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Question 24069
NEET Chemistry · Hard

Equilibrium: Buffer relation MCQ

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Buffer relationHardQuestion 24069

Question

For a basic buffer, pOH is given by:

  1. A
    pOH=pKb+log⁡[salt][base]pOH=pK_b+\log\dfrac{[salt]}{[base]}
    Correct
  2. B
    pOH=pKb−log⁡[salt][base]pOH=pK_b-\log\dfrac{[salt]}{[base]}
  3. C
    pOH=14−pKbpOH=14-pK_b
  4. D
    pOH=log⁡KbpOH=\log K_b

Correct answer

pOH=pKb+log⁡[salt][base]pOH=pK_b+\log\dfrac{[salt]}{[base]}

Explanation

This is the analogous expression for basic buffers.

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