When 30 mL of 0.2 M NH_4OH is added to 30 | vedclass

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(D) 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 $pOH$

Vedclass · Accepted Answer

$4.8$

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(D) 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: $pOH = pK_b + \log \frac{[Salt]}{[Base]}$.First,calculate the $pOH$ from the given $pH$: $pOH = 14 - pH = 14 - 8.2 = 5.8$.Since the volumes of both solutions are equal $(30 \ mL)$,the ratio of concentrations $\frac{[Salt]}{[Base]}$ is equal to the ratio of their molarities: $\frac{[NH_4Cl]}{[NH_4OH]} = \frac{2 \ M}{0.2 \ M} = 10$.Substitute these values into the Henderson-Hasselbalch equation: $5.8 = pK_b + \log(10)$.Since $\log(10) = 1$,we have $5.8 = pK_b + 1$.Therefore,$pK_b = 5.8 - 1 = 4.8$.

When $30 \ mL$ of $0.2 \ M \ NH_4OH$ is added to $30 \ mL$ of $2