Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
30 ml of 0.2 M NaOH is added with 50 ml 0.2 M CH3COOH solution. The extra volume of 0.2 M NaOH required to make the pH of the solution 5.00 is 10x .The value of x is. The ionisation constant of CH3COOH=2×10-5 .
ChemistryIonic EquilibriumJEE Main
Solution:
2176 Upvotes Verified Answer
The correct answer is: 3
Milli.moles of NaOH=30×0.2=6

m.m of CH3COOH=50×0.2=10

Now:



pH=-log(2×10-5) +log64

=4.87

Suppose 'v'mL of NaOH is added then

m.mol of CH3COONa=6+v×0.2

m.mol of CH3COOH=4-v×0.2

pH= pK a +log Salt Acid
5=-log(2×10-5) +log6+0.2v4-0.2v

0.3010=log6+0.2v4-0.2v

So 6+0.2v4-0.2v=2v=3.33 mL=103mL

So x=3

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.