Numerical/Formula
1. TITLE
How to solve pH of Strong Acid-Base Mixtures in NEET — Step-by-step
2. PROBLEM TYPE INTRO (50 words)
This numerical category requires calculating the final pH when strong acids and strong bases are mixed. You will use neutralization and limiting reagent formulas to find the net $H^+$ or $OH^-$ left. The most typical mistake students make is forgetting to use the total mixed volume for the final concentration.
3. MCQ (Full numerical problem)
Calculate the pH of the resultant mixture when $10 \text{ mL}$ of $0.1 \text{ M } H_2SO_4$ is mixed with $10 \text{ mL}$ of $0.1 \text{ M } KOH$.
(a) $1.30$
(b) $1.00$
(c) $7.00$
(d) $12.70$
4. STEP-BY-STEP SOLUTION
SOLUTION
Given:
* Volume of $H_2SO_4$ ($V_1$) = $10 \text{ mL}$
* Molarity of $H_2SO_4$ ($M_1$) = $0.1 \text{ M}$
* Volume of $KOH$ ($V_2$) = $10 \text{ mL}$
* Molarity of $KOH$ ($M_2$) = $0.1 \text{ M}$
To Find:
Final $pH$ of the resultant mixture.
Formula:
* Moles of $H^+$ = $n \times M \times V$ (where n is basicity of the acid)
* Moles of $OH^-$ = $n \times M \times V$ (where n is acidity of the base)
* Final $[H^+] = \frac{\text{Net } H^+ \text{ millimoles}}{\text{Total Volume } (V_1 + V_2)}$
* $pH = -\log[H^+]$
Explain when to use this formula: Use this neutralization formula whenever mixing strong acids and strong bases, checking to see which ion ($H^+$ or $OH^-$) remains in excess after they react in a 1:1 ratio.
Step 1: Calculate the initial millimoles of $H^+$ and $OH^-$
Since $H_2SO_4$ is a diprotic acid (gives 2 $H^+$ per molecule):
Millimoles of $H^+ = 2 \times 0.1 \text{ M} \times 10 \text{ mL} = 2 \text{ mmol}$
Since $KOH$ is a monoacidic base:
Millimoles of $OH^- = 1 \times 0.1 \text{ M} \times 10 \text{ mL} = 1 \text{ mmol}$
Step 2: Determine the limiting reagent and final concentration
$H^+$ and $OH^-$ neutralize each other ($1 \text{ mmol}$ of $OH^-$ neutralizes $1 \text{ mmol}$ of $H^+$).
Leftover $H^+ = 2 \text{ mmol} - 1 \text{ mmol} = 1 \text{ mmol}$
Total volume of mixture = $10 \text{ mL} + 10 \text{ mL} = 20 \text{ mL}$
Final $[H^+] = \frac{1 \text{ mmol}}{20 \text{ mL}} = 0.05 \text{ M}$
Step 3: Calculate the final pH
$pH = -\log[H^+]$
$pH = -\log(0.05) = -\log(5 \times 10^{-2})$
$pH = 2 - \log(5)$
Using $\log(5) \approx 0.70$:
$pH = 2 - 0.70 = 1.30$
Answer: $1.30$ (Option A)
5. WHERE STUDENTS GO WRONG
List 2-3 common mistakes:
- ❌ Mistake 1: Forgetting the n-factor (basicity) of the acid. Students often calculate $H^+$ as $1 \times 0.1 \times 10 = 1 \text{ mmol}$, leading to a false complete neutralization and picking Option (c) $7.00$.
- ❌ Mistake 2: Forgetting to add the volumes. Students divide the leftover $1 \text{ mmol}$ of $H^+$ by the initial $10 \text{ mL}$ volume instead of the total $20 \text{ mL}$, getting $[H^+] = 0.1 \text{ M}$ and falsely picking Option (b) $1.00$.
- ❌ Mistake 3: Finding pOH instead of pH. If a calculation sign error occurs or students confuse the leftover ion, they might subtract $1.30$ from $14$, ending up with Option (d) $12.70$.
6. QUICK FORMULA REFERENCE
Formulas Used:
Formula 1 — $[H^+]_{\text{net}} = \frac{N_aV_a - N_bV_b}{V_a + V_b}$
Use when strong acid is in excess ($N_aV_a > N_bV_b$).
Formula 2 — $[OH^-]_{\text{net}} = \frac{N_bV_b - N_aV_a}{V_a + V_b}$
Use when strong base is in excess ($N_bV_b > N_aV_a$), then calculate $pOH = -\log[OH^-]$ and find $pH = 14 - pOH$.
7. SIMILAR PROBLEM HINT
"If NEET changes the values, the method stays the same. Practice with: Mixing $10 \text{ mL}$ of $0.2 \text{ M } Ca(OH)_2$ with $25 \text{ mL}$ of $0.1 \text{ M } HCl$ (Base will be in excess!)".
8. NCERT REFERENCE
Based on: NCERT Class 11, Chapter 6, Exercise 6.66.
9. CTA
More solved numericals → Equilibrium Complete Guide
Practice problems → [Topic Test: Equilibrium]