| ⇦ | 
| ⇨ |
In a buffer solution containing equal concentration of B⁻ and HB, the Kb for B⁻ is 10⁻¹⁰.The pH of buffer solution is
Options
(a) 10
(b) 7
(c) 6
(d) 4
Correct Answer:
4
Explanation:
We know, pOH = pK(b) + log{[B⁻]/[HB]},
Since, [B⁻] = [HB] (given).
Therefore pOH = pK(b) ⇒ pOH = 10,
Therefore pH = 14 – 10 = 4.
Related Questions: - Neutral divalent carbon species released as reaction intermediate in reaction
- The potassium ferricyanide produces on ionisation
- Rectified spirit has ethanol
- Square planar complex of the type MABXL(where A,B,X and L are unidentates) show
- Solubility of iodine in water may be increased by adding
Topics: Equilibrium
(104)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Neutral divalent carbon species released as reaction intermediate in reaction
- The potassium ferricyanide produces on ionisation
- Rectified spirit has ethanol
- Square planar complex of the type MABXL(where A,B,X and L are unidentates) show
- Solubility of iodine in water may be increased by adding
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