The pH of a solution containing 0.1 N NH_4OH | vedclass

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(B) The given solution is a basic buffer consisting of a weak base $(NH_4OH)$ and its salt with a strong acid $(NH_4Cl)$.For a basic buffer,the $pOH$

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$4.75$

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(B) The given solution is a basic buffer consisting of a weak base $(NH_4OH)$ and its salt with a strong acid $(NH_4Cl)$.For a basic buffer,the $pOH$ is calculated using the Henderson-Hasselbalch equation: $pOH = pK_b + \log \frac{[Salt]}{[Base]}$.Given: $pH = 9.25$,$[Salt] = 0.1 \ N$,$[Base] = 0.1 \ N$.First,calculate $pOH$: $pOH = 14 - pH = 14 - 9.25 = 4.75$.Substitute the values into the equation: $4.75 = pK_b + \log \frac{0.1}{0.1}$.Since $\log(1) = 0$,we get $4.75 = pK_b + 0$.Therefore,$pK_b = 4.75$.

The $pH$ of a solution containing $0.1 \ N$ $NH_4OH$ and $0.1 \ N$ $NH_4Cl$ is $9.25$. What is the $pK_b$ for $NH_4OH$?

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