{K_p > K_c}$ (at $RT > 1$, e.g., $PCl_5 \rightleftharpoons PCl_3 + Cl_2, \Delta n_g = +1$).
If $\mathbf{\Delta n_g < 0} \implies \mathbf{K_p < K_c}$ (e.g., $N_2 + 3H_2 \rightleftharpoons 2NH_3, \Delta n_g = -2$).

6.3

PCl₅, PCl₃ and Cl₂ are at equilibrium at 500 K with concentrations 1.59 M PCl₃, 1.59 M Cl₂ and 1.41 M PCl₅. Calculate Kc for PCl₅ ⇌ PCl₃ + Cl₂.

$$K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]} = \frac{1.59 \times 1.59}{1.41} = \frac{2.5281}{1.41} = \mathbf{1.79\text{ M}}$$

Kc = 1.79 mol L⁻¹.

6.5

For the equilibrium 2NOCl(g) ⇌ 2NO(g) + Cl₂(g), Kc = 3.75 × 10⁻⁶ at 1069 K. Calculate Kp for the reaction at this temperature.

$\Delta n_g = (2 + 1) - 2 = \mathbf{+1}$.
$K_p = K_c (RT)^{\Delta n_g} = (3.75 \times 10^{-6}) \times (0.0831 \times 1069)^1 = 3.75 \times 10^{-6} \times 88.834 = \mathbf{0.0333}\text{ (or } 3.33 \times 10^{-2}\text{)}$.

Kp = 0.0333 bar.

6.5 Heterogeneous Equilibria

When reactants and products exist in different physical phases, it is termed heterogeneous equilibrium.

Pure Solids and Pure Liquids Rule

The molar concentration (density divided by molar mass) of a pure solid or liquid is constant and independent of the amount present: $$[\text{Pure Solid}] = \text{constant}, \quad [\text{Pure Liquid}] = \text{constant}$$ Therefore, their concentrations are omitted from the equilibrium expression!

Thermal decomposition of limestone: $$CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)} \implies \mathbf{K_c = [CO_2]}, \quad \mathbf{K_p = p_{CO_2}}$$ At any given temperature, there is a fixed constant pressure of $CO_2$ in equilibrium with $CaCO_3$ and $CaO$.
Nickel purification (Mond process): $$Ni_{(s)} + 4CO_{(g)} \rightleftharpoons Ni(CO)_{4(g)} \implies \mathbf{K_c = \frac{[Ni(CO)_4]}{[CO]^4}}$$

6.6

The value of Kp for CO₂(g) + C(s) ⇌ 2CO(g) is 3.0 at 1000 K. If initially p_CO₂ = 0.48 bar and p_CO = 0 bar with pure graphite present, calculate the equilibrium partial pressures of CO and CO₂.

At equilibrium: $p_{CO_2} = 0.48 - x$ bar, $p_{CO} = 2x$ bar.
$$K_p = \frac{p_{CO}^2}{p_{CO_2}} = \frac{(2x)^2}{0.48 - x} = 3.0 \implies 4x^2 = 1.44 - 3x \implies 4x^2 + 3x - 1.44 = 0$$ Solving quadratic: $x = \frac{-3 + \sqrt{9 - 4(4)(-1.44)}}{8} = \frac{-3 + \sqrt{32.04}}{8} = \frac{-3 + 5.66}{8} = \mathbf{0.33\text{ bar}}$.
$p_{CO} = 2x = 2(0.33) = \mathbf{0.66\text{ bar}}$.
$p_{CO_2} = 0.48 - 0.33 = \mathbf{0.15\text{ bar}}$.

p_CO = 0.66 bar; p_CO₂ = 0.15 bar.

6.6 Applications: Extent, Direction ($Q_c$ vs $K_c$), and $\Delta G^\circ$

Extent of Reaction as a Function of Equilibrium Constant (K)

Predicting Reaction Direction: The Reaction Quotient ($Q$)

The reaction quotient $Q_c$ is calculated using the exact same mathematical formula as $K_c$, but using non-equilibrium / instantaneous concentrations at any given arbitrary time $t$:

If $Q_c < K_c$: Ratio of products to reactants is less than at equilibrium $\implies$ Net reaction proceeds forward ($\rightarrow$).
If $Q_c = K_c$: System is at dynamic equilibrium ($\rightleftharpoons$).
If $Q_c > K_c$: Excess products present $\implies$ Net reaction proceeds backward ($\leftarrow$).

Relation between Equilibrium Constant ($K$), Reaction Quotient ($Q$) and $\Delta G$

$$\Delta G = \Delta G^\circ + RT \ln Q$$ $$\text{At equilibrium, } \Delta G = 0 \text{ and } Q = K \implies \mathbf{\Delta_r G^\circ = - RT \ln K = - 2.303 \, RT \log K}$$ $$\mathbf{K = e^{-\Delta G^\circ / RT}}$$

6.7

The value of Kc for 2A ⇌ B + C is 2 × 10⁻³. At a given time, [A] = [B] = [C] = 3 × 10⁻⁴ M. In which direction will the reaction proceed?

$$Q_c = \frac{[B][C]}{[A]^2} = \frac{(3 \times 10^{-4})(3 \times 10^{-4})}{(3 \times 10^{-4})^2} = \mathbf{1}$$ Since $Q_c (1) > K_c (2 \times 10^{-3})$, the reaction will proceed in the reverse direction ($\leftarrow$) to form more reactants.

Qc = 1 > Kc. Reaction proceeds in the reverse (backward) direction.

6.10

The value of ΔG° for phosphorylation of glucose in glycolysis is 13.8 kJ/mol. Find the value of Kc at 298 K.

$\ln K_c = \frac{-\Delta G^\circ}{RT} = \frac{-13800\text{ J mol}^{-1}}{8.314 \times 298} = -5.569$.
$K_c = e^{-5.569} = \mathbf{3.81 \times 10^{-3}}$.

Kc = 3.81 × 10⁻³ (Positive ΔG° indicates reactants predominate at equilibrium).

6.11

Hydrolysis of sucrose gives Sucrose + H₂O ⇌ Glucose + Fructose. Kc = 2 × 10¹³ at 300 K. Calculate ΔG° at 300 K.

$\Delta G^\circ = - RT \ln K_c = - 8.314\text{ J K}^{-1}\text{mol}^{-1} \times 300\text{ K} \times \ln(2 \times 10^{13})$
$= - 2494.2 \times 30.627 = \mathbf{-7.64 \times 10^4\text{ J mol}^{-1}} = \mathbf{-76.4\text{ kJ mol}^{-1}}$.

ΔG° = −76.4 kJ mol⁻¹.

6.8 Le Chatelier's Principle

Le Chatelier's Principle:
"If a system at equilibrium is subjected to a change in concentration, pressure, volume, or temperature, the system shifts in a direction that tends to counteract or nullify the effect of the imposed change."

Le Chatelier's Principle: Summary of Equilibrium Shifts

Industrial Application: Haber Process for $NH_3$

$$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)} \qquad \Delta_r H = -92.38\text{ kJ mol}^{-1}$$ To maximise ammonia yield according to Le Chatelier's principle:

High Pressure (~200 atm): Shifts reaction forward because $4\text{ moles gas} \rightarrow 2\text{ moles gas}$ ($\Delta n_g = -2$).
Optimum Moderate Temperature (~500 °C / 773 K): Since the reaction is exothermic, low temperature favours equilibrium yield, but slows kinetics. $500^\circ\text{C}$ provides an optimum economic rate.
Iron Catalyst with $K_2O / Al_2O_3$ promoter: Rapidly establishes equilibrium without altering yield.
Continuous Liquefaction & Removal of $NH_3$: Keeps $Q_c < K_c$, permanently driving the reaction forward.

6.9 & 6.10 Acid-Base Theories: Arrhenius, Brönsted-Lowry & Lewis

Theory	Definition of Acid	Definition of Base	Limitation / Scope
Arrhenius	Produces $H^+_{(aq)}$ / $H_3O^+_{(aq)}$ in water	Produces $OH^-_{(aq)}$ in water	Restricted to aqueous medium; fails to explain basicity of $NH_3$.
Brönsted-Lowry	Proton ($H^+$) Donor	Proton ($H^+$) Acceptor	Applies to non-aqueous media; introduces conjugate acid-base pairs.
Lewis	Electron-pair Acceptor (electrophile)	Electron-pair Donor (nucleophile)	Broadest concept; includes non-protonic acids ($BF_3, AlCl_3$).

Brönsted-Lowry Conjugate Acid-Base Pairs & Amphiprotic Nature of Water

6.12

What will be the conjugate bases for the following Brönsted acids: HF, H₂SO₄ and HCO₃⁻?

Remove one proton ($H^+$) from each acid:

$HF - H^+ \implies \mathbf{F^-}$
$H_2SO_4 - H^+ \implies \mathbf{HSO_4^-}$
$HCO_3^- - H^+ \implies \mathbf{CO_3^{2-}}$

Conjugate bases: F⁻, HSO₄⁻, CO₃²⁻.

6.13

Write the conjugate acids for the following Brönsted bases: NH₂⁻, NH₃ and HCOO⁻.

Add one proton ($H^+$) to each base:

$NH_2^- + H^+ \implies \mathbf{NH_3}$
$NH_3 + H^+ \implies \mathbf{NH_4^+}$
$HCOO^- + H^+ \implies \mathbf{HCOOH}$

Conjugate acids: NH₃, NH₄⁺, HCOOH.

6.15

Classify the following species into Lewis acids and Lewis bases: (a) OH⁻ (b) F⁻ (c) H⁺ (d) BCl₃.

(a) OH⁻: Lewis Base (donates an electron lone pair $:OH^-$).
(b) F⁻: Lewis Base (has 4 lone pairs to donate).
(c) H⁺: Lewis Acid (has vacant orbital to accept an electron pair).
(d) BCl₃: Lewis Acid (electron deficient central B atom with incomplete sextet).

Lewis bases: OH⁻, F⁻; Lewis acids: H⁺, BCl₃.

6.11 Ionization of Water, pH Scale, and Weak Electrolytes

Ionic Product of Water ($K_w$)

Water undergoes auto-protolysis: $H_2O_{(l)} + H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)} + OH^-_{(aq)}$

$$\mathbf{K_w = [H^+][OH^-] = 1.0 \times 10^{-14} \quad (\text{at } 298\text{ K})}$$ $$\mathbf{pH + pOH = pK_w = 14.00 \quad (\text{at } 298\text{ K})}$$ Note: Auto-ionization is endothermic. As $T \uparrow$, $K_w \uparrow$, and neutral water pH drops below 7 (e.g., at $310\text{ K}$, $K_w = 2.7 \times 10^{-14} \implies \text{neutral pH} = 6.78$).

The Complete pH Scale (0 to 14) at 298 K with Common Substances

Ionization of Weak Acids & Ostwald's Dilution Law

For a weak acid $HA \rightleftharpoons H^+ + A^-$ with initial concentration $c$ and degree of dissociation $\alpha$: $$K_a = \frac{c\alpha \cdot c\alpha}{c(1 - \alpha)} = \frac{c\alpha^2}{1 - \alpha}$$ For weak electrolytes ($\alpha \ll 1 \implies 1 - \alpha \approx 1$):

$$\mathbf{\alpha = \sqrt{\frac{K_a}{c}} = \sqrt{K_a \cdot V}}$$ $$\mathbf{[H^+] = c\alpha = \sqrt{K_a \cdot c}}$$ $$\mathbf{pH = \frac{1}{2}(pK_a - \log c)}$$

Conjugate Acid-Base Relation

$$\mathbf{K_a \times K_b = K_w = 1.0 \times 10^{-14}}$$ $$\mathbf{pK_a + pK_b = pK_w = 14.00}$$

6.16

The concentration of hydrogen ion in a sample of soft drink is 3.8 × 10⁻³ M. What is its pH?

$$pH = -\log[H^+] = -\log(3.8 \times 10^{-3}) = -[\log 3.8 + \log 10^{-3}] = -(0.58 - 3.0) = \mathbf{2.42}$$

pH = 2.42 (Acidic soft drink).

6.17

Calculate pH of a 1.0 × 10⁻⁸ M solution of HCl.

