| ⇦ | 
| ⇨ |
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: - Heat of combustion ΔH for C(s) , H₂(g)and CH₄(g) are -94, -68 and -213 kcal/mol.
- The correct order of ionisation energy for comparing carbon, nitrogen and oxygen
- At 25⁰ C , the highest osmotic pressure is exhibited by 0.1 M solution of
- A secondary amine could be prepared readily from the starting material
- At equilibrium, the value of change in Gibb’s free energy (ΔG) is
Topics: Equilibrium
(104)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Heat of combustion ΔH for C(s) , H₂(g)and CH₄(g) are -94, -68 and -213 kcal/mol.
- The correct order of ionisation energy for comparing carbon, nitrogen and oxygen
- At 25⁰ C , the highest osmotic pressure is exhibited by 0.1 M solution of
- A secondary amine could be prepared readily from the starting material
- At equilibrium, the value of change in Gibb’s free energy (ΔG) is
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.
Share this page with your friends
Leave a Reply