The pOH of a basic buffer (e.g.,NH_4OH/NH_4Cl | vedclass

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Question

(A) The Henderson-Hasselbalch equation for a basic buffer is given by: $pOH = pK_b + \log \frac{[Salt]}{[Base]}$.Initially,$5 = pK_b + \log \frac{[Sal

Vedclass · Accepted Answer

$5.48$

Vedclass · Answer

(A) The Henderson-Hasselbalch equation for a basic buffer is given by: $pOH = pK_b + \log \frac{[Salt]}{[Base]}$.Initially,$5 = pK_b + \log \frac{[Salt]}{[Base]}$.When the concentration of the salt is tripled,the new concentration becomes $3 	imes [Salt]$.The new $pOH$ is: $pOH_{new} = pK_b + \log \frac{3 	imes [Salt]}{[Base]}$.This can be written as: $pOH_{new} = (pK_b + \log \frac{[Salt]}{[Base]}) + \log 3$.Substituting the initial values: $pOH_{new} = 5 + 0.48 = 5.48$.

The $pOH$ of a basic buffer (e.g.,$NH_4OH/NH_4Cl$) is $5$. If the concentration of the salt is tripled while the concentration of the base remains constant,find the new $pOH$ value. (Given: $\log 3 = 0.48$)

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