Buffer Formation Question – Ionic Equilibrium JEE Main 2026 PYQ

Medium 26 Mar 2026
Nishant Kumar Gupta
Nishant Kumar Gupta
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Weak Base Buffer Volume Calculation – Ionic Equilibrium JEE Main 2026 PYQ

Quick Summary

Question Type: Buffer solution and Henderson equation calculation
Chapter: Ionic Equilibrium – Buffer Solution
Difficulty: ⭐⭐⭐ Medium
Time to Solve: 3-5 minutes
Key Formula: pOH = pKb + log([Salt]/[Base])
Correct Answer: (B) x = 14.3, y = 85.7
Why: Since pH = 9, pOH = 5. Using pOH = pKb + log([Salt]/[Base]) gives [Salt]/[Base] = 1/5. Then x/(y – x) = 1/5 and x + y = 100, so x = 14.3 mL and y = 85.7 mL.

The Question

JEE Main 2026 – Ionic Equilibrium

Consider a weak base B of pKb = 5.699. x mL of 0.02 M HCl and y mL of 0.02 M weak base B are mixed to make 100 mL of a buffer of pH 9 at 25°C. The values of x and y respectively are:

Given: log 2 = 0.3010, log 3 = 0.4771, log 5 = 0.699

(A) x = 42.7, y = 57.3
(B) x = 14.3, y = 85.7
(C) x = 85.7, y = 14.3
(D) x = 11.1, y = 88.9

Quick Answer

Correct Option: (B) x = 14.3, y = 85.7

Reasoning: This is a basic buffer formed by mixing a weak base with a strong acid. Since pH = 9, we first calculate pOH = 14 – 9 = 5. Then:

  • pOH = pKb + log([Salt]/[Base])
  • 5 = 5.699 + log([Salt]/[Base])
  • log([Salt]/[Base]) = -0.699
  • [Salt]/[Base] = 1/5

Now salt formed is proportional to x, and base remaining is proportional to y – x.

  • x / (y – x) = 1/5
  • y = 6x
  • x + y = 100
  • 7x = 100
  • x = 14.3 mL, y = 85.7 mL

Video Solution

If you want the full explanation in a clear step-by-step teaching format, watch the video solution below:

Watch Full Video Solution on YouTube

Understanding the Concept

Why This Is a Basic Buffer Problem

When a weak base is partially neutralized by a strong acid, a basic buffer is formed. The final solution contains:

  • unreacted weak base
  • its salt formed by reaction with strong acid

For such a buffer, we use the Henderson equation:

pOH = pKb + log([Salt]/[Base])

So the whole problem reduces to finding the ratio of salt to base and then relating that ratio to the given volumes.

The Key Principle

  1. Convert pH into pOH
  2. Use the Henderson equation for a basic buffer
  3. Express salt and base in terms of x and y
  4. Use total volume = 100 mL

Detailed Step-by-Step Solution

Step 1: Calculate pOH

  • Given pH = 9
  • So pOH = 14 – 9 = 5

Step 2: Apply Henderson Equation for Basic Buffer

Handwritten solution explaining weak base buffer volume calculation for JEE Main 2026 Ionic Equilibrium question


Download Handwritten Solution

  • Use: pOH = pKb + log([Salt]/[Base])
  • 5 = 5.699 + log([Salt]/[Base])
  • log([Salt]/[Base]) = -0.699

Step 3: Find the Ratio of Salt to Base

  • [Salt]/[Base] = 10-0.699
  • Since log 5 = 0.699, 10-0.699 = 1/5
  • Therefore, [Salt]/[Base] = 1/5

Step 4: Relate the Ratio to x and y

  • Salt is formed from HCl added, so salt ∝ x
  • Weak base remaining after neutralization ∝ y – x
  • So x / (y – x) = 1/5

Solving:

  • 5x = y – x
  • y = 6x

Step 5: Use Total Volume Condition

  • x + y = 100
  • x + 6x = 100
  • 7x = 100
  • x = 14.3 mL
  • y = 85.7 mL

Final Answer

Option (B): x = 14.3, y = 85.7 ✓

The correct values are x = 14.3 mL and y = 85.7 mL.

