Ionic Equilibrium: pH of Acid–Base Mixtures Explained
Mixing a Strong Acid with a Strong Base causes a neutralization reaction. If you mix exact, identical amounts of both, they perfectly cancel out, yielding salt and water with a pH of 7.
But examiners rarely give you perfect mixtures. What happens if you mix 100 mL of an acid with 50 mL of a base? One of them will overpower the other! To find the exact pH of the final leftover solution, we must master the Milli-Equivalents (N × V) Method.
Video Tutorial: The Neutralization Method
Watch Abhishek Sengar sir from CHEMCA break down the N × V logic and apply it to a step-by-step numerical example for JEE and NEET.
1. The Secret Weapon: Normality × Volume
To figure out which chemical "wins" the neutralization battle, you cannot just look at Molarity or Volume alone. You must calculate the number of milli-equivalents (meq) for both the acid and the base.
Remember: Normality (N) = Molarity (M) × n-factor
Let's define our two competitors:
- Acid's power: NaVa
- Base's power: NbVb
2. The Three Battle Scenarios
Once you calculate NaVa and NbVb, simply compare the two numbers.
Fig: Always subtract the smaller N×V value from the larger one. The winner dictates whether the solution is acidic or basic!
3. Solving a Classic Numerical
Example: Mix 100 mL of 0.2 M HCl with 100 mL of 0.1 M NaOH. Find the final pH.
Step 1: Calculate N×V
For HCl (Acid): n-factor = 1. So, Normality = 0.2 N.
NaVa = 0.2 × 100 = 20 meq.
For NaOH (Base): n-factor = 1. So, Normality = 0.1 N.
NbVb = 0.1 × 100 = 10 meq.
Step 2: Compare and Calculate
Since 20 > 10, the Acid wins. The solution is Acidic.
Excess Acid = 20 - 10 = 10 meq.
Total Volume = Va + Vb = 100 + 100 = 200 mL.
[H+] = Excess meq / Total Volume = 10 / 200 = 1/20 M.
Step 3: Calculate pH
pH = -log(1/20) = log(20) = log(2 × 10) = log(2) + log(10)
pH = 0.301 + 1 = 1.301.
Practice Questions for JEE & NEET
Let's test your ability to avoid the two most common traps examiners set in these types of numericals!
Question 1 (The n-factor Trap): 50 mL of 0.1 M H2SO4 is mixed with 50 mL of 0.1 M NaOH. What is the final pH of the mixture? (Given log 5 = 0.7).
Answer: pH = 1.3
Reasoning:
This is a major trap! H2SO4 is a dibasic acid, meaning its n-factor is 2. If you forget to multiply Molarity by the n-factor, your whole answer will be wrong.
1. Normality of Acid = 0.1 M × 2 = 0.2 N.
NaVa = 0.2 × 50 = 10 meq.
2. Normality of Base (NaOH) = 0.1 M × 1 = 0.1 N.
NbVb = 0.1 × 50 = 5 meq.
3. Acid > Base. Excess Acid = 10 - 5 = 5 meq.
4. Total Volume = 50 + 50 = 100 mL.
5. [H+] = 5 / 100 = 0.05 M = 5 × 10-2 M.
6. pH = -log(5 × 10-2) = 2 - log(5) = 2 - 0.7 = 1.3.
Question 2 (The Volume Trap): In the formula to find the final concentration, why is it absolutely mandatory to divide the excess milli-equivalents by the sum of the volumes (Va + Vb)? Why can't we just divide by the acid's volume?
Answer: Because Concentration is defined as Moles per TOTAL Volume of the solution.
Reasoning:
When you physically pour 100 mL of Acid and 100 mL of Base into the same beaker, the liquids combine. The new, final solution has a total volume of 200 mL.
Even though only the acid "won" the reaction, those leftover acid molecules are now swimming around in a 200 mL pool, not a 100 mL pool! Forgetting to add the volumes together is the number one reason students fail neutralization calculations. Always divide by the Total Mixed Volume!
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