The Henderson-Hasselbalch equation for an aci | vedclass

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Question

(B) The Henderson-Hasselbalch equation for an acidic buffer is: $pH = pK_a + \log \frac{[	ext{Conjugate Base}]}{[	ext{Acid}]}$ Rearranging this,we g

Vedclass · Accepted Answer

$[	ext{Acid}] 	imes 10^{-5} = [	ext{Conjugate Base}]$

Vedclass · Answer

(B) The Henderson-Hasselbalch equation for an acidic buffer is: $pH = pK_a + \log \frac{[	ext{Conjugate Base}]}{[	ext{Acid}]}$ Rearranging this,we get: $pH - pK_a = \log \frac{[	ext{Conjugate Base}]}{[	ext{Acid}]}$ Given that $pH - pK_a = 5$,we substitute this into the equation: $5 = \log \frac{[	ext{Conjugate Base}]}{[	ext{Acid}]}$ Taking the antilog on both sides: $10^5 = \frac{[	ext{Conjugate Base}]}{[	ext{Acid}]}$ Therefore,$[	ext{Conjugate Base}] = [	ext{Acid}] 	imes 10^5$ Or,$[	ext{Acid}] = [	ext{Conjugate Base}] 	imes 10^{-5}$ Comparing this with the given options,option $B$ is correct.

The Henderson-Hasselbalch equation for an acidic buffer is given by $pH - pK_a = 5$. This condition holds true when:

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