In extremely dilute solutions ($c \le 10^{-6}\text{ M}$), $H^+$ from water auto-ionization must be included!
$[H^+]_{\text{total}} = 10^{-8} + x$, where $x = [OH^-] = [H^+]_{\text{water}}$.
$$K_w = [H^+][OH^-] = (10^{-8} + x)(x) = 10^{-14} \implies x^2 + 10^{-8}x - 10^{-14} = 0$$ $$x = \frac{-10^{-8} + \sqrt{10^{-16} + 4(10^{-14})}}{2} = 9.5 \times 10^{-8}\text{ M}$$ $[H^+]_{\text{total}} = 1.0 \times 10^{-8} + 9.5 \times 10^{-8} = \mathbf{1.05 \times 10^{-7}\text{ M}}$.
$$pH = -\log(1.05 \times 10^{-7}) = 7 - 0.021 = \mathbf{6.98}$$ (A dilute acid solution can never have $pH > 7$!)

pH = 6.98.

6.12 Salt Hydrolysis and Buffer Solutions

Salt Hydrolysis Types and pH Formulas

Salt Type	Undergoes Hydrolysis	Nature of Solution	pH Formula at 298 K	Key Example
Strong Acid + Strong Base	No hydrolysis (hydration only)	Neutral ($\text{pH} = 7$)	$$\mathbf{pH = 7}$$	$NaCl, KNO_3, BaCl_2$
Weak Acid + Strong Base	Anion hydrolysis ($A^- + H_2O \rightleftharpoons HA + OH^-$)	Basic ($\text{pH} > 7$)	$$\mathbf{pH = 7 + \frac{1}{2}(pK_a + \log c)}$$	$CH_3COONa, KCN$
Strong Acid + Weak Base	Cation hydrolysis ($B^+ + H_2O \rightleftharpoons BOH + H^+$)	Acidic ($\text{pH} < 7$)	$$\mathbf{pH = 7 - \frac{1}{2}(pK_b + \log c)}$$	$NH_4Cl, (NH_4)_2SO_4$
Weak Acid + Weak Base	Both Cation & Anion hydrolysis	Depends on $pK_a - pK_b$	$$\mathbf{pH = 7 + \frac{1}{2}(pK_a - pK_b)}$$	$CH_3COONH_4$

Buffer Solutions & Henderson-Hasselbalch Equations

A buffer solution resists changes in pH upon the addition of small quantities of strong acid, strong base, or upon dilution.

Mechanism of Acidic Buffer Action (CH₃COOH + CH₃COONa)

On Addition of Strong Acid (H⁺): CH₃COO⁻ + H⁺ ⇌ CH₃COOH (Weak, unionized!) pH remains ~constant On Addition of Strong Base (OH⁻): CH₃COOH + OH⁻ ⇌ CH₃COO⁻ + H₂O pH remains ~constant

Henderson-Hasselbalch Equation for Acidic Buffer: $$\mathbf{pH = pK_a + \log \frac{[\text{Salt}]}{[\text{Acid}]}}$$ For Basic Buffer (e.g., $NH_4OH + NH_4Cl$): $$\mathbf{pOH = pK_b + \log \frac{[\text{Salt}]}{[\text{Base}]}} \implies \mathbf{pH = 14 - pOH}$$ When $[\text{Salt}] = [\text{Acid}]$, $\mathbf{pH = pK_a}$ (Maximum buffer capacity). Dilution does not alter buffer pH!

6.25

The pKa of acetic acid and pKb of ammonium hydroxide are 4.76 and 4.75 respectively. Calculate the pH of ammonium acetate solution.

$$pH = 7 + \frac{1}{2}[pK_a - pK_b] = 7 + \frac{1}{2}[4.76 - 4.75] = 7 + \frac{1}{2}(0.01) = \mathbf{7.005}$$

pH = 7.005 (Practically neutral).

6.13 Solubility Equilibria & Solubility Product ($K_{sp}$)

For a sparingly soluble salt $M_x X_y$ dissolving with molar solubility $S\text{ mol L}^{-1}$: $$M_x X_{y(s)} \rightleftharpoons x M^{p+}_{(aq)} + y X^{q-}_{(aq)}$$ $$[M^{p+}] = xS, \quad [X^{q-}] = yS$$

$$\mathbf{K_{sp} = [M^{p+}]^x [X^{q-}]^y = (xS)^x (yS)^y = x^x y^y \cdot S^{(x+y)}}$$ $$\mathbf{S = \left(\frac{K_{sp}}{x^x y^y}\right)^{\frac{1}{x+y}}}$$

Precipitation Criteria: Comparing Ionic Product (Q_sp) with Solubility Product (K_sp)

Common Ion Effect on Solubility

In accordance with Le Chatelier's principle, adding a soluble salt that provides a common ion suppresses the ionization and drastically reduces the solubility of a sparingly soluble salt: $$\text{Example: } AgCl_{(s)} \rightleftharpoons Ag^+_{(aq)} + Cl^-_{(aq)}$$ Adding $NaCl$ increases $[Cl^-]$, driving the equilibrium to the left and precipitating $AgCl$.

6.26

Calculate the solubility of A₂X₃ in pure water, assuming neither ion reacts with water. The solubility product Ksp = 1.1 × 10⁻²³.

$A_2X_3 \rightleftharpoons 2A^{3+} + 3X^{2-}$.
$K_{sp} = (2S)^2 (3S)^3 = 4S^2 \times 27S^3 = \mathbf{108 S^5}$.
$$108 S^5 = 1.1 \times 10^{-23} \implies S^5 = \frac{1.1 \times 10^{-23}}{108} = 1.018 \times 10^{-25} \approx 1.0 \times 10^{-25}$$ $$S = (1.0 \times 10^{-25})^{1/5} = \mathbf{1.0 \times 10^{-5}\text{ mol L}^{-1}}$$

S = 1.0 × 10⁻⁵ mol L⁻¹.

6.28

Calculate the molar solubility of Ni(OH)₂ in 0.10 M NaOH. (Ksp of Ni(OH)₂ = 2.0 × 10⁻¹⁵).

$[OH^-] = 0.10\text{ M}$ (from strong base $NaOH$; $2S \ll 0.10$).
$$K_{sp} = [Ni^{2+}][OH^-]^2 = S \times (0.10)^2 = 2.0 \times 10^{-15}$$ $$S = \frac{2.0 \times 10^{-15}}{0.01} = \mathbf{2.0 \times 10^{-13}\text{ M}}$$ (Notice that solubility in pure water was $7.9 \times 10^{-6}\text{ M}$, reduced by a factor of 40 million due to the common ion effect!)

Solubility = 2.0 × 10⁻¹³ M.

Equilibrium Conceptual Quiz (25 MCQs)

Test your mastery of physical and chemical equilibria, Kp and Kc relations, Le Chatelier's principle, acid-base theories, pH, salt hydrolysis, buffers, and solubility product. Select an option to see instant explanation feedback.

1 At equilibrium, which of the following conditions is always satisfied?

Correct: Dynamic equilibrium is characterised by equal forward and reverse reaction rates.

Incorrect: Concentrations become constant, not necessarily equal.

Incorrect: The process is dynamic, not static.

Incorrect: Equilibrium constant has a definite non-zero value.

2 The relation between Kp and Kc for a gaseous reaction is given by:

Incorrect: Exponent should be +Δng.

Correct: Derived from ideal gas equation: p = cRT ⇒ Kp = Kc(RT)^Δng.

Incorrect: Kc = Kp(RT)^−Δng.

Incorrect: Only true if Δng = −1.

3 For the synthesis of ammonia N₂(g) + 3H₂(g) ⇌ 2NH₃(g), the relation between Kp and Kc is:

Incorrect: Δng is not zero.

Correct: Δng = 2 − (1 + 3) = −2. Hence Kp = Kc(RT)⁻².

Incorrect: Δng is −2, not +2.

Incorrect: Δng = −2.

4 Which factor changes the numerical value of an equilibrium constant (Kc or Kp)?

Incorrect: Concentration changes shift equilibrium position without changing K.

Incorrect: Inert gas does not affect K.

Incorrect: A catalyst speeds up both rates equally without altering K.

Correct: The numerical value of the equilibrium constant depends solely on temperature!

5 If the reaction quotient Qc < Kc, in which direction will the reaction proceed?

Incorrect: Reverse reaction occurs when Qc > Kc.

Correct: When Qc < Kc, product concentration is lower than equilibrium value, so forward reaction is favored.

Incorrect: At equilibrium Qc = Kc.

Incorrect: Net forward reaction occurs.

6 According to Le Chatelier's principle, an increase in pressure on the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g) will:

Correct: High pressure shifts equilibrium to the side with fewer gaseous moles (4 mol → 2 mol).

Incorrect: That would increase moles.

Incorrect: Δng ≠ 0, so pressure has a strong effect.

Incorrect: Rate actually increases due to higher concentration.

7 For an endothermic reaction (ΔH > 0), increasing temperature causes the equilibrium constant to:

Incorrect: That happens for exothermic reactions.

Incorrect: Temperature alters K.

Correct: Endothermic reactions absorb heat; higher T favors the forward direction, increasing Kc.

Incorrect: K is never zero.

8 What is the conjugate base of HCO₃⁻?

Incorrect: H₂CO₃ is the conjugate acid of HCO₃⁻.

Correct: Removing H⁺ from HCO₃⁻ gives the conjugate base CO₃²⁻.

Incorrect: CO₂ is an anhydride.

Incorrect: Unrelated conjugate species.

9 Which of the following is a Lewis acid?

Incorrect: NH₃ has a lone pair and is a Lewis base.

Incorrect: H₂O is a Lewis base.

Correct: BF₃ has an incomplete octet (electron deficient sextet) and accepts an electron pair.

Incorrect: OH⁻ is a Lewis base.

10 The ionic product of water (Kw) at 298 K is:

Incorrect: 10⁻⁷ is the concentration of H⁺ or OH⁻.

Correct: Kw = [H⁺][OH⁻] = (10⁻⁷)(10⁻⁷) = 1.0 × 10⁻¹⁴ M² at 298 K.

Incorrect: Arithmetic error.

Incorrect: 7 is the neutral pH.

11 What is the pH of a 0.001 M HCl solution at 298 K?

Incorrect: That is 0.1 M HCl.

Incorrect: That is 0.01 M HCl.

Correct: HCl is strong: [H⁺] = 10⁻³ M ⇒ pH = −log(10⁻³) = 3.

Incorrect: 11 is basic.

12 What is the pH of a 1.0 × 10⁻⁸ M solution of HCl at 298 K?

Incorrect: An acid solution cannot have a basic pH (> 7)!

Incorrect: Pure water is 7.00.

Correct: Including water auto-ionization (10⁻⁸ + 9.5×10⁻⁸ = 1.05×10⁻⁷ M), pH = −log(1.05×10⁻⁷) = 6.98.

Incorrect: Far too acidic.

13 For a conjugate acid-base pair, the relation between Ka, Kb, and Kw is:

Incorrect: That is for logarithmic pK values.

Incorrect: That is not the product relation.

Correct: Ka × Kb = Kw = 1.0 × 10⁻¹⁴ (and pKa + pKb = 14).

Incorrect: Product equals Kw, not 1.

14 According to Ostwald's dilution law, the degree of dissociation (α) of a weak electrolyte is proportional to:

Incorrect: Dissociation increases with dilution, not concentration.

Incorrect: α is inversely proportional to √c.

Correct: α = √(Ka / c) = √(Ka × V), so α ∝ √V.

Incorrect: Proportional to √V, not V.

15 An aqueous solution of sodium acetate (CH₃COONa) is:

Incorrect: Cation from strong base NaOH does not hydrolyse.

Correct: Acetate anion hydrolyses: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻, producing excess OH⁻ (pH > 7).

Incorrect: Weak acid salt undergoes hydrolysis.

Incorrect: Solution is alkaline.

16 The pH of a salt of weak acid and weak base (e.g., CH₃COONH₄) is given by:

Incorrect: That formula is for weak acid + strong base.

Incorrect: That formula is for strong acid + weak base.

Correct: pH = 7 + ½(pKa − pKb), completely independent of concentration c.

Incorrect: Missing the 7 term.

17 The Henderson-Hasselbalch equation for an acidic buffer is:

Incorrect: Inverted ratio.

Correct: pH = pKa + log([Salt]/[Acid]).

Incorrect: Mixed pH and pOH.

Incorrect: Sign before log is positive.

18 When equal volumes of 0.1 M CH₃COOH and 0.1 M CH₃COONa are mixed, the pH of the buffer is equal to:

Incorrect: 7.00 is neutral.

Correct: When [Salt] = [Acid], log([Salt]/[Acid]) = log(1) = 0, so pH = pKa = 4.76.

Incorrect: That applies to basic buffers.

Incorrect: Wrong relation.

19 For a sparingly soluble salt of type AB₂ (e.g., PbCl₂), the solubility product Ksp in terms of solubility S is:

Incorrect: That is for AB type (e.g., AgCl).

