Numerical/Formula

1. TITLE

How to solve Buffer pH (Henderson-Hasselbalch) numericals in NEET — Step-by-step

2. PROBLEM TYPE INTRO

This is an ionic equilibrium numerical focused on acidic buffer solutions, which are designed to resist changes in pH. You will need the Henderson-Hasselbalch equation to evaluate the pH. A typical mistake students make is flipping the salt and acid concentration ratio inside the logarithm, leading to an incorrect sign in the final answer.

3. MCQ

Problem: Calculate the pH of an acidic buffer solution containing $0.1 \text{ M}$ acetic acid ($CH_3COOH$) and $1.0 \text{ M}$ sodium acetate ($CH_3COONa$). (Given: $pK_a$ of acetic acid = $4.76$, $\log 10 = 1$)

(A) $5.76$
(B) $3.76$
(C) $4.76$
(D) $6.76$

4. STEP-BY-STEP SOLUTION

SOLUTION

Given:
Concentration of Acid $[\text{HA}]$ (Acetic acid) = $0.1 \text{ M}$
Concentration of Salt $[\text{A}^-]$ (Sodium acetate) = $1.0 \text{ M}$
$pK_a$ of acetic acid = $4.76$
$\log 10 = 1$

To Find:
The pH of the acidic buffer solution.

Formula:
$\text{pH} = \text{pK}_a + \log \frac{[\text{Salt}]}{[\text{Acid}]}$
Use this formula (the Henderson-Hasselbalch equation) to find the pH of an acidic buffer. If you take equal molar concentrations of the acid and the salt, the $\log$ term vanishes (since $\log 1 = 0$), and the pH becomes exactly equal to the $pK_a$.

Step 1: Identify the ratio of Salt to Acid
$\text{Ratio} = \frac{[\text{Salt}]}{[\text{Acid}]} = \frac{1.0 \text{ M}}{0.1 \text{ M}} = 10$

Step 2: Calculate the log of the ratio
$\log(10) = 1$

Step 3: Apply the Henderson-Hasselbalch equation
$\text{pH} = 4.76 + 1$
$\text{pH} = 5.76$

Answer: (A) 5.76

5. WHERE STUDENTS GO WRONG

  • ❌ Mistake 1 (Sign Error -> Option B): Flipping the log ratio to $\log\frac{[\text{Acid}]}{[\text{Salt}]}$. This results in subtracting the log value instead of adding it (calculating $4.76 - 1 = 3.76$).
  • ❌ Mistake 2 (Common Calculation Mistake -> Option C): Assuming that because it's a buffer, the pH is always equal to the $pK_a$ ($4.76$), completely ignoring the concentration variations.
  • ❌ Mistake 3 (Unit/Math Error -> Option D): Making a basic arithmetic error or randomly adding the concentrations directly to the $pK_a$ without using the log scale.

6. QUICK FORMULA REFERENCE

Formulas Used:

$\text{pH} = \text{pK}_a + \log \frac{[\text{Salt}]}{[\text{Acid}]}$ — Use when calculating the pH of an acidic buffer (a weak acid + salt formed with a strong base).
$\text{pOH} = \text{pK}_b + \log \frac{[\text{Conjugate acid}, BH^+]}{[\text{Base}, B]}$ — Use when calculating the pOH of a basic buffer (a weak base + salt of its conjugate acid). (Note: calculate pH afterward using $\text{pH} + \text{pOH} = 14$)

7. SIMILAR PROBLEM HINT

"If NEET changes the values, the method stays the same. Practice with: A basic buffer variation (like Ammonia and Ammonium Chloride) where you must use $pK_b$ to find the pOH first, and then subtract from 14 to get the final pH."

8. NCERT REFERENCE

Based on: NCERT Class 11, Chapter 6 (Equilibrium), Section 6.12.1 (Designing Buffer Solution).

9. CTA

More solved numericals → Equilibrium Complete Guide
Practice problems → [Topic Test: Equilibrium]