Q.81 Chemistry ⭐ 360° Option Elimination Active

NEET 2021 Chemistry Question 81 Solution

Topic: Equilibrium · Official NTA Booklet Code M1

Question

The pKb of dimethylamine and pKa of acetic acid are 3.27 and 4.77 respectively at T (K). The correct option for the pH of dimethylammonium acetate solution is:

(1)8.50
(2)5.50
(3)7.75
(4)6.25

✓

Verified Against NTA Final Answer Key

Option (3) : 7.75

📚 Equilibrium Easy

📋 Given Parameters & Problem Setup

Salt Dimethylammonium acetate (CH3COO- NH2+(CH3)2)

pKb of Dimethylamine 3.27

pKa of Acetic Acid 4.77

⚡

The 30-Second Exam Hall Shortcut: ⚡ Weak Acid + Weak Base Formula: pH = 7 + ½(pKa - pKb)

⏱️ Save ~20 seconds

Dimethylammonium acetate is the salt of a weak acid ( $CH_3COOH$ ) and weak base ( $(CH_3)_2NH$ ). The universal Henderson-hydrolysis formula is $pH = 7 + \frac{1}{2}(pK_a - pK_b) = 7 + \frac{1}{2}(4.77 - 3.27) = 7 + \frac{1}{2}(1.50) = 7 + 0.75 = 7.75$ ! Direct 10-second calculation matching Option (3)!

1 Identify salt type: Weak acid (

CH_3COOH

pK_a = 4.77

) + Weak base (

(CH_3)_2NH

pK_b = 3.27

2 Use standard salt hydrolysis formula:

pH = \frac{1}{2}pK_w + \frac{1}{2}pK_a - \frac{1}{2}pK_b

3 Substitute standard

pK_w = 14

pH = 7 + \frac{1}{2}(4.77 - 3.27)

4 Calculate:

pH = 7 + 0.75 = 7.75

5 Match Option (3).

🛡️

Why You Can 100% Trust This Shortcut: Canonical NCERT Ionic Equilibrium hydrolysis formula for salt of weak acid and weak base.

📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Classifying the Electrolyte Components of the Salt

Dimethylammonium acetate ( $CH_3COONH_2(CH_3)_2$ ) is formed by the neutralization of acetic acid ( $CH_3COOH$ , a weak acid with $pK_a = 4.77$ ) and dimethylamine ( $(CH_3)_2NH$ , a weak base with $pK_b = 3.27$ ). Both the cation and the anion undergo simultaneous hydrolysis in aqueous solution.

Step 2

Formulating the Hydrolysis Equilibrium and pH Equation

The hydrolysis equilibrium is: $CH_3COO^- + (CH_3)_2NH_2^+ + H_2O \rightleftharpoons CH_3COOH + (CH_3)_2NH$ The hydronium ion concentration in such a solution is independent of salt concentration and given by: $[H_3O^+] = \sqrt{\frac{K_w \cdot K_a}{K_b}}$ Taking negative logarithms on both sides:

pH = \frac{1}{2} pK_w + \frac{1}{2} pK_a - \frac{1}{2} pK_b = 7 + \frac{1}{2}(pK_a - pK_b)

Step 3

Substituting the Numerical Values

Substitute $pK_a = 4.77$ and $pK_b = 3.27$ :

pH = 7 + \frac{1}{2}(4.77 - 3.27) = 7 + \frac{1}{2}(1.50) = 7 + 0.75 = 7.75

⭐

360° Option Elimination Engine Key Feature

Diagnose every choice: Know exactly why the right option works and why distractors fail.

4-Option Deep Scan

1 1

✗ Incorrect

Verdict & Scientific Reason:

Result of adding $1.50$ directly to 7 without the factor of $1/2$ ( $7 + 1.50 = 8.50$ ).

⚠️

Student Fallacy (Why candidates pick this trap): Forgetting to divide

(pK_a - pK_b)

by 2.

2 2

✗ Incorrect

Verdict & Scientific Reason:

Result of subtracting $1.50$ from 7 without factor of $1/2$ ( $7 - 1.50 = 5.50$ ).

⚠️

Student Fallacy (Why candidates pick this trap): Sign inversion and omission of half-factor.

3 3

✓ 100% Correct

Verdict & Scientific Reason:

Exact pH obtained by proper application of salt hydrolysis formula: $7 + 0.75 = 7.75$ .

💡

Why this is airtight: None; textbook ionic equilibrium calculation.

4 4

✗ Incorrect

Verdict & Scientific Reason:

Result of inverting signs: $7 - 0.75 = 6.25$ (as if base were weaker than acid).

⚠️

Student Fallacy (Why candidates pick this trap): Swapping

pK_a

and

pK_b

in the numerator.

⚠️

NTA Trap Alert: Sign Inversion and Factor-of-Half Traps

Two traps exist here: 1. Forgetting the factor of $1/2$ , leading to $7 + 1.50 = 8.50$ (Option 1). 2. Inverting $pK_a$ and $pK_b$ to get $7 - 0.75 = 6.25$ (Option 4). Since $pK_b (3.27) < pK_a (4.77)$ , the base is STRONGER than the acid, meaning the solution must be slightly basic ( $pH > 7$ ).

🛡️ How to avoid the -1 mark: Sanity check: Dimethylamine (

pK_b = 3.27

) is stronger as a base than acetic acid (

pK_a = 4.77

) is as an acid

\implies

Solution must be basic (

pH > 7

). Only

7.75

and

8.50

are

> 7

. With the

1/2

factor,

pH = 7.75

📖

Core NEET Syllabus Standard

NCERT Chemistry · Equilibrium Standard Benchmark

Core Scientific Principle Tested:

"For a salt of weak acid and weak base, $pH = 7 + \frac{1}{2}(pK_a - pK_b)$ , independent of salt concentration."

Syllabus Focus: Hydrolysis of Salts and the pH of their Solutions

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

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NEET 2021 Chemistry Question 81 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Weak Acid + Weak Base Formula: pH = 7 + ½(pKa - pKb)

Step-by-Step Rigorous Workflow & Mathematical Proof

Classifying the Electrolyte Components of the Salt

Formulating the Hydrolysis Equilibrium and pH Equation

Substituting the Numerical Values

360° Option Elimination Engine Key Feature

NTA Trap Alert: Sign Inversion and Factor-of-Half Traps

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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