When 10 mL of 0.1 M acetic acid (pKa = 5.0 ) is titrated against 10mL of 0.1 M

When 10 Ml Of 01 M Acetic Acid Pka Chemistry Question

When 10 mL of 0.1 M acetic acid (pKa = 5.0 ) is titrated against 10mL of 0.1 M ammonia solution (pKb = 5.0),the equivalence point occurs at pH

Options

(a) 5
(b) 6
(c) 7
(d) 9

Correct Answer:

7

Explanation:

pKₐ = -logKₐ : pK(b) = -logK(b),
pH = -1/2[logKₐ + log K(w) – logK(b),
-1/2[-5 + log(1*10⁻¹⁴)-(-5)],
-1/2[-5-14+5]=-1/2(-14)=7.

Related Questions:

  1. A dibromo derivative of an alkane reacts with sodium metal to form an alicyclic
  2. Which of the following is isoelectronic?
  3. Which of the following is dihydric alcohol
  4. Which one is not equal to zero for an ideal solution
  5. The half-life of a reaction is inversely proportional to the square of the

Topics: Equilibrium (104)
Subject: Chemistry (2512)

Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score

18000+ students are using NEETLab to improve their score. What about you?

Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.

NEETLab Mobile App

Share this page with your friends

Be the first to comment

Leave a Reply

Your email address will not be published.


*