Incorrect: Incorrect stoichiometry.

Correct: AB₂ ⇌ A²⁺ + 2B⁻ ⇒ Ksp = (S)(2S)² = 4S³.

Incorrect: That is for AB₃ type.

20 A precipitate forms when the ionic product (Qsp) and solubility product (Ksp) satisfy:

Incorrect: Solution is unsaturated; no precipitation.

Incorrect: Solution is saturated at dynamic equilibrium.

Correct: Precipitation occurs when solution is supersaturated (Qsp > Ksp).

Incorrect: Pure solvent condition.

21 Passing dry HCl gas into a saturated solution of NaCl causes precipitation of pure NaCl due to:

Incorrect: Thermal change is not the primary cause.

Correct: Dissociation of HCl supplies high [Cl⁻], exceeding Qsp > Ksp and precipitating pure NaCl via the common ion effect.

Incorrect: Water does not evaporate.

Incorrect: No acid-base neutralization occurs.

22 Which pair represents a buffer solution?

Incorrect: Strong acid + salt does not buffer.

Incorrect: Strong base + salt does not buffer.

Correct: Weak acid (CH₃COOH) + its conjugate salt (CH₃COONa) forms an acidic buffer.

Incorrect: Strong acid system.

23 The molar solubility of a sparingly soluble salt A₂X₃ with solubility product Ksp is given by:

Correct: A₂X₃ ⇌ 2A³⁺ + 3X²⁻ ⇒ Ksp = (2S)²(3S)³ = 108S⁵ ⇒ S = (Ksp / 108)^(1/5).

Incorrect: That is for AB₂.

Incorrect: That is for AB₃.

Incorrect: 2² × 3³ = 4 × 27 = 108, not 36.

24 At 310 K (human body temperature), Kw = 2.7 × 10⁻¹⁴. The pH of neutral pure water at 310 K is:

Incorrect: 7.00 is neutral pH at 298 K only.

Correct: [H⁺] = √(2.7 × 10⁻¹⁴) = 1.64 × 10⁻⁷ M ⇒ pH = −log(1.64 × 10⁻⁷) = 6.78.

Incorrect: pH decreases because auto-ionization is endothermic.

Incorrect: That is pKw at 298 K.

25 Which of the following salts produces an acidic solution when dissolved in water?

Incorrect: NaCl is neutral.

Incorrect: K₂CO₃ is basic (strong base KOH + weak acid H₂CO₃).

Correct: NH₄Cl is a salt of strong acid HCl + weak base NH₄OH; NH₄⁺ hydrolyses to give excess H⁺ (pH < 7).

Incorrect: CH₃COONa is basic.

Quiz Results

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NCERT Textbook Exercises (6.1 to 6.73 Fully Solved)

Complete step-by-step solutions for all 73 end-of-chapter questions from CBSE / NCERT Class 11 Chemistry Chapter 6 (Equilibrium).

6.1

A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased. (a) What is the initial effect on vapour pressure? (b) How do rates of evaporation and condensation change initially? (c) What happens when equilibrium is restored finally and what will be the final vapour pressure?

(a) Initial effect on vapour pressure: When the container volume is suddenly increased, the same number of vapour molecules occupy a larger volume, so the vapour pressure initially decreases.
(b) Initial rate changes: The rate of evaporation depends on liquid surface area and temperature, which remain constant, so evaporation rate is initially unchanged. However, with fewer vapour molecules per unit volume colliding with the liquid surface, the rate of condensation initially decreases.
(c) Final equilibrium state: Since evaporation rate exceeds condensation rate, more liquid evaporates until the vapour concentration reaches the original saturation point. Once equilibrium is re-established (Rate_evap = Rate_cond), the final vapour pressure restores to its exact original equilibrium vapour pressure at that fixed temperature!

(a) Decreases initially; (b) Rate of evaporation unchanged, rate of condensation decreases; (c) Final vapour pressure returns to original value.

6.2

What is Kc for the following equilibrium when the equilibrium concentrations are: [SO₂] = 0.60 M, [O₂] = 0.82 M and [SO₃] = 1.90 M? Reaction: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g).

$$K_c = \frac{[SO_3]^2}{[SO_2]^2 [O_2]} = \frac{(1.90)^2}{(0.60)^2 \times 0.82} = \frac{3.61}{0.36 \times 0.82} = \frac{3.61}{0.2952} = \mathbf{12.228\text{ L mol}^{-1}}$$

Kc = 12.23 L mol⁻¹ (or 12.228 M⁻¹).

6.3

At a certain temperature and total pressure of 10⁵ Pa, iodine vapour contains 40% by volume of I atoms: I₂(g) ⇌ 2I(g). Calculate Kp for the equilibrium.

Total pressure $p_{\text{total}} = 10^5\text{ Pa} = 1.0\text{ bar}$.
Partial pressure of I atoms: $p_I = \frac{40}{100} \times 10^5\text{ Pa} = 0.40 \times 10^5\text{ Pa} = 0.40\text{ bar}$.
Partial pressure of $I_2$: $p_{I_2} = \frac{60}{100} \times 10^5\text{ Pa} = 0.60 \times 10^5\text{ Pa} = 0.60\text{ bar}$.
$$K_p = \frac{(p_I)^2}{p_{I_2}} = \frac{(0.40 \times 10^5)^2}{0.60 \times 10^5} = \frac{0.16 \times 10^{10}}{0.60 \times 10^5} = \mathbf{2.67 \times 10^4\text{ Pa}} = \mathbf{0.267\text{ bar}}$$

Kp = 2.67 × 10⁴ Pa (or 0.267 bar).

6.4

Write the expression for the equilibrium constant Kc for each of the following reactions: (i) 2NOCl(g) ⇌ 2NO(g) + Cl₂(g); (ii) 2Cu(NO₃)₂(s) ⇌ 2CuO(s) + 4NO₂(g) + O₂(g); (iii) CH₃COOC₂H₅(aq) + H₂O(l) ⇌ CH₃COOH(aq) + C₂H₅OH(aq); (iv) Fe³⁺(aq) + 3OH⁻(aq) ⇌ Fe(OH)₃(s); (v) I₂(s) + 5F₂(g) ⇌ 2IF₅(g).

(i) $K_c = \frac{[NO]^2 [Cl_2]}{[NOCl]^2}$
(ii) Pure solids $Cu(NO_3)_2$ and $CuO$ are omitted: $K_c = [NO_2]^4 [O_2]$
(iii) Pure liquid water is omitted: $K_c = \frac{[CH_3COOH][C_2H_5OH]}{[CH_3COOC_2H_5]}$
(iv) Pure solid $Fe(OH)_3$ has constant concentration: $K_c = \frac{1}{[Fe^{3+}][OH^-]^3}$
(v) Pure solid $I_2$ is omitted: $K_c = \frac{[IF_5]^2}{[F_2]^5}$

Pure solids and liquids are omitted from Kc expressions as their active masses are constant.

6.5

Find out the value of Kc for each of the following equilibria from the value of Kp: (i) 2NOCl(g) ⇌ 2NO(g) + Cl₂(g); Kp = 1.8 × 10⁻² at 500 K; (ii) CaCO₃(s) ⇌ CaO(s) + CO₂(g); Kp = 167 at 1073 K.

Using $K_p = K_c (RT)^{\Delta n_g} \implies K_c = \frac{K_p}{(RT)^{\Delta n_g}}$ with $R = 0.0831\text{ bar L mol}^{-1}\text{K}^{-1}$:
(i) $\Delta n_g = (2 + 1) - 2 = 1$.
$$K_c = \frac{1.8 \times 10^{-2}}{(0.0831 \times 500)^1} = \frac{1.8 \times 10^{-2}}{41.55} = \mathbf{4.33 \times 10^{-4}\text{ mol L}^{-1}}$$
(ii) $\Delta n_g = 1 - 0 = 1$.
$$K_c = \frac{167}{(0.0831 \times 1073)^1} = \frac{167}{89.166} = \mathbf{1.87\text{ mol L}^{-1}}$$

(i) Kc = 4.33 × 10⁻⁴ mol L⁻¹; (ii) Kc = 1.87 mol L⁻¹.

6.6

For the equilibrium NO(g) + O₃(g) ⇌ NO₂(g) + O₂(g), Kc = 6.3 × 10¹⁴ at 1000 K. What is Kc for the reverse reaction?

For the reverse reaction $NO_{2(g)} + O_{2(g)} \rightleftharpoons NO_{(g)} + O_{3(g)}$: $$K'_c = \frac{1}{K_c} = \frac{1}{6.3 \times 10^{14}} = \mathbf{1.59 \times 10^{-15}}$$

K'_c = 1.59 × 10⁻¹⁵.

6.7

Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression?

The active mass (molar concentration) of a pure substance is: $$\text{Molar Concentration} = \frac{\text{Moles}}{\text{Volume}} = \frac{\text{Mass} / \text{Molar Mass}}{\text{Volume}} = \frac{\text{Density}}{\text{Molar Mass}}$$ For pure solids and liquids, both density and molar mass are intrinsic intensive properties that remain strictly constant at a given temperature, regardless of the quantity present. Their constant values are incorporated into the equilibrium constant itself. Hence, pure liquids and solids are omitted.

Because their molar concentrations (Density / Molar Mass) are constant at a given temperature.

6.8

Reaction between N₂ and O₂ takes place as: 2N₂(g) + O₂(g) ⇌ 2N₂O(g). If 0.482 mol of N₂ and 0.933 mol of O₂ are placed in a 10 L vessel at a temperature where Kc = 2.0 × 10⁻³⁷, determine the equilibrium composition.

Initial concentrations in 10 L: $[N_2] = \frac{0.482}{10} = 0.0482\text{ M}$, $[O_2] = \frac{0.933}{10} = 0.0933\text{ M}$.
Since $K_c = 2.0 \times 10^{-37}$ is extraordinarily small, only a negligible amount of $N_2O$ is formed ($x \ll 0.0482$):
$[N_2]_{\text{eq}} \approx 0.0482\text{ M}$, $[O_2]_{\text{eq}} \approx 0.0933\text{ M}$.
$$K_c = \frac{[N_2O]^2}{[N_2]^2 [O_2]} \implies 2.0 \times 10^{-37} = \frac{[N_2O]^2}{(0.0482)^2 (0.0933)} = \frac{[N_2O]^2}{2.168 \times 10^{-4}}$$
$$[N_2O]^2 = 2.0 \times 10^{-37} \times 2.168 \times 10^{-4} = 4.336 \times 10^{-41} = 43.36 \times 10^{-42}$$
$$[N_2O] = \sqrt{43.36 \times 10^{-42}} = \mathbf{6.58 \times 10^{-21}\text{ M}}$$

[N₂] = 0.0482 M, [O₂] = 0.0933 M, [N₂O] = 6.58 × 10⁻²¹ M.

6.9

Nitric oxide reacts with Br₂: 2NO(g) + Br₂(g) ⇌ 2NOBr(g). When 0.087 mol of NO and 0.0437 mol of Br₂ are mixed, 0.0518 mol of NOBr is obtained at equilibrium. Calculate equilibrium amount of NO and Br₂.

From stoichiometry, 2 mol of NOBr formed consumes 2 mol of NO and 1 mol of Br₂:
Moles of NO consumed $= 0.0518\text{ mol}$.
Moles of NO remaining at equilibrium $= 0.087 - 0.0518 = \mathbf{0.0352\text{ mol}}$.
Moles of Br₂ consumed $= \frac{0.0518}{2} = 0.0259\text{ mol}$.
Moles of Br₂ remaining at equilibrium $= 0.0437 - 0.0259 = \mathbf{0.0178\text{ mol}}$.

At equilibrium: n(NO) = 0.0352 mol, n(Br₂) = 0.0178 mol.

6.10

At 450 K, Kp = 2.0 × 10¹⁰ bar⁻¹ for: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g). What is Kc at this temperature?

$\Delta n_g = 2 - (2 + 1) = -1$.
$$K_c = K_p (RT)^{-\Delta n_g} = K_p (RT)^1 = (2.0 \times 10^{10}) \times (0.0831 \times 450) = 2.0 \times 10^{10} \times 37.395 = \mathbf{7.48 \times 10^{11}\text{ L mol}^{-1}}$$

Kc = 7.48 × 10¹¹ L mol⁻¹.

6.11

A sample of HI(g) is placed in a flask at 0.2 atm. At equilibrium, partial pressure of HI is 0.04 atm. What is Kp for 2HI(g) ⇌ H₂(g) + I₂(g)?

