Arrange the following in increasing order of solubility product:

Q: Arrange the following in increasing order of solubility product:

The solubility product constant (Ksp) reflects the equilibrium position of a sparingly soluble salt in water. Based on standard literature Ksp values at 298 K: - HgS: approx 4 x 10-53 (extremely insoluble, Group IIB cation analysis) - PbS: approx 8 x 10-28 (highly insoluble, Group IIA cation analysis) - AgBr: approx 5 x 10-13 (sparingly soluble halide salt) - Ca(OH)2: approx 5.5 x 10-6 (moderately soluble base) Comparing these Ksp orders: Hence, the correct increasing sequence is: HgS < PbS < Ag

Year	2026	2025	2024	Total
Questions	10	17	8	35

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

The solubility product constant ( $Kₛₚ$ $K_{sp}$) reflects the equilibrium position of a sparingly soluble salt in water.

Core Logic

Based on standard literature $Kₛₚ$ $K_{sp}$ values at $298 K$ $298\text{ K}$:

$HgS$ $HgS$: $\approx 4 \times 10⁻⁵³$ $\approx 4 \times 10^{-53}$ (extremely insoluble, Group IIB cation analysis)
$PbS$ $PbS$: $\approx 8 \times 10⁻²⁸$ $\approx 8 \times 10^{-28}$ (highly insoluble, Group IIA cation analysis)
$AgBr$ $AgBr$: $\approx 5 \times 10⁻¹³$ $\approx 5 \times 10^{-13}$ (sparingly soluble halide salt)
$Ca(OH)₂$ $Ca(OH)_2$: $\approx 5.5 \times 10⁻⁶$ $\approx 5.5 \times 10^{-6}$ (moderately soluble base)

Step 1: Arrangement

Comparing these $Kₛₚ$ $K_{sp}$ orders:

4 \times 10⁻⁵³ < 8 \times 10⁻²⁸ < 5 \times 10⁻¹³ < 5.5 \times 10⁻⁶

Hence, the correct increasing sequence is: $HgS < PbS < AgBr < Ca(OH)₂$ $HgS < PbS < AgBr < Ca(OH)_2$.

Pattern Recognition

Sulphides of heavy transition metals like $Hg²⁺$ $Hg^{2+}$ and $Pb²⁺$ $Pb^{2+}$ have exceptionally small $Kₛₚ$ $K_{sp}$ values compared to halides or hydroxides. Among sulphides, $HgS$ $HgS$ is famously known to have one of the lowest solubility products found in inorganic qualitative analysis.

Chapter Mix

Class 11 Chemistry: Equilibrium

More Equilibrium Previous-Year Questions — Page 7

Q84 jee_main_2024_29_jan_morning Kp and Kc Relationship

For the reaction

N₂O₄(g) leftharpoons 2NO₂(g)

Kₚ = 0.492

atm at

300K

Kc

for the reaction at same temperature is

______ \times 10⁻²

(Given: R = 0.082 L atm mol⁻¹ K⁻¹)

Numerical Answer. Answer: 2 to 2

Solution

Related Formula

Kₚ = Kc \cdot (RT)Δ ng

Core Logic

For the given gaseous equilibrium reaction:

N₂O₄(g) leftharpoons 2NO₂(g)

First, find the change in the number of moles of gas ( $Δ ng$ $\Delta n_g$):

Δ ng = nₚ - nᵣ = 2 - 1 = 1

Step 1: Calculation

Substitute the given values into the $Kₚ - Kc$ $K_p - K_c$ relationship: $Kₚ = 0.492$ $K_p = 0.492$ $R = 0.082$ $R = 0.082$ $T = 300 K$ $T = 300\text{ K}$

0.492 = Kc \cdot (0.082 \times 300)¹

Kc = (0.492)/(0.082 \times 300)

Kc = (0.492)/(24.6)

$Kc = 0.02$ $K_c = 0.02$

Converting to the requested format ( $x \times 10⁻²$ $x \times 10^{-2}$):

Kc = 2 \times 10⁻²

So, the value is 2.

