MENU

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. Ratio of Cᵥ and Cᵥ of the gas X is 1.4. The number of atoms of the gas X present
  2. The most effective electrolyte for the coagulation of As₂S₃ sol is
  3. P₂O₅ is an anhydride of
  4. Many transition metals and their compounds exhibit
  5. The function of ”Sodium pump” is a biological process operating in each

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Your email address will not be published.


*