Initial: $p_{HI} = 0.2\text{ atm}$.
At equilibrium: $p_{HI} = 0.04\text{ atm} \implies$ decrease in pressure $= 0.2 - 0.04 = 0.16\text{ atm}$.
From reaction $2HI \rightleftharpoons H_2 + I_2$, $p_{H_2} = p_{I_2} = \frac{0.16}{2} = 0.08\text{ atm}$.
$$K_p = \frac{p_{H_2} \cdot p_{I_2}}{p_{HI}^2} = \frac{0.08 \times 0.08}{(0.04)^2} = \frac{0.0064}{0.0016} = \mathbf{4.0}$$

Kp = 4.0.

6.12

A mixture of 1.57 mol N₂, 1.92 mol H₂ and 8.13 mol NH₃ is introduced into a 20 L vessel at 500 K. Kc for N₂(g) + 3H₂(g) ⇌ 2NH₃(g) is 1.7 × 10². Is the reaction at equilibrium? If not, what is the net direction?

Molar concentrations in 20 L: $[N_2] = \frac{1.57}{20} = 0.0785\text{ M}$, $[H_2] = \frac{1.92}{20} = 0.096\text{ M}$, $[NH_3] = \frac{8.13}{20} = 0.4065\text{ M}$.
Reaction quotient $Q_c$: $$Q_c = \frac{[NH_3]^2}{[N_2][H_2]^3} = \frac{(0.4065)^2}{(0.0785)(0.096)^3} = \frac{0.1652}{0.0785 \times 8.847 \times 10^{-4}} = \frac{0.1652}{6.945 \times 10^{-5}} = \mathbf{2.38 \times 10^3}$$ Since $Q_c (2.38 \times 10^3) > K_c (1.7 \times 10^2)$, the mixture is NOT at equilibrium. Net reaction proceeds in the reverse direction (right to left).

Qc = 2.38 × 10³ > Kc. Net reaction proceeds in reverse direction (towards reactants).

6.13

The equilibrium constant expression for a gas reaction is Kc = [NH₃]⁴ [O₂]⁵ / [NO]⁴ [H₂O]⁶. Write the balanced chemical equation.

Species in numerator are products; species in denominator are reactants:
$$\mathbf{4NO(g) + 6H_2O(g) \rightleftharpoons 4NH_3(g) + 5O_2(g)}$$

4NO(g) + 6H₂O(g) ⇌ 4NH₃(g) + 5O₂(g).

6.14

One mole of H₂O and one mole of CO are taken in 10 L vessel and heated to 725 K. At equilibrium, 40% of water (by mass) reacts with CO: H₂O(g) + CO(g) ⇌ H₂(g) + CO₂(g). Calculate Kc.

Initial moles: $n(H_2O) = 1.0$, $n(CO) = 1.0$.
40% reacts $\implies x = 0.40\text{ mol}$.
At equilibrium: $n(H_2O) = 0.60$, $n(CO) = 0.60$, $n(H_2) = 0.40$, $n(CO_2) = 0.40$.
Since $\Delta n_g = 0$, volume cancels out: $$K_c = \frac{x^2}{(1 - x)^2} = \frac{(0.40)^2}{(0.60)^2} = \frac{0.16}{0.36} = \mathbf{0.444}$$

Kc = 0.444.

6.15

At 700 K, Kc for H₂(g) + I₂(g) ⇌ 2HI(g) is 54.8. If 0.5 mol L⁻¹ of HI(g) is present at equilibrium at 700 K, what are the concentrations of H₂(g) and I₂(g)?

Since the system was started from pure HI, $[H_2] = [I_2] = x$.
$$K_c = \frac{[HI]^2}{[H_2][I_2]} \implies 54.8 = \frac{(0.5)^2}{x^2} \implies x^2 = \frac{0.25}{54.8} = 4.562 \times 10^{-3}$$
$$x = \sqrt{4.562 \times 10^{-3}} = \mathbf{0.0675\text{ M}}$$

[H₂] = [I₂] = 0.068 M (0.0675 mol L⁻¹).

6.16

What is the equilibrium concentration of each substance in 2ICl(g) ⇌ I₂(g) + Cl₂(g) when initial [ICl] was 0.78 M and Kc = 0.14?

At equilibrium: $[ICl] = 0.78 - 2x$, $[I_2] = x$, $[Cl_2] = x$.
$$K_c = \frac{x^2}{(0.78 - 2x)^2} = 0.14 \implies \frac{x}{0.78 - 2x} = \sqrt{0.14} = 0.374$$
$$x = 0.374(0.78 - 2x) = 0.2917 - 0.748x \implies 1.748x = 0.2917 \implies x = \mathbf{0.167\text{ M}}$$
$[I_2] = [Cl_2] = \mathbf{0.167\text{ M}}$, $[ICl] = 0.78 - 2(0.167) = 0.78 - 0.334 = \mathbf{0.446\text{ M}}$.

[ICl] = 0.446 M, [I₂] = 0.167 M, [Cl₂] = 0.167 M.

6.17

Kp = 0.04 atm at 899 K for C₂H₆(g) ⇌ C₂H₄(g) + H₂(g). What is the equilibrium pressure of C₂H₆ when initial pressure is 4.0 atm?

At equilibrium: $p(C_2H_6) = 4.0 - p$, $p(C_2H_4) = p$, $p(H_2) = p$.
$$K_p = \frac{p^2}{4.0 - p} = 0.04 \implies p^2 + 0.04p - 0.16 = 0$$
$$p = \frac{-0.04 + \sqrt{(0.04)^2 - 4(1)(-0.16)}}{2} = \frac{-0.04 + \sqrt{0.0016 + 0.64}}{2} = \frac{-0.04 + 0.801}{2} = \mathbf{0.38\text{ atm}}$$
$$p(C_2H_6) = 4.0 - 0.38 = \mathbf{3.62\text{ atm}}$$
Concentration: $c = \frac{p}{RT} = \frac{3.62}{0.0821 \times 899} = \mathbf{0.049\text{ mol L}^{-1}}$.

Equilibrium partial pressure = 3.62 atm (or [C₂H₆] = 0.049 M).

6.18

Ethyl acetate equilibrium: CH₃COOH(l) + C₂H₅OH(l) ⇌ CH₃COOC₂H₅(l) + H₂O(l). (i) Write Qc; (ii) If 1.00 mol acid + 0.18 mol ethanol gives 0.171 mol ester at equilibrium at 293 K, find Kc; (iii) Starting with 0.5 mol ethanol and 1.0 mol acid, 0.214 mol ester is found. Has equilibrium been reached?

(i) $Q_c = \frac{[CH_3COOC_2H_5][H_2O]}{[CH_3COOH][C_2H_5OH]}$ (water is a product, not excess solvent).
(ii) At equilibrium: $n(\text{ester}) = n(H_2O) = 0.171$, $n(\text{acid}) = 1.00 - 0.171 = 0.829$, $n(\text{ethanol}) = 0.18 - 0.171 = 0.009$.
$$K_c = \frac{(0.171)(0.171)}{(0.829)(0.009)} = \frac{0.02924}{0.00746} = \mathbf{3.92}$$
(iii) Current moles: ester = $0.214$, $H_2O = 0.214$, acid = $1.0 - 0.214 = 0.786$, ethanol = $0.5 - 0.214 = 0.286$.
$$Q_c = \frac{(0.214)(0.214)}{(0.786)(0.286)} = \frac{0.0458}{0.2248} = \mathbf{0.204}$$
Since $Q_c (0.204) < K_c (3.92)$, equilibrium has NOT been reached; reaction continues forward!

(i) Qc as written; (ii) Kc = 3.92; (iii) No, Qc = 0.204 < Kc, forward reaction continues.

6.19

A sample of pure PCl₅ was introduced into an evacuated vessel at 473 K. At equilibrium, [PCl₅] = 0.5 × 10⁻¹ mol L⁻¹. If Kc = 8.3 × 10⁻³, what are [PCl₃] and [Cl₂]?

PCl₅ ⇌ PCl₃ + Cl₂ $\implies [PCl_3] = [Cl_2] = x$.
$$K_c = \frac{x^2}{[PCl_5]} \implies 8.3 \times 10^{-3} = \frac{x^2}{0.05} \implies x^2 = 4.15 \times 10^{-4}$$
$$x = \sqrt{4.15 \times 10^{-4}} = \mathbf{0.0204\text{ mol L}^{-1}} = \mathbf{2.04 \times 10^{-2}\text{ M}}$$

[PCl₃] = [Cl₂] = 0.0204 M (2.04 × 10⁻² mol L⁻¹).

6.20

FeO(s) + CO(g) ⇌ Fe(s) + CO₂(g); Kp = 0.265 atm at 1050 K. Initial pressures: p_CO = 1.4 atm, p_CO₂ = 0.80 atm. Find equilibrium partial pressures.

Pure solids FeO and Fe are omitted: $K_p = \frac{p_{CO_2}}{p_{CO}} = 0.265$.
Initial $Q_p = \frac{0.80}{1.4} = 0.571 > 0.265 \implies$ reverse reaction occurs:
At equilibrium: $p_{CO_2} = 0.80 - p$, $p_{CO} = 1.4 + p$.
$$\frac{0.80 - p}{1.4 + p} = 0.265 \implies 0.80 - p = 0.371 + 0.265p \implies 1.265p = 0.429 \implies p = 0.339\text{ atm}$$
$$p_{CO} = 1.4 + 0.339 = \mathbf{1.74\text{ atm}}, \quad p_{CO_2} = 0.80 - 0.339 = \mathbf{0.46\text{ atm}}$$

p(CO) = 1.74 atm, p(CO₂) = 0.46 atm.

6.21

Kc for N₂(g) + 3H₂(g) ⇌ 2NH₃(g) at 500 K is 0.061. Reaction mixture contains 3.0 M N₂, 2.0 M H₂ and 0.5 M NH₃. Is the system at equilibrium? In which direction does it proceed?

$$Q_c = \frac{[NH_3]^2}{[N_2][H_2]^3} = \frac{(0.5)^2}{(3.0)(2.0)^3} = \frac{0.25}{24} = \mathbf{0.0104}$$
Since $Q_c (0.0104) < K_c (0.061)$, the reaction is not at equilibrium and proceeds in the forward direction to form more NH₃.

Qc = 0.0104 < Kc. Net reaction proceeds in the forward direction.

6.22

2BrCl(g) ⇌ Br₂(g) + Cl₂(g); Kc = 32 at 500 K. Initial [BrCl] = 3.3 × 10⁻³ mol L⁻¹. What is its molar concentration at equilibrium?

At equilibrium: $[BrCl] = 3.3 \times 10^{-3} - 2x$, $[Br_2] = [Cl_2] = x$.
$$K_c = \frac{x^2}{(3.3 \times 10^{-3} - 2x)^2} = 32 \implies \frac{x}{3.3 \times 10^{-3} - 2x} = \sqrt{32} = 5.657$$
$$x = 5.657(3.3 \times 10^{-3}) - 11.314x = 0.01867 - 11.314x \implies 12.314x = 0.01867 \implies x = 1.516 \times 10^{-3}\text{ M}$$
$$[BrCl] = 3.3 \times 10^{-3} - 2(1.516 \times 10^{-3}) = 3.3 \times 10^{-3} - 3.032 \times 10^{-3} = \mathbf{2.68 \times 10^{-4}\text{ M}}$$

[BrCl] = 2.68 × 10⁻⁴ mol L⁻¹.

6.23

At 1127 K and 1 atm, a gaseous mixture of CO and CO₂ in equilibrium with solid carbon contains 90.55% CO by mass: C(s) + CO₂(g) ⇌ 2CO(g). Calculate Kc.

Let total mass = 100 g $\implies$ mass CO = 90.55 g, mass CO₂ = 9.45 g.
Moles of CO $= \frac{90.55}{28} = 3.234\text{ mol}$; Moles of CO₂ $= \frac{9.45}{44} = 0.215\text{ mol}$. Total moles $= 3.449\text{ mol}$.
$p_{CO} = \frac{3.234}{3.449} \times 1\text{ atm} = 0.938\text{ atm}$; $p_{CO_2} = \frac{0.215}{3.449} \times 1\text{ atm} = 0.062\text{ atm}$.
$$K_p = \frac{p_{CO}^2}{p_{CO_2}} = \frac{(0.938)^2}{0.062} = \frac{0.8798}{0.062} = 14.19\text{ atm}$$
$\Delta n_g = 2 - 1 = 1 \implies K_c = \frac{K_p}{RT} = \frac{14.19}{0.0821 \times 1127} = \mathbf{0.153\text{ mol L}^{-1}}$.