Chapter Mix

Class 11 Chemistry: Equilibrium

Q85 jee_main_2024_30_january_evening Buffer Solutions

The pH of an aqueous solution containing 1M benzoic acid

(pKₐ = 4.20)

and 1M sodium benzoate is 4.5. The volume of benzoic acid solution in 300 mL of this buffer solution is ______ mL.

Numerical Answer. Answer: 100 to 100

Solution

Related Formula

Henderson-Hasselbalch Equation for Acidic Buffers:

pH = pKₐ + ( [Salt][Acid] )

Core Logic

Let the volume of 1M Benzoic acid be $Vₐ$ $V_a$ mL and the volume of 1M Sodium benzoate be $Vₛ$ $V_s$ mL. Total volume = $Vₛ + Vₐ = 300 mL$ $V_s + V_a = 300\,\text{mL}$.

Millimoles of acid = $1 \times Vₐ = Vₐ$ $1 \times V_a = V_a$ Millimoles of salt = $1 \times Vₛ = Vₛ$ $1 \times V_s = V_s$

Applying Henderson's Equation:

4.5 = 4.2 + ((Vₛ)/(Vₐ))

Step 1: Calculate Volume Ratio

((Vₛ)/(Vₐ)) = 4.5 - 4.2 = 0.3

Since $2 \approx 0.3$ $\log 2 \approx 0.3$, we have:

(Vₛ)/(Vₐ) = 2

$Vₛ = 2 Vₐ$ $V_s = 2 V_a$

Step 2: Substitute and Solve

We know $Vₛ + Vₐ = 300$ $V_s + V_a = 300$ Substituting $Vₛ = 2 Vₐ$ $V_s = 2 V_a$:

2 Vₐ + Vₐ = 300

$3 Vₐ = 300$ $3 V_a = 300$

Vₐ = 100 mL

Chapter Mix

Class 11 Chemistry: Equilibrium

Q82 jee_main_2024_30_jan_morning Solubility Product

The pH at which

Mg(OH)₂

[Kₛₚ=1\times 10⁻¹¹]

begins to precipitate from a solution containing

0.10 M Mg²⁺

ions is

Numerical Answer. Answer: 9 to 9

Solution

Related Formula

Kₛₚ = [Mg²⁺][OH^-]²

pOH = - [OH^-]

$pH + pOH = 14$ $pH + pOH = 14$

Core Logic

Precipitation begins just when the ionic product equals the solubility product ( $Qₛₚ = Kₛₚ$ $Q_{sp} = K_{sp}$).

Step 1: Calculating required [OH-]

[Mg²⁺][OH^-]² = 10⁻¹¹

Given $[Mg²⁺] = 0.10 M$ $[Mg^{2+}] = 0.10 \text{ M}$

0.10 \times [OH^-]² = 10⁻¹¹

[OH^-]² = 10⁻¹⁰

[OH^-] = 10⁻⁵ M

Step 2: Finding pH

pOH = - (10⁻⁵) = 5

$pH = 14 - pOH$ $pH = 14 - pOH$ $pH = 14 - 5 = 9$ $pH = 14 - 5 = 9$

Chapter Mix

Class 11 Chemistry: Equilibrium

Q71 jee_main_2024_31_jan_evening Equilibrium Constants (Kp and Kc)

A(g) leftharpoons B(g) + (C)/(2)(g)

The correct relationship between

KP

α

and equilibrium pressure

P

A. $KP = α(1)/(2)P(1)/(2)(2 + α)(1)/(2)$
B. $KP = α(3)/(2)P(1)/(2)(2 + α)(1)/(2)(1 - α)$
C. $KP = α(1)/(2)P(3)/(2)(2 + α)(3)/(2)$
D. $KP = α(1)/(2)P(1)/(2)(2 + α)(3)/(2)$

Solution

Related Formula

KP = PB \cdot (PC)(1)/(2)PA

where $Pᵢ$ $P_i$ is the partial pressure of component $i$ $i$.