Essential Formulas for This Topic

Primary Equations

  1. Basic Buffer Formula:
    • pOH = pKb + log([Salt]/[Base])
    • Used for weak base + strong acid buffer
    • Always use pOH, not pH directly
  2. pH-pOH Relation:
    • pH + pOH = 14
    • At 25°C, pOH = 14 – pH
    • Here pOH = 5
  3. Neutralization Logic:
    • Salt formed = moles of HCl added
    • Base remaining = initial base – consumed base
    • These values are used in the buffer ratio

Important Constants

  • Given: log 5 = 0.699
  • At 25°C, pH + pOH = 14
  • A buffer requires weak base and its salt together
  • Strong acid partially neutralizes the weak base to form the buffer

Common Mistakes to Avoid

❌ Mistake 1: Using pH Directly

Wrong Thinking: “Put pH = 9 directly into the base buffer formula.”

Correct Approach: For a basic buffer, convert pH to pOH first. Here pOH = 5.

❌ Mistake 2: Reversing the Ratio

Wrong Approach: Using [Base]/[Salt] instead of [Salt]/[Base]

Correct Approach: The correct relation is pOH = pKb + log([Salt]/[Base]).

❌ Mistake 3: Taking y as Base Remaining

Common Error:

  • Writing x / y = 1/5
  • Ignoring that some base reacts with HCl

Correct Approach: Base remaining is y – x, not y.

❌ Mistake 4: Forgetting x + y = 100

Wrong Thinking: “After finding y = 6x, the problem is finished.”

Correct Understanding:

  • The final mixture volume is 100 mL
  • So x + y = 100
  • This is required to get the actual numerical values

Key Concept Summary

What You Must Remember

  1. Weak base + strong acid gives a basic buffer: Use pOH, not pH directly
  2. Henderson equation is essential: pOH = pKb + log([Salt]/[Base])
  3. Salt comes from acid added: Base remaining is after neutralization
  4. Use the ratio carefully: x / (y – x) = salt/base
  5. Apply the total volume condition: x + y = 100

The Golden Rule for Basic Buffer Questions

“Convert pH to pOH first, then use the salt/base ratio after neutralization.”

Frequently Asked Questions

Q1: Why do we use pOH here instead of pH?

A: Because the buffer is made from a weak base and its salt, so the Henderson equation is written in terms of pOH and pKb.

Q2: Why is salt proportional to x?

A: Because the strong acid HCl forms the salt by reacting with the weak base, so its amount depends on x.

Q3: Why is the remaining base equal to y – x?

A: Because the amount of base consumed is equal to the amount of HCl added, so unreacted base is y – x.

Q4: How do we convert -0.699 into 1/5?

A: Since log 5 = 0.699, 10-0.699 = 1/5.

Q5: What are the final values of x and y?

A: x = 14.3 mL and y = 85.7 mL.

Prerequisites to Solve This Question

Before attempting this problem, you should understand:

  1. Buffer solutions: Acidic and basic buffer formation
  2. Henderson equation: For weak acid and weak base buffers
  3. Neutralization concept: Mole relation between acid and base
  4. pH and pOH conversion: At 25°C
  5. Basic logarithms: Especially powers and standard log values

After Solving This, You Can:

✅ Solve weak base buffer numerical questions confidently
✅ Use Henderson equation in JEE Main equilibrium problems
✅ Convert pH to pOH correctly for basic buffers
✅ Relate acid-base reaction moles to buffer composition
✅ Avoid common mistakes in salt/base ratio questions
✅ Handle similar Ionic Equilibrium PYQs faster

Study Tips for This Topic

For JEE Main:

  1. Memorize Henderson equations: For both acidic and basic buffers
  2. Identify the buffer type first: Weak acid buffer or weak base buffer
  3. Track neutralization carefully: Always see how much weak base remains
  4. Use log values smartly: They are often given to simplify the calculation

Common JEE Variants:

  • Find pH of a buffer
  • Find salt/base ratio
  • Find volumes required to prepare a buffer
  • Buffer calculation after partial neutralization

Difficulty Rating & Exam Frequency

Difficulty Level: ⭐⭐⭐ (3/5) – Medium
JEE Main Frequency: High – Buffer solution questions are common in Ionic Equilibrium
JEE Advanced Frequency: Medium – Often combined with salt hydrolysis and equilibrium concepts
Topic Importance: Very High – Buffer concept is a core part of Ionic Equilibrium

Written by Nishant Kumar
Chemistry Educator with 10+ Years of Experience Teaching JEE Aspirants
Founder – PadhoLikhoJEE

Last Updated: March 2026
Question Source: JEE Main 2026 PYQ
Topic: Ionic Equilibrium – Buffer Solution

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