Kc = 0.153 mol L⁻¹ (Kp = 14.19 atm).

6.24

Calculate (a) ΔG° and (b) equilibrium constant for NO(g) + ½ O₂(g) ⇌ NO₂(g) at 298 K, given Δf G°(NO₂) = 52.0 kJ/mol, Δf G°(NO) = 87.0 kJ/mol, Δf G°(O₂) = 0.

(a) $\Delta_r G^\circ = \Delta_f G^\circ(NO_2) - [\Delta_f G^\circ(NO) + \frac{1}{2}\Delta_f G^\circ(O_2)] = 52.0 - 87.0 = \mathbf{-35.0\text{ kJ mol}^{-1}}$.
(b) $\log K = \frac{-\Delta_r G^\circ}{2.303 RT} = \frac{-(-35000)}{2.303 \times 8.314 \times 298} = \frac{35000}{5705.8} = 6.134$.
$$K = 10^{6.134} = \mathbf{1.36 \times 10^6}$$

(a) ΔG° = −35.0 kJ mol⁻¹; (b) K = 1.36 × 10⁶.

6.25

Does the number of moles of reaction products increase, decrease or remain same when pressure is decreased (volume increased)? (a) PCl₅(g) ⇌ PCl₃(g) + Cl₂(g); (b) CaO(s) + CO₂(g) ⇌ CaCO₃(s); (c) 3Fe(s) + 4H₂O(g) ⇌ Fe₃O₄(s) + 4H₂(g).

Decreasing pressure shifts equilibrium to the side with MORE moles of gas:
(a) $1\text{ mol} \rightleftharpoons 2\text{ mol}$: shifts forward $\implies$ Products increase.
(b) $1\text{ mol} \rightleftharpoons 0\text{ mol}$: shifts backward $\implies$ Products decrease.
(c) $4\text{ mol} \rightleftharpoons 4\text{ mol}$ ($\Delta n_g = 0$): Remains unchanged.

(a) Increases; (b) Decreases; (c) Remains same.

6.26

Which of the following reactions will get affected by increasing pressure? Mention forward or backward: (i) COCl₂(g) ⇌ CO(g) + Cl₂(g); (ii) CH₄(g) + 2S₂(g) ⇌ CS₂(g) + 2H₂S(g); (iii) CO₂(g) + C(s) ⇌ 2CO(g); (iv) 2H₂(g) + CO(g) ⇌ CH₃OH(g); (v) CaCO₃(s) ⇌ CaO(s) + CO₂(g); (vi) 4NH₃(g) + 5O₂(g) ⇌ 4NO(g) + 6H₂O(g).

Increasing pressure shifts equilibrium to the side with fewer gaseous moles:
(i) $1 \rightarrow 2 \implies$ Backward.
(ii) $3 \rightarrow 3 \implies$ No effect ($\Delta n_g = 0$).
(iii) $1 \rightarrow 2 \implies$ Backward.
(iv) $3 \rightarrow 1 \implies$ Forward.
(v) $0 \rightarrow 1 \implies$ Backward.
(vi) $9 \rightarrow 10 \implies$ Backward.

(i) Backward; (ii) Not affected; (iii) Backward; (iv) Forward; (v) Backward; (vi) Backward.

6.27

Kc = 1.6 × 10⁵ at 1024 K for H₂(g) + Br₂(g) ⇌ 2HBr(g). Find equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a sealed container.

For reverse dissociation: $2HBr \rightleftharpoons H_2 + Br_2$, $K'_p = \frac{1}{K_p} = \frac{1}{1.6 \times 10^5} = 6.25 \times 10^{-6}$.
$p(H_2) = p(Br_2) = p$, $p(HBr) = 10.0 - 2p \approx 10.0\text{ bar}$.
$$K'_p = \frac{p^2}{(10.0)^2} = 6.25 \times 10^{-6} \implies p^2 = 6.25 \times 10^{-4} \implies p = \mathbf{2.5 \times 10^{-2}\text{ bar}} = \mathbf{0.025\text{ bar}}$$
$p(H_2) = p(Br_2) = \mathbf{0.025\text{ bar}}$, $p(HBr) = 10.0 - 0.05 = \mathbf{9.95\text{ bar}}$.

p(H₂) = p(Br₂) = 0.025 bar; p(HBr) = 9.95 bar.

6.28

CH₄(g) + H₂O(g) ⇌ CO(g) + 3H₂(g), ΔH > 0. (a) Write Kp; (b) Effect on Kp and composition by: (i) increasing pressure, (ii) increasing temperature, (iii) catalyst?

(a) $K_p = \frac{p_{CO} \cdot p_{H_2}^3}{p_{CH_4} \cdot p_{H_2O}}$
(b) (i) Increasing pressure: $K_p$ unchanged; equilibrium shifts backward (toward fewer moles).
(ii) Increasing temperature: Reaction is endothermic $\implies K_p$ increases and equilibrium shifts forward.
(iii) Catalyst: $K_p$ and composition unchanged; equilibrium reached faster.

(a) Kp expression; (b) (i) Kp unchanged, shifts back; (ii) Kp increases, shifts forward; (iii) No effect on Kp or composition.

6.29

Describe the effect of: (a) addition of H₂, (b) addition of CH₃OH, (c) removal of CO, (d) removal of CH₃OH on the equilibrium: 2H₂(g) + CO(g) ⇌ CH₃OH(g).

(a) Addition of $H_2$ (reactant): shifts Forward.
(b) Addition of $CH_3OH$ (product): shifts Backward.
(c) Removal of $CO$ (reactant): shifts Backward.
(d) Removal of $CH_3OH$ (product): shifts Forward.

(a) Forward; (b) Backward; (c) Backward; (d) Forward.

6.30

At 473 K, Kc for PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) is 8.3 × 10⁻³, Δr H° = 124.0 kJ/mol. (a) Write Kc expression; (b) Kc for reverse reaction; (c) Effect on Kc if (i) more PCl₅ added, (ii) pressure increased, (iii) temperature increased?

(a) $K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]}$
(b) $K'_c = \frac{1}{8.3 \times 10^{-3}} = \mathbf{120.48}$
(c) (i) Adding $PCl_5$: $K_c$ is unchanged.
(ii) Increasing pressure: $K_c$ is unchanged.
(iii) Increasing temperature: Since $\Delta H > 0$ (endothermic), $K_c$ increases.

(a) Formula; (b) K'_c = 120.5; (c) (i) No effect, (ii) No effect, (iii) Kc increases.

6.31

CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g); Kp = 10.1 at 400 °C. If charged with p_CO = p_H₂O = 4.0 bar, what is partial pressure of H₂ at equilibrium?

At equilibrium: $p_{CO} = p_{H_2O} = 4.0 - p$, $p_{CO_2} = p_{H_2} = p$.
$$K_p = \frac{p^2}{(4.0 - p)^2} = 10.1 \implies \frac{p}{4.0 - p} = \sqrt{10.1} = 3.178$$
$$p = 3.178(4.0 - p) = 12.712 - 3.178p \implies 4.178p = 12.712 \implies p = \mathbf{3.04\text{ bar}}$$

p(H₂) = 3.04 bar.

6.32

Predict which of the following will have appreciable concentration of both reactants and products: (a) Cl₂(g) ⇌ 2Cl(g), Kc = 5 × 10⁻³⁹; (b) Cl₂(g) + 2NO(g) ⇌ 2NOCl(g), Kc = 3.7 × 10⁸; (c) Cl₂(g) + 2NO₂(g) ⇌ 2NO₂Cl(g), Kc = 1.8.

Appreciable concentrations of both occur when $10^{-3} \le K_c \le 10^3$:
(a) $K_c = 5 \times 10^{-39} \ll 10^{-3}$ (hardly proceeds).
(b) $K_c = 3.7 \times 10^8 \gg 10^3$ (proceeds nearly to completion).
(c) $K_c = 1.8$ falls squarely in the range $10^{-3}$ to $10^3$!

Reaction (c) has appreciable concentrations of both.

6.33

Kc for 3O₂(g) ⇌ 2O₃(g) is 2.0 × 10⁻⁵⁰ at 25 °C. If [O₂] = 1.6 × 10⁻² M, what is [O₃]?

$$K_c = \frac{[O_3]^2}{[O_2]^3} \implies 2.0 \times 10^{-50} = \frac{[O_3]^2}{(1.6 \times 10^{-2})^3} = \frac{[O_3]^2}{4.096 \times 10^{-6}}$$
$$[O_3]^2 = 2.0 \times 10^{-50} \times 4.096 \times 10^{-6} = 8.192 \times 10^{-56}$$
$$[O_3] = \sqrt{8.192 \times 10^{-56}} = \mathbf{2.86 \times 10^{-28}\text{ M}}$$

[O₃] = 2.86 × 10⁻²⁸ M.

6.34

CO(g) + 3H₂(g) ⇌ CH₄(g) + H₂O(g) at 1300 K in a 1 L flask contains 0.30 mol CO, 0.10 mol H₂, 0.02 mol H₂O and unknown CH₄. Kc = 3.90. Determine [CH₄].

$$K_c = \frac{[CH_4][H_2O]}{[CO][H_2]^3} \implies 3.90 = \frac{[CH_4](0.02)}{(0.30)(0.10)^3} = \frac{0.02 [CH_4]}{3.0 \times 10^{-4}}$$
$$[CH_4] = \frac{3.90 \times 3.0 \times 10^{-4}}{0.02} = \frac{1.17 \times 10^{-3}}{0.02} = \mathbf{0.0585\text{ M}}$$

[CH₄] = 0.0585 M (or 5.85 × 10⁻² mol L⁻¹).

6.35

What is meant by conjugate acid-base pair? Find conjugate acid/base for: HNO₂, CN⁻, HClO₄, F⁻, OH⁻, CO₃²⁻, S²⁻.

A conjugate acid-base pair differs by exactly one proton ($H^+$):
• $HNO_2$ (acid) $\rightarrow$ conjugate base: $\mathbf{NO_2^-}$
• $CN^-$ (base) $\rightarrow$ conjugate acid: $\mathbf{HCN}$
• $HClO_4$ (acid) $\rightarrow$ conjugate base: $\mathbf{ClO_4^-}$
• $F^-$ (base) $\rightarrow$ conjugate acid: $\mathbf{HF}$
• $OH^-$ (amphiprotic) $\rightarrow$ conjugate acid: $\mathbf{H_2O}$, conjugate base: $\mathbf{O^{2-}}$
• $CO_3^{2-}$ (base) $\rightarrow$ conjugate acid: $\mathbf{HCO_3^-}$
• $S^{2-}$ (base) $\rightarrow$ conjugate acid: $\mathbf{HS^-}$

Conjugate species identified by adding or subtracting one proton (H⁺).

6.36

Which of the following are Lewis acids? H₂O, BF₃, H⁺, NH₄⁺.

Lewis acids are electron-pair acceptors:
• $BF_3$: Lewis acid (incomplete octet on B).
• $H^+$: Lewis acid (empty orbital).
• $H_2O$: Lewis base (two lone pairs).
• $NH_4^+$: Neither (octet complete, cannot accept electron pair).

BF₃ and H⁺ are Lewis acids.

6.37

What will be the conjugate bases for Brönsted acids: HF, H₂SO₄ and HCO₃⁻?

Remove $H^+$:
• $HF \rightarrow \mathbf{F^-}$
• $H_2SO_4 \rightarrow \mathbf{HSO_4^-}$
• $HCO_3^- \rightarrow \mathbf{CO_3^{2-}}$

F⁻, HSO₄⁻, CO₃²⁻.

6.38

Write the conjugate acids for Brönsted bases: NH₂⁻, NH₃ and HCOO⁻.

Add $H^+$:
• $NH_2^- \rightarrow \mathbf{NH_3}$
• $NH_3 \rightarrow \mathbf{NH_4^+}$
• $HCOO^- \rightarrow \mathbf{HCOOH}$

NH₃, NH₄⁺, HCOOH.

6.39

Species H₂O, HCO₃⁻, HSO₄⁻ and NH₃ can act both as Brönsted acids and bases. Give conjugate acid and base for each.

