JEE Main · Chemistry ↓ Falling

Equilibrium appeared 36 times across 3 years — 4.2% of Chemistry. This question is from Solubility Product.

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Questions 10 18 8 36
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The pH at which Mg(OH)₂ [Kₛₚ=1× 10⁻¹¹] begins to precipitate from a solution containing 0.10 M Mg²⁺ ions is

Numerical Answer Type:
Enter a numerical value Answer: 9 to 9 +4 marks

Solution & Explanation

Related Formula
Kₛₚ = [Mg²⁺][OH^-]² pOH = - [OH^-]

pH + pOH = 14

Core Logic

Precipitation begins just when the ionic product equals the solubility product (Qₛₚ = Kₛₚ).

Step 1: Calculating required [OH-]
[Mg²⁺][OH^-]² = 10⁻¹¹

Given [Mg²⁺] = 0.10 M

0.10 × [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 - 5 = 9

Chapter Mix

Class 11 Chemistry: Equilibrium

Reference Study Guides

Equilibrium

More Equilibrium Previous-Year Questions — Page 8

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

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