How to solve pH of Weak Acids (Approximation vs Quadratic) in NEET — Step-by-step

Numerical/Formula

1. TITLE

How to solve pH of Weak Acids (Approximation vs Quadratic) in NEET — Step-by-step

2. PROBLEM TYPE INTRO (50 words)

This category of numerical belongs to ionic equilibrium, specifically calculating the pH of weak acids. We will need the acid dissociation formula $K_a = \frac{c\alpha^2}{1-\alpha}$. The typical mistake students make is blindly using the approximation method ($K_a \approx c\alpha^2$) even when the degree of ionization ($\alpha$) is large, leading to incorrect pH values.

3. MCQ (Full numerical problem)

Question:
The ionization constant ($K_a$) of hydrofluoric acid (HF) is $3.2 \times 10^{-4}$. Calculate the pH of a $0.02 \text{ M}$ solution of HF. (Given: $\log 2.4 = 0.38$)

(A) 2.62
(B) 2.60
(C) 3.50
(D) 0.92

4. STEP-BY-STEP SOLUTION

SOLUTION
Given:
Initial concentration ($c$) = $0.02 \text{ M}$
Ionization constant ($K_a$) = $3.2 \times 10^{-4}$

To Find:
pH of the weak acid solution

Formula:
$K_a = \frac{c\alpha^2}{1-\alpha}$
Use this exact formula when $\alpha$ is not negligible. If $\alpha \ll 1$ (typically less than $0.05$), the denominator $1-\alpha \approx 1$, and you can use the approximation $K_a = c\alpha^2$.

Step 1: Check if the approximation is valid
Let's tentatively assume $\alpha \ll 1$ and use $\alpha = \sqrt{K_a / c}$.
$\alpha = \sqrt{3.2 \times 10^{-4} / 0.02} = \sqrt{160 \times 10^{-4}} \approx 0.126$
Since $0.126 > 0.05$ (or 12.6% > 5%), the approximation is invalid. We must use the quadratic method.

Step 2: Apply the full quadratic equation
$K_a = \frac{c\alpha^2}{1-\alpha}$
$3.2 \times 10^{-4} = \frac{0.02\alpha^2}{1-\alpha}$
$0.016(1-\alpha) = \alpha^2$
$\alpha^2 + 0.016\alpha - 0.016 = 0$
Solving the quadratic equation gives $\alpha = 0.12$ (rejecting the negative root).

Step 3: Calculate $[H^+]$ and pH
$[H^+] = c\alpha = 0.02 \times 0.12 = 2.4 \times 10^{-3} \text{ M}$
$\text{pH} = -\log[H^+] = -\log(2.4 \times 10^{-3})$
$\text{pH} = 3 - \log(2.4) = 3 - 0.38 = 2.62$

Answer: 2.62 (Option A)

5. WHERE STUDENTS GO WRONG

• ❌ Mistake 1: Blindly using the approximation $[H^+] = \sqrt{K_a \cdot c}$. This yields a pH of 2.60 (Option B), which is incorrect because $\alpha$ is too large to ignore.
• ❌ Mistake 2: Forgetting to multiply by concentration. Students find $\alpha = 0.12$ and mistakenly calculate $-\log(\alpha)$ to get 0.92 (Option D) instead of finding $[H^+] = c\alpha$.
• ❌ Mistake 3: Confusing pH with $pK_a$. Students simply calculate $-\log(K_a) = -\log(3.2 \times 10^{-4}) = 3.50$ (Option C).

6. QUICK FORMULA REFERENCE

Formulas Used:

$K_a = \frac{c\alpha^2}{1-\alpha}$ — Use when the degree of ionization $\alpha > 0.05$.
$[H^+] = \sqrt{K_a \cdot c}$ — Use when the degree of ionization $\alpha \ll 1$ (safe to use when $\alpha \le 0.05$).
$\text{pH} = -\log[H^+]$ — Use to find the final pH from the hydronium ion concentration.

7. SIMILAR PROBLEM HINT

"If NEET changes the values, the method stays the same. Practice with: Hypochlorous acid (HOCl) with $K_a = 2.5 \times 10^{-5}$ and $c = 0.08 \text{ M}$, where the approximation method does work flawlessly because $\alpha$ is very small ($1.41 \times 10^{-3}$)!"

8. NCERT REFERENCE

Based on: NCERT Class 11, Chapter 6, Example 6.18.

9. CTA

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