• $H_2O$: Conjugate acid = $\mathbf{H_3O^+}$, Conjugate base = $\mathbf{OH^-}$
• $HCO_3^-$: Conjugate acid = $\mathbf{H_2CO_3}$, Conjugate base = $\mathbf{CO_3^{2-}}$
• $HSO_4^-$: Conjugate acid = $\mathbf{H_2SO_4}$, Conjugate base = $\mathbf{SO_4^{2-}}$
• $NH_3$: Conjugate acid = $\mathbf{NH_4^+}$, Conjugate base = $\mathbf{NH_2^-}$

Amphiprotic species form conjugate acids with +H⁺ and conjugate bases with −H⁺.

6.40

Classify into Lewis acids and bases: (a) OH⁻, (b) F⁻, (c) H⁺, (d) BCl₃.

(a) $OH^-$: Lewis base (donates lone pair).
(b) $F^-$: Lewis base (donates lone pair).
(c) $H^+$: Lewis acid (accepts electron pair).
(d) $BCl_3$: Lewis acid (electron deficient).

Lewis bases: OH⁻, F⁻; Lewis acids: H⁺, BCl₃.

6.41

The concentration of hydrogen ion in a soft drink is 3.8 × 10⁻³ M. What is its pH?

$$pH = -\log(3.8 \times 10^{-3}) = 3 - \log 3.8 = 3 - 0.58 = \mathbf{2.42}$$

pH = 2.42.

6.42

The pH of vinegar is 3.76. Calculate the concentration of hydrogen ion in it.

$$[H^+] = 10^{-pH} = 10^{-3.76} = 10^{0.24} \times 10^{-4} = \mathbf{1.74 \times 10^{-4}\text{ M}}$$

[H⁺] = 1.74 × 10⁻⁴ M.

6.43

Ka of HF, HCOOH and HCN at 298 K are 6.8 × 10⁻⁴, 1.8 × 10⁻⁴ and 4.8 × 10⁻⁹. Calculate Kb of their conjugate bases.

Using $K_b = \frac{K_w}{K_a} = \frac{10^{-14}}{K_a}$:
• For $F^-$: $K_b = \frac{10^{-14}}{6.8 \times 10^{-4}} = \mathbf{1.47 \times 10^{-11}}$
• For $HCOO^-$: $K_b = \frac{10^{-14}}{1.8 \times 10^{-4}} = \mathbf{5.56 \times 10^{-11}}$
• For $CN^-$: $K_b = \frac{10^{-14}}{4.8 \times 10^{-9}} = \mathbf{2.08 \times 10^{-6}}$

Kb(F⁻) = 1.47 × 10⁻¹¹; Kb(HCOO⁻) = 5.56 × 10⁻¹¹; Kb(CN⁻) = 2.08 × 10⁻⁶.

6.44

Ionization constant of phenol is 1.0 × 10⁻¹⁰. What is [C₆H₅O⁻] in 0.05 M phenol? What is degree of ionization if solution is also 0.01 M sodium phenolate?

In pure 0.05 M phenol: $[C_6H_5O^-] = \sqrt{K_a \cdot c} = \sqrt{1.0 \times 10^{-10} \times 0.05} = \sqrt{5.0 \times 10^{-12}} = \mathbf{2.24 \times 10^{-6}\text{ M}}$.
With 0.01 M sodium phenolate ($[C_6H_5O^-] \approx 0.01\text{ M}$):
$$K_a = \frac{[H^+][C_6H_5O^-]}{[C_6H_5OH]} = \frac{(c\alpha)(0.01)}{0.05} \implies 1.0 \times 10^{-10} = \alpha \times 0.20 \implies \alpha = \mathbf{5.0 \times 10^{-10}}$$

In pure solution: [C₆H₅O⁻] = 2.24 × 10⁻⁶ M; with sodium phenolate: α = 5.0 × 10⁻¹⁰ (common ion effect).

6.45

Ka1 for H₂S is 9.1 × 10⁻⁸. Calculate [HS⁻] in 0.1 M solution. How will it be affected if solution is 0.1 M in HCl? If Ka2 = 1.2 × 10⁻¹³, calculate [S²⁻] in both conditions.

In pure 0.1 M $H_2S$: $[HS^-] = \sqrt{K_{a1} \cdot c} = \sqrt{9.1 \times 10^{-8} \times 0.1} = \mathbf{9.54 \times 10^{-5}\text{ M}}$.
In pure solution, $[S^{2-}] = K_{a2} = \mathbf{1.2 \times 10^{-13}\text{ M}}$.
In 0.1 M HCl ($[H^+] = 0.1\text{ M}$):
$[HS^-] = \frac{K_{a1} [H_2S]}{[H^+]} = \frac{9.1 \times 10^{-8} \times 0.1}{0.1} = \mathbf{9.1 \times 10^{-8}\text{ M}}$.
$$[S^{2-}] = \frac{K_{a1} K_{a2} [H_2S]}{[H^+]^2} = \frac{(9.1 \times 10^{-8})(1.2 \times 10^{-13})(0.1)}{(0.1)^2} = \mathbf{1.09 \times 10^{-19}\text{ M}}$$

Pure H₂S: [HS⁻] = 9.54 × 10⁻⁵ M, [S²⁻] = 1.2 × 10⁻¹³ M. In 0.1 M HCl: [HS⁻] = 9.1 × 10⁻⁸ M, [S²⁻] = 1.09 × 10⁻¹⁹ M.

6.46

Ka of acetic acid is 1.74 × 10⁻⁵. Calculate degree of dissociation in 0.05 M solution, [CH₃COO⁻] and pH.

$$\alpha = \sqrt{\frac{K_a}{c}} = \sqrt{\frac{1.74 \times 10^{-5}}{0.05}} = \sqrt{3.48 \times 10^{-4}} = \mathbf{1.866 \times 10^{-2}} \quad (1.87\%)$$
$$[CH_3COO^-] = c\alpha = 0.05 \times 0.01866 = \mathbf{9.33 \times 10^{-4}\text{ M}}$$
$$pH = -\log(9.33 \times 10^{-4}) = 4 - 0.97 = \mathbf{3.03}$$

α = 0.0187 (1.87%), [CH₃COO⁻] = 9.33 × 10⁻⁴ M, pH = 3.03.

6.47

The pH of 0.01 M organic acid is 4.15. Calculate anion concentration, Ka, and pKa.

$$[A^-] = [H^+] = 10^{-4.15} = 10^{0.85} \times 10^{-5} = \mathbf{7.08 \times 10^{-5}\text{ M}}$$
$$K_a = \frac{[H^+][A^-]}{c} = \frac{(7.08 \times 10^{-5})^2}{0.01} = \frac{5.01 \times 10^{-9}}{10^{-2}} = \mathbf{5.01 \times 10^{-7}}$$
$$pK_a = -\log(5.01 \times 10^{-7}) = 7 - 0.70 = \mathbf{6.30}$$

[A⁻] = 7.08 × 10⁻⁵ M, Ka = 5.01 × 10⁻⁷, pKa = 6.30.

6.48

Assuming complete dissociation, calculate pH: (a) 0.003 M HCl; (b) 0.005 M NaOH; (c) 0.002 M HBr; (d) 0.002 M KOH.

(a) $[H^+] = 3 \times 10^{-3}\text{ M} \implies pH = -\log(3 \times 10^{-3}) = 3 - 0.477 = \mathbf{2.52}$
(b) $[OH^-] = 5 \times 10^{-3}\text{ M} \implies pOH = 3 - 0.699 = 2.30 \implies pH = 14 - 2.30 = \mathbf{11.70}$
(c) $[H^+] = 2 \times 10^{-3}\text{ M} \implies pH = 3 - 0.301 = \mathbf{2.70}$
(d) $[OH^-] = 2 \times 10^{-3}\text{ M} \implies pOH = 2.70 \implies pH = 14 - 2.70 = \mathbf{11.30}$

(a) 2.52; (b) 11.70; (c) 2.70; (d) 11.30.

6.49

Calculate pH: (a) 2 g TlOH in 2 L; (b) 0.3 g Ca(OH)₂ in 500 mL; (c) 0.3 g NaOH in 200 mL; (d) 1 mL 13.6 M HCl diluted to 1 L.

(a) TlOH (molar mass = 221.4): $n = 2/221.4 = 0.00903\text{ mol}$ in 2 L $\implies [OH^-] = 4.52 \times 10^{-3}\text{ M} \implies pOH = 2.35 \implies \mathbf{pH = 11.65}$.
(b) $Ca(OH)_2$ (74.1 g/mol): $n = 0.3/74.1 = 0.00405\text{ mol}$ in 0.5 L $\implies [Ca(OH)_2] = 0.0081\text{ M} \implies [OH^-] = 0.0162\text{ M} \implies pOH = 1.79 \implies \mathbf{pH = 12.21}$.
(c) NaOH (40 g/mol): $n = 0.3/40 = 0.0075\text{ mol}$ in 0.2 L $\implies [OH^-] = 0.0375\text{ M} \implies pOH = 1.43 \implies \mathbf{pH = 12.57}$.
(d) HCl: $M_1 V_1 = M_2 V_2 \implies [H^+] = \frac{13.6 \times 1}{1000} = 0.0136\text{ M} \implies \mathbf{pH = 1.87}$.

(a) 11.65; (b) 12.21; (c) 12.57; (d) 1.87.

6.50

Degree of ionization of 0.1 M bromoacetic acid is 0.132. Calculate pH and pKa.

$$[H^+] = c\alpha = 0.1 \times 0.132 = \mathbf{0.0132\text{ M}} \implies pH = -\log(0.0132) = \mathbf{1.88}$$
$$K_a = \frac{c\alpha^2}{1 - \alpha} = \frac{0.1 \times (0.132)^2}{1 - 0.132} = \frac{0.001742}{0.868} = \mathbf{2.01 \times 10^{-3}}$$
$$pK_a = -\log(2.01 \times 10^{-3}) = 3 - 0.303 = \mathbf{2.70}$$

pH = 1.88, pKa = 2.70.

6.51

The pH of 0.005 M codeine is 9.95. Calculate its ionization constant Kb and pKb.

$pOH = 14 - 9.95 = 4.05 \implies [OH^-] = 10^{-4.05} = 10^{0.95} \times 10^{-5} = \mathbf{8.91 \times 10^{-5}\text{ M}}$.
$$K_b = \frac{[OH^-]^2}{c} = \frac{(8.91 \times 10^{-5})^2}{0.005} = \frac{7.94 \times 10^{-9}}{5 \times 10^{-3}} = \mathbf{1.59 \times 10^{-6}}$$
$$pK_b = -\log(1.59 \times 10^{-6}) = 6 - 0.20 = \mathbf{5.80}$$

Kb = 1.59 × 10⁻⁶, pKb = 5.80.

6.52

What is the pH of 0.001 M aniline (Kb = 4.27 × 10⁻¹⁰)? Calculate degree of ionization and Ka of conjugate acid.

$$\alpha = \sqrt{\frac{K_b}{c}} = \sqrt{\frac{4.27 \times 10^{-10}}{10^{-3}}} = \sqrt{4.27 \times 10^{-7}} = \mathbf{6.53 \times 10^{-4}}$$
$$[OH^-] = c\alpha = 10^{-3} \times 6.53 \times 10^{-4} = 6.53 \times 10^{-7}\text{ M} \implies pOH = 6.18 \implies \mathbf{pH = 7.82}$$
$$K_a = \frac{K_w}{K_b} = \frac{10^{-14}}{4.27 \times 10^{-10}} = \mathbf{2.34 \times 10^{-5}}$$

pH = 7.82, α = 6.53 × 10⁻⁴, Ka(conjugate acid) = 2.34 × 10⁻⁵.

6.53

Degree of ionization of 0.05 M acetic acid (pKa = 4.74). How is it affected by: (a) 0.01 M HCl, (b) 0.1 M HCl?

$K_a = 10^{-4.74} = 1.82 \times 10^{-5}$.
In pure solution: $\alpha = \sqrt{\frac{1.82 \times 10^{-5}}{0.05}} = \mathbf{0.0191}$ (1.91%).
(a) In 0.01 M HCl: $[H^+] \approx 0.01\text{ M} \implies \alpha = \frac{K_a}{[H^+]} = \frac{1.82 \times 10^{-5}}{0.01} = \mathbf{1.82 \times 10^{-3}}$ (0.18%).
(b) In 0.1 M HCl: $[H^+] \approx 0.1\text{ M} \implies \alpha = \frac{1.82 \times 10^{-5}}{0.1} = \mathbf{1.82 \times 10^{-4}}$ (0.018%).

Pure: α = 0.0191; (a) α = 1.82 × 10⁻³; (b) α = 1.82 × 10⁻⁴.

6.54

Kb of dimethylamine is 5.4 × 10⁻⁴. Calculate degree of ionization in 0.02 M solution. What % is ionized in 0.1 M NaOH?

