NIOS Class 12 Chemistry Chapter 12 Ionic Equilibrium Solutions English Medium As Per New Syllabus to each chapter is provided in the list so that you can easily browse throughout different chapters NIOS Class 12 Chemistry Chapter 12 Ionic Equilibrium Notes in English and select need one. NIOS Class 12 Chemistry Solutions English Medium Download PDF. NIOS Study Material of Class 12 Chemistry Notes Paper Code: 313.
NIOS Class 12 Chemistry Chapter 12 Ionic Equilibrium
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Ionic Equilibrium
Chapter: 12
| Module – V: Chemical Dynamics |
INTEXT QUESTIONS 12.1
1. Define Arrhenius acid and give two examples.
Ans: According to Arrhenius concept an acid is defined as a substance that is capable of producing hydrogen ion (H⁺) by ionisation in aqueous solution. For example, HCl and CH₃COOH.
2. What are the limitations of Arrhenius definition?
Ans: Arrhenius definition has the following drawbacks:
z It is limited to only aqueous solutions and requires ionisation of the substance.
z It does not explain the acidic or basic behaviour of some substances which lack a hydrogen (or a hydroxide) ion.
For example, AlCl₃ and Na₃CO₃ which lack a hydroxide.
3. How does a Brønsted- Lowry base differ from an Arrhenius base?
Ans: In the Bornsted-Lowry concept, any molecule or ion that can accept a proton is a base whereas in Arrhenius concept a base is the one which provides hydroxide ions in solution.
4. Classify the following into Brønsted- Lowry acid or Brønsted- Lowry base.
HCl, NH₃, H3O⁺, CN⁻
Ans: Acids HCl, H₃O⁺
Bases NH₃, CN⁻
5. The degree of dissociation of two electrolytes X and Y are 1.0 × 10⁻⁴ and 0.96. Comment on the nature of these electrolytes.
Ans: X is a weak electrolyte and Y is a strong electrolyte.
INTEXT QUESTIONS 12.2
1. HF is a weak acid in water. Write down the expression for Ka for the dissociation of HF.
Ans: The ionisation of weak acid, HF, can be represented as:
HF(aq) + H₂O H₃O⁺ (aq) + F⁻ (aq)
The expression for Kₐ would be,
kₐ = ([H₃O⁺ ][F⁻ ])/[HF]
2. Consider a weak base BOH which partially dissociates in its aqueous solutions as per the following equilibrium:
B + H2O ⇌ BH⁺ + OH⁻
Ans: For a weak base BOH which partially dissociates in aqueous solutions, and has a degree of dissociation as α we can write:
B + H₂O ⇌ BH⁺ + OH⁻
The equilibrium constant expression or base dissociation constant can be written as:
K = [BH⁺][OH⁻]/ [HO][B] = [cα] [cα]/{c[1- α] 55}
rearranging we get,
⇒ 55K K = kb = {[cα ] [cα ]}\ {c[α-1] =
(c²α²)/{c[1- α] (1-α)
Since the acid B is very weak, α <<1; we can neglected in comparison to 1in the denominator to get
Kb ≈ cα² or α² = (kb)/(c) or α = √kb/c
3. A sample of lime juice has a hydronium ion concentration of 6.3 × 10⁻² M. Calculate its pH.
Ans: Given hydronium ion concentration, [H₃O+] = 6.3 × 10⁻²M
As per definition pH = –log [H₃O⁺]
⇒ pH = – log 6.3 × 10⁻²
⇒ pH = –(0.7993 – 2.0000)
⇒ pH = –(–1.2007) = 1.2007
4. Calculate the pH of 1.0 M aqueous solution of amino acid glycine – a weak acid. The Kₐ = 1.67 ×10⁻¹⁰
Ans: Given: Concentration of glycine = 1.0M
Ka = 1.67 × 10⁻¹⁰.
For a weak acid α = √Ka/c = α = √1.67 × 10⁻¹⁰ = 1.29 × 10⁻⁵
⇒ [H₃O+] = 1 × 1.29 × 10⁻⁵ = 1.29 × 10⁻⁵M
pH = –log [H₃O+] = –log [1.29 × 10⁻⁵]
= – (–4.8894) = 4.8894.
INTEXT QUESTIONS 12.3
1. Calculate the pH of a solution containing 0.05 M benzoic acid and 0.025 M sodium benzoate. Benzoic acid has a pKa of 4.2.
Ans: Here, [Acid] = 0.05M and [Salt] = 0.025M; and pKa = 4.2
Substituting the values in Handerson equation, we get
pH = 4.2 + log (0.05/0.025) = 4.2 + log 2 = 4.2 + 0.3010 = 4.5010
2. Calculate the solubility product for Ag₂SO₄ if [SO₄²⁻] = 2.5 ´ 10⁻² M.
Ans: Let the solubility of Ag₂SO₄ be ‘s’ mol dm⁻³
The concentrations of the Ag⁺ and the SO₄⁻² would be ‘2s’ mol dm⁻³ and ‘s’ mol dm⁻³ respectively, and Ksp = [Ag⁺]²
[SO₄²⁻]
Given [SO₄⁻] = 2.5 × 10⁻²M ⇒ [Ag⁺] = 2 × 2.5 ×10⁻²M = 5 ×10⁻²M
Substituting the values in the expression of Ksp we get,
Ksp = [5 × 10⁻²]² × [2.5 × 10⁻²] = 6.25
× 10⁻⁵ mol³ dm⁻⁹
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