Home/Question #6829
COMMON (JEE/NEET)Easy

Q.The Henderson-Hasselbalch equation for a basic buffer is:

a
pOH=pKb+log⁡([Salt]/[Base])pOH = pKb + \log([Salt]/[Base])
b
pH=pKa+log⁡([Salt]/[Acid])pH = pKa + \log([Salt]/[Acid])
c
pOH=pKb+log⁡([Base]/[Salt])pOH = pKb + \log([Base]/[Salt])
d
pH=pKb+log⁡([Salt]/[Base])pH = pKb + \log([Salt]/[Base])

Correct Answer: Option A

The correct solution involves applying the fundamental concept to derive the final value step by step...

Unlock Detailed Solution

Register for Free

Already a member? Login here