In pure 0.02 M solution: $\alpha = \sqrt{\frac{K_b}{c}} = \sqrt{\frac{5.4 \times 10^{-4}}{0.02}} = \sqrt{0.027} = \mathbf{0.164}$ (16.4%).
In 0.1 M NaOH: $[OH^-] \approx 0.1\text{ M} \implies \alpha = \frac{K_b}{[OH^-]} = \frac{5.4 \times 10^{-4}}{0.1} = \mathbf{5.4 \times 10^{-3}} = \mathbf{0.54\%}$.

Pure: 16.4%; in 0.1 M NaOH: 0.54%.

6.55

Calculate [H⁺] in biological fluids: (a) Muscle fluid, 6.83; (b) Stomach fluid, 1.2; (c) Blood, 7.38; (d) Saliva, 6.4.

$[H^+] = 10^{-pH}$:
(a) $10^{-6.83} = 10^{0.17} \times 10^{-7} = \mathbf{1.48 \times 10^{-7}\text{ M}}$
(b) $10^{-1.2} = 10^{0.80} \times 10^{-2} = \mathbf{6.31 \times 10^{-2}\text{ M}}$
(c) $10^{-7.38} = 10^{0.62} \times 10^{-8} = \mathbf{4.17 \times 10^{-8}\text{ M}}$
(d) $10^{-6.4} = 10^{0.60} \times 10^{-7} = \mathbf{3.98 \times 10^{-7}\text{ M}}$

(a) 1.48 × 10⁻⁷ M; (b) 6.31 × 10⁻² M; (c) 4.17 × 10⁻⁸ M; (d) 3.98 × 10⁻⁷ M.

6.56

pH of milk (6.8), black coffee (5.0), tomato juice (4.2), lemon juice (2.2) and egg white (7.8). Calculate [H⁺].

• Milk: $10^{-6.8} = \mathbf{1.58 \times 10^{-7}\text{ M}}$
• Black coffee: $10^{-5.0} = \mathbf{1.0 \times 10^{-5}\text{ M}}$
• Tomato juice: $10^{-4.2} = \mathbf{6.31 \times 10^{-5}\text{ M}}$
• Lemon juice: $10^{-2.2} = \mathbf{6.31 \times 10^{-3}\text{ M}}$
• Egg white: $10^{-7.8} = \mathbf{1.58 \times 10^{-8}\text{ M}}$

Calculated from [H⁺] = 10^(−pH).

6.57

0.561 g KOH dissolved in water to give 200 mL solution at 298 K. Calculate [K⁺], [H⁺], [OH⁻] and pH.

Molar mass KOH = 56.1 g/mol $\implies n = 0.561/56.1 = 0.01\text{ mol}$.
$[KOH] = [K^+] = [OH^-] = \frac{0.01}{0.20} = \mathbf{0.05\text{ M}} = \mathbf{5.0 \times 10^{-2}\text{ M}}$.
$[H^+] = \frac{10^{-14}}{0.05} = \mathbf{2.0 \times 10^{-13}\text{ M}}$.
$$pH = -\log(2.0 \times 10^{-13}) = 13 - 0.301 = \mathbf{12.70}$$

[K⁺] = [OH⁻] = 0.05 M, [H⁺] = 2.0 × 10⁻¹³ M, pH = 12.70.

6.58

Solubility of Sr(OH)₂ at 298 K is 19.23 g/L. Calculate [Sr²⁺], [OH⁻] and pH.

Molar mass of $Sr(OH)_2 = 87.6 + 34 = 121.6\text{ g/mol}$.
Molar solubility $S = \frac{19.23}{121.6} = \mathbf{0.158\text{ M}}$.
$[Sr^{2+}] = S = \mathbf{0.158\text{ M}}$, $[OH^-] = 2S = 2(0.158) = \mathbf{0.316\text{ M}}$.
$pOH = -\log(0.316) = 0.50 \implies pH = 14 - 0.50 = \mathbf{13.50}$.

[Sr²⁺] = 0.158 M, [OH⁻] = 0.316 M, pH = 13.50.

6.59

Ka of propanoic acid is 1.32 × 10⁻⁵. Calculate α in 0.05 M and pH. What is α in 0.01 M HCl?

Pure solution: $\alpha = \sqrt{\frac{1.32 \times 10^{-5}}{0.05}} = \mathbf{1.625 \times 10^{-2}}$ (1.63%).
$[H^+] = c\alpha = 0.05 \times 0.01625 = 8.12 \times 10^{-4}\text{ M} \implies \mathbf{pH = 3.09}$.
In 0.01 M HCl ($[H^+] = 0.01\text{ M}$): $\alpha = \frac{1.32 \times 10^{-5}}{0.01} = \mathbf{1.32 \times 10^{-3}}$.

Pure: α = 0.0163, pH = 3.09. In 0.01 M HCl: α = 1.32 × 10⁻³.

6.60

pH of 0.1 M cyanic acid (HCNO) is 2.34. Calculate Ka and degree of ionization.

$[H^+] = 10^{-2.34} = 10^{0.66} \times 10^{-3} = \mathbf{4.57 \times 10^{-3}\text{ M}}$.
$$\alpha = \frac{[H^+]}{c} = \frac{4.57 \times 10^{-3}}{0.1} = \mathbf{0.0457} \quad (4.57\%)$$
$$K_a = \frac{[H^+]^2}{c} = \frac{(4.57 \times 10^{-3})^2}{0.1} = \frac{2.088 \times 10^{-5}}{0.1} = \mathbf{2.09 \times 10^{-4}}$$

Ka = 2.09 × 10⁻⁴, α = 0.0457 (4.57%).

6.61

Ka of nitrous acid is 4.5 × 10⁻⁴. Calculate pH of 0.04 M NaNO₂ and degree of hydrolysis.

Salt of weak acid + strong base: $K_h = \frac{K_w}{K_a} = \frac{10^{-14}}{4.5 \times 10^{-4}} = 2.22 \times 10^{-11}$.
Degree of hydrolysis $h = \sqrt{\frac{K_h}{c}} = \sqrt{\frac{2.22 \times 10^{-11}}{0.04}} = \mathbf{2.36 \times 10^{-5}}$.
$$pH = 7 + \frac{1}{2}[pK_a + \log c] = 7 + \frac{1}{2}[3.35 + \log(0.04)] = 7 + \frac{1}{2}[3.35 - 1.40] = 7 + 0.975 = \mathbf{7.98}$$

h = 2.36 × 10⁻⁵, pH = 7.98.

6.62

0.02 M pyridinium hydrochloride has pH = 3.44. Calculate ionization constant of pyridine.

Salt of weak base + strong acid: $[H^+] = 10^{-3.44} = 3.63 \times 10^{-4}\text{ M}$.
$K_h = \frac{[H^+]^2}{c} = \frac{(3.63 \times 10^{-4})^2}{0.02} = \frac{1.318 \times 10^{-7}}{0.02} = 6.59 \times 10^{-6}$.
$$K_b = \frac{K_w}{K_h} = \frac{10^{-14}}{6.59 \times 10^{-6}} = \mathbf{1.52 \times 10^{-9}}$$

Kb(pyridine) = 1.52 × 10⁻⁹.

6.63

Predict if solutions are neutral, acidic or basic: NaCl, KBr, NaCN, NH₄NO₃, NaNO₂, KF.

• NaCl: Neutral (strong acid + strong base).
• KBr: Neutral (strong acid + strong base).
• NaCN: Basic (weak acid HCN + strong base NaOH).
• NH₄NO₃: Acidic (strong acid HNO₃ + weak base NH₄OH).
• NaNO₂: Basic (weak acid HNO₂ + strong base NaOH).
• KF: Basic (weak acid HF + strong base KOH).

Neutral: NaCl, KBr; Acidic: NH₄NO₃; Basic: NaCN, NaNO₂, KF.

6.64

Ka of chloroacetic acid is 1.35 × 10⁻³. What will be pH of 0.1 M acid and 0.1 M sodium salt?

For 0.1 M acid: $[H^+] = \sqrt{K_a \cdot c} = \sqrt{1.35 \times 10^{-3} \times 0.1} = \sqrt{1.35 \times 10^{-4}} = 1.162 \times 10^{-2}\text{ M} \implies \mathbf{pH = 1.93}$.
For 0.1 M salt: $pK_a = 3 - \log 1.35 = 2.87$.
$$pH = 7 + \frac{1}{2}[pK_a + \log c] = 7 + \frac{1}{2}[2.87 + \log(0.1)] = 7 + \frac{1}{2}[2.87 - 1] = 7 + 0.935 = \mathbf{7.94}$$

Acid pH = 1.93; Salt pH = 7.94.

6.65

Ionic product of water at 310 K is 2.7 × 10⁻¹⁴. What is the pH of neutral water at this temperature?

In neutral water: $[H^+] = [OH^-] = \sqrt{K_w} = \sqrt{2.7 \times 10^{-14}} = 1.643 \times 10^{-7}\text{ M}$.
$$pH = -\log(1.643 \times 10^{-7}) = 7 - 0.216 = \mathbf{6.78}$$

pH = 6.78.

6.66

Calculate pH of resultant mixtures: (a) 10 mL 0.2 M Ca(OH)₂ + 25 mL 0.1 M HCl; (b) 10 mL 0.01 M H₂SO₄ + 10 mL 0.01 M Ca(OH)₂; (c) 10 mL 0.1 M H₂SO₄ + 10 mL 0.1 M KOH.

(a) Millimoles: $OH^- = 10 \times 0.2 \times 2 = 4.0\text{ mmol}$; $H^+ = 25 \times 0.1 = 2.5\text{ mmol}$.
Excess $OH^- = 4.0 - 2.5 = 1.5\text{ mmol}$ in 35 mL $\implies [OH^-] = \frac{1.5}{35} = 0.0429\text{ M} \implies pOH = 1.37 \implies \mathbf{pH = 12.63}$.
(b) Millimoles: $H^+ = 10 \times 0.01 \times 2 = 0.20\text{ mmol}$; $OH^- = 10 \times 0.01 \times 2 = 0.20\text{ mmol}$. Completely neutral $\implies \mathbf{pH = 7.00}$.
(c) Millimoles: $H^+ = 10 \times 0.1 \times 2 = 2.0\text{ mmol}$; $OH^- = 10 \times 0.1 = 1.0\text{ mmol}$.
Excess $H^+ = 2.0 - 1.0 = 1.0\text{ mmol}$ in 20 mL $\implies [H^+] = \frac{1.0}{20} = 0.05\text{ M} \implies \mathbf{pH = 1.30}$.

(a) 12.63; (b) 7.00; (c) 1.30.

6.67

Determine solubilities and ion concentrations at 298 K from Table 6.9: Ag₂CrO₄ (1.1 × 10⁻¹²), BaCrO₄ (1.2 × 10⁻¹⁰), Fe(OH)₃ (1.0 × 10⁻³⁸), PbCl₂ (1.6 × 10⁻⁵).

• $Ag_2CrO_4$: $K_{sp} = 4S^3 = 1.1 \times 10^{-12} \implies S = 6.5 \times 10^{-5}\text{ M}$. $[Ag^+] = 1.3 \times 10^{-4}\text{ M}$, $[CrO_4^{2-}] = 6.5 \times 10^{-5}\text{ M}$.
• $BaCrO_4$: $K_{sp} = S^2 = 1.2 \times 10^{-10} \implies S = 1.1 \times 10^{-5}\text{ M}$. $[Ba^{2+}] = [CrO_4^{2-}] = 1.1 \times 10^{-5}\text{ M}$.
• $Fe(OH)_3$: $K_{sp} = 27S^4 = 1.0 \times 10^{-38} \implies S = 1.39 \times 10^{-10}\text{ M}$. $[Fe^{3+}] = 1.39 \times 10^{-10}\text{ M}$, $[OH^-] = 4.16 \times 10^{-10}\text{ M}$.
• $PbCl_2$: $K_{sp} = 4S^3 = 1.6 \times 10^{-5} \implies S = 1.59 \times 10^{-2}\text{ M}$. $[Pb^{2+}] = 1.59 \times 10^{-2}\text{ M}$, $[Cl^-] = 3.18 \times 10^{-2}\text{ M}$.

Solubilities and individual ion molarities computed via stoichiometric Ksp equations.

6.68

Ksp of Ag₂CrO₄ and AgBr are 1.1 × 10⁻¹² and 5.0 × 10⁻¹³. Calculate the ratio of molarities of their saturated solutions.