Step 1: Setting up the ICE Table

For the reaction $A(g) leftharpoons B(g) + (1)/(2) C(g)$

Let initial moles of $A = 1$ $A = 1$. At equilibrium: Moles of $A = 1 - α$ $A = 1 - \alpha$ Moles of $B = α$ $B = \alpha$ Moles of $C = (α)/(2)$ $C = \frac{\alpha}{2}$

Total moles at equilibrium = $(1 - α) + α + (α)/(2) = 1 + (α)/(2) = (2 + α)/(2)$

Step 2: Calculating Partial Pressures

Using mole fraction $\times$ $\times$ Total Pressure ( $P$ $P$): $PA = (1 - α)/(1 + (α)/(2)) \cdot P$ $PB = (α)/(1 + (α)/(2)) \cdot P$ $PC = ((α)/(2))/(1 + (α)/(2)) \cdot P$

Step 3: Calculating Kp

KP = PB \cdot (PC)(1)/(2)PA

KP = ( (α)/(1 + α/2) P ) \cdot ( (α/2)/(1 + α/2) P )1/2(1 - α)/(1 + α/2) P

KP = α \cdot (α/2)1/2 \cdot P3/2(1 + α/2)3/2 \cdot (1 + α/2)/((1 - α) P)

KP = α3/2 \cdot P1/2\sqrt(2) \cdot (1 + α/2)1/2 \cdot (1 - α)

Since $1 + α/2 = (2+α)/(2)$ $1 + \alpha/2 = \frac{2+\alpha}{2}$, the $\sqrt(2)$ $\sqrt{2}$ in denominator cancels out perfectly leaving:

KP = α(3)/(2) P(1)/(2)(2 + α)(1)/(2)(1 - α)

Chapter Mix

Class 11 Chemistry: Equilibrium

Q62 jee_main_2024_31_jan_morning Equilibrium Constant

For the given reaction, choose the correct expression of

KC

from the following :-

Fe(aq)³⁺ + SCN(aq)⁻ leftharpoons (FeSCN)(aq)²⁺

A. $KC = [FeSCN²⁺][Fe³⁺][SCN⁻]$
B. $KC = [Fe³⁺][SCN⁻][FeSCN²⁺]$
C. $KC = [FeSCN²⁺][Fe³⁺]²[SCN⁻]²$
D. $KC = [FeSCN²⁺]²[Fe³⁺][SCN⁻]$

Solution

Related Formula

KC = [Products][Reactants]

Core Logic

KC = Products ion conc.Reactants ion conc.

KC = [FeSCN²⁺][Fe³⁺][SCN⁻]

Chapter Mix

Class 11 Chemistry: Equilibrium

Arrange the following in increasing order of solubility product: Ca(OH)₂, AgBr, PbS, HgS$$Ca(OH)_{2},\ AgBr,\ PbS,\ HgS$$

Solution & Explanation

Related Formula

Core Logic

Step 1: Arrangement

Pattern Recognition

Chapter Mix

Reference Study Guides

More Equilibrium Previous-Year Questions — Page 7

Solution

Related Formula

Core Logic

Step 1: Calculation

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Volume Ratio

Step 2: Substitute and Solve

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculating required [OH-]

Step 2: Finding pH

Chapter Mix

Solution

Related Formula

Step 1: Setting up the ICE Table

Step 2: Calculating Partial Pressures

Step 3: Calculating Kp

Chapter Mix

Solution

Related Formula

Core Logic

Chapter Mix

More Equilibrium Questions — jee_main_2025_28_jan_evening

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

Welcome back

Arrange the following in increasing order of solubility product: $Ca(OH)₂, AgBr, PbS, HgS$ $$Ca(OH)_{2},\ AgBr,\ PbS,\ HgS$$