For $Ag_2CrO_4$: $S_1 = (K_{sp}/4)^{1/3} = (1.1 \times 10^{-12} / 4)^{1/3} = 6.50 \times 10^{-5}\text{ M}$.
For $AgBr$: $S_2 = \sqrt{K_{sp}} = \sqrt{5.0 \times 10^{-13}} = 7.07 \times 10^{-7}\text{ M}$.
$$\text{Ratio} = \frac{S_1}{S_2} = \frac{6.50 \times 10^{-5}}{7.07 \times 10^{-7}} = \mathbf{91.9}$$

Ratio = 91.9 (Ag₂CrO₄ is ~92 times more soluble than AgBr).

6.69

Equal volumes of 0.002 M sodium iodate and cupric chlorate are mixed. Will cupric iodate precipitate? (Ksp = 7.4 × 10⁻⁸).

When equal volumes mix, concentrations halve:
$[Cu^{2+}] = 0.001\text{ M} = 10^{-3}\text{ M}$, $[IO_3^-] = 0.001\text{ M} = 10^{-3}\text{ M}$.
$$Q_{sp} = [Cu^{2+}][IO_3^-]^2 = (10^{-3})(10^{-3})^2 = 1.0 \times 10^{-9}$$
Since $Q_{sp} (1.0 \times 10^{-9}) < K_{sp} (7.4 \times 10^{-8})$, NO precipitation will occur!

Qsp = 1.0 × 10⁻⁹ < Ksp. No precipitation occurs.

6.70

Ka of benzoic acid = 6.46 × 10⁻⁵ and Ksp of silver benzoate = 2.5 × 10⁻¹³. How many times is silver benzoate more soluble in a buffer of pH 3.19 compared to pure water?

In pure water: $S_0 = \sqrt{K_{sp}} = \sqrt{2.5 \times 10^{-13}} = 5.0 \times 10^{-7}\text{ M}$.
At pH 3.19: $[H^+] = 10^{-3.19} = 6.46 \times 10^{-4}\text{ M}$.
Using $S = \sqrt{K_{sp} \left(1 + \frac{[H^+]}{K_a}\right)}$:
$$\frac{[H^+]}{K_a} = \frac{6.46 \times 10^{-4}}{6.46 \times 10^{-5}} = 10 \implies 1 + \frac{[H^+]}{K_a} = 11$$
$$S = \sqrt{2.5 \times 10^{-13} \times 11} = \sqrt{2.75 \times 10^{-12}} = 1.658 \times 10^{-6}\text{ M}$$
$$\text{Ratio} = \frac{1.658 \times 10^{-6}}{5.0 \times 10^{-7}} = \mathbf{3.32}$$

Silver benzoate is 3.32 times more soluble at pH 3.19.

6.71

What is the maximum concentration of equimolar solutions of FeSO₄ and Na₂S so that on mixing equal volumes, there is no precipitation? (Ksp(FeS) = 6.3 × 10⁻¹⁸).

Let equimolar concentration before mixing be $c$. On mixing equal volumes, concentrations halve: $[Fe^{2+}] = [S^{2-}] = c/2$.
For no precipitation: $Q_{sp} \le K_{sp} \implies \left(\frac{c}{2}\right)^2 \le 6.3 \times 10^{-18} \implies \frac{c^2}{4} \le 6.3 \times 10^{-18}$
$$c^2 \le 25.2 \times 10^{-18} \implies c \le \mathbf{5.02 \times 10^{-9}\text{ M}}$$

Maximum concentration = 5.02 × 10⁻⁹ M.

6.72

What is the minimum volume of water required to dissolve 1 g of CaSO₄ at 298 K? (Ksp = 9.1 × 10⁻⁶).

Molar mass of $CaSO_4 = 40 + 32 + 64 = 136\text{ g mol}^{-1}$.
Molar solubility $S = \sqrt{K_{sp}} = \sqrt{9.1 \times 10^{-6}} = 3.017 \times 10^{-3}\text{ mol L}^{-1}$.
Solubility in g/L $= 3.017 \times 10^{-3} \times 136 = \mathbf{0.410\text{ g L}^{-1}}$.
Volume to dissolve 1 g $= \frac{1\text{ g}}{0.410\text{ g L}^{-1}} = \mathbf{2.44\text{ L}}$.

Minimum volume of water required = 2.44 L.

6.73

[S²⁻] in 0.1 M HCl saturated with H₂S is 1.0 × 10⁻¹⁹ M. If 10 mL of this is added to 5 mL of 0.04 M FeSO₄, MnCl₂, ZnCl₂ and CdCl₂, in which will precipitation take place? (Ksp: FeS = 6.3×10⁻¹⁸, MnS = 2.5×10⁻¹³, ZnS = 1.6×10⁻²⁴, CdS = 8.0×10⁻²⁸).

After mixing: Total volume $= 15\text{ mL}$.
$[M^{2+}] = 0.04 \times \frac{5}{15} = 1.33 \times 10^{-2}\text{ M}$.
$[S^{2-}] = 1.0 \times 10^{-19} \times \frac{10}{15} = 6.67 \times 10^{-20}\text{ M}$.
$$Q_{sp} = [M^{2+}][S^{2-}] = (1.33 \times 10^{-2})(6.67 \times 10^{-20}) = \mathbf{8.87 \times 10^{-22}}$$
Compare $Q_{sp}$ with $K_{sp}$ of each sulphide:

FeS: $K_{sp} = 6.3 \times 10^{-18} > Q_{sp} \implies$ No precipitation.
MnS: $K_{sp} = 2.5 \times 10^{-13} > Q_{sp} \implies$ No precipitation.
ZnS: $K_{sp} = 1.6 \times 10^{-24} < Q_{sp} \implies$ Precipitates!
CdS: $K_{sp} = 8.0 \times 10^{-28} < Q_{sp} \implies$ Precipitates!

Precipitation will take place in ZnCl₂ and CdCl₂ solutions only.

Equilibrium High-Yield Revision Matrix

Essential master formulas, Le Chatelier decision matrix, salt hydrolysis classification, Henderson-Hasselbalch equations, and solubility product relations for rapid CBSE Board, JEE Main & NEET review.

1. Master Chemical Equilibrium Formulas

Concept / Relationship	Mathematical Formula	Key Notes & Conditions
Equilibrium Constant ($K_c$)	$K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}$	Pure solids and pure liquids are omitted (active mass = 1).
Gaseous Equilibrium ($K_p$)	$K_p = K_c (RT)^{\Delta n_g}$	$\Delta n_g = \sum n_{p(g)} - \sum n_{r(g)}$; $R = 0.0831\text{ bar L K}^{-1}\text{mol}^{-1}$.
Standard Gibbs Free Energy ($\Delta G^\circ$)	$\Delta_r G^\circ = -RT \ln K = -2.303 RT \log K$	$K = e^{-\Delta G^\circ / RT}$; at equilibrium $\Delta G = 0$.
Direction Prediction ($Q$ vs $K$)	$Q < K \implies \text{Forward } (\rightarrow)$ $Q = K \implies \text{Equilibrium } (\rightleftharpoons)$ $Q > K \implies \text{Backward } (\leftarrow)$	$Q$ uses non-equilibrium / instantaneous concentrations.
Temperature Dependence (van 't Hoff)	$\log\left(\frac{K_2}{K_1}\right) = \frac{\Delta H^\circ}{2.303 R}\left[\frac{T_2 - T_1}{T_1 T_2}\right]$	Endothermic ($\Delta H > 0$): $T \uparrow \implies K \uparrow$. Exothermic ($\Delta H < 0$): $T \uparrow \implies K \downarrow$.

2. Master Ionic Equilibrium Formulas

System	Governing Equation	pH Formula at 298 K
Ionic Product of Water ($K_w$)	$K_w = [H^+][OH^-] = 10^{-14}$	$pH + pOH = 14.00$
Weak Acid ($HA$)	$\alpha = \sqrt{\frac{K_a}{c}}, \quad [H^+] = c\alpha = \sqrt{K_a \cdot c}$	$pH = \frac{1}{2}[pK_a - \log c]$
Weak Base ($BOH$)	$\alpha = \sqrt{\frac{K_b}{c}}, \quad [OH^-] = c\alpha = \sqrt{K_b \cdot c}$	$pOH = \frac{1}{2}[pK_b - \log c] \implies pH = 14 - pOH$
Conjugate Acid-Base Pair	$K_a \times K_b = K_w = 1.0 \times 10^{-14}$	$pK_a + pK_b = 14.00$
Weak Acid + Strong Base Salt	$CH_3COONa: K_h = \frac{K_w}{K_a}$	$pH = 7 + \frac{1}{2}[pK_a + \log c] \quad (\text{Basic})$
Strong Acid + Weak Base Salt	$NH_4Cl: K_h = \frac{K_w}{K_b}$	$pH = 7 - \frac{1}{2}[pK_b + \log c] \quad (\text{Acidic})$
Weak Acid + Weak Base Salt	$CH_3COONH_4: K_h = \frac{K_w}{K_a K_b}$	$pH = 7 + \frac{1}{2}[pK_a - pK_b] \quad (\text{Independent of } c)$
Acidic Buffer ($HA + NaA$)	Henderson-Hasselbalch Equation	$pH = pK_a + \log\frac{[\text{Salt}]}{[\text{Acid}]}$
Basic Buffer ($BOH + BCl$)	Henderson-Hasselbalch Equation	$pOH = pK_b + \log\frac{[\text{Salt}]}{[\text{Base}]}$
Sparingly Soluble Salt $A_x B_y$	$K_{sp} = x^x y^y S^{(x+y)}$	$S = \left(\frac{K_{sp}}{x^x y^y}\right)^{\frac{1}{x+y}}$

Class 11 Chemistry Chapter 6: Equilibrium Tests

3 Graded Test Levels: Foundation (CBSE Board essentials), Intermediate (Numericals & Buffer calculations), and Advanced (NEET / JEE Main competitive multi-concept problems). Select answers to test yourself.

Level 1: Foundation Test (CBSE Board Level)

F1Which of the following is true for a reversible reaction at equilibrium?

Reversible reactions never reach 100% completion.

Correct: Defining characteristic of dynamic chemical equilibrium.

Kc varies widely depending on temperature and nature of reactants.

Equilibrium can only be established in a closed system.

F2For CaCO₃(s) ⇌ CaO(s) + CO₂(g), the equilibrium constant Kp is:

Pure solids are omitted.

Incorrect.

Correct: Concentrations/partial pressures of pure solids CaCO₃ and CaO are constant.

Incorrect.

F3What is the pH of 0.01 M NaOH solution?

Correct: [OH⁻] = 10⁻² M ⇒ pOH = 2 ⇒ pH = 14 − 2 = 12.

2 is the pOH, not pH!

Incorrect calculation.

NaOH is strongly basic.

Level 2: Intermediate Test (Numerical & Application Level)

I1For 2NO₂(g) ⇌ N₂O₄(g), Kp / Kc is equal to:

Incorrect exponent.

Δng is negative.

Correct: Δng = 1 − 2 = −1 ⇒ Kp = Kc(RT)⁻¹ ⇒ Kp/Kc = (RT)⁻¹ = 1/(RT).

Only true when Δng = 0.

I2The pH of a buffer solution containing 0.1 M CH₃COOH (pKa = 4.76) and 0.1 M CH₃COONa is:

7.00 is neutral.

Correct: pH = pKa + log([Salt]/[Acid]) = 4.76 + log(0.1/0.1) = 4.76 + 0 = 4.76.

Occurs when [Salt]/[Acid] = 10.

Occurs when [Acid]/[Salt] = 10.

Level 3: Advanced Test (JEE Main / NEET Competitive Level)

A1Solubility product of Ag₂CrO₄ is 1.08 × 10⁻¹². What is the solubility of Ag₂CrO₄ in 0.1 M AgNO₃ solution?

Correct: [Ag⁺] ≈ 0.1 M (from common ion AgNO₃). Ksp = [Ag⁺]²[CrO₄²⁻] ⇒ 1.08×10⁻¹² = (0.1)² · S ⇒ S = 1.08×10⁻¹⁰ M.

That is the solubility in pure water.

Arithmetic error in (0.1)².

Incorrect formula.

A2An aqueous solution of 0.1 M ammonium acetate (CH₃COONH₄) has pKa(acetic acid) = 4.76 and pKb(ammonia) = 4.75. The pH of this solution is:

Incorrect.

Correct: pH = 7 + ½(pKa − pKb) = 7 + ½(4.76 − 4.75) = 7 + 0.005 = 7.005. Notice concentration 0.1 M has no effect!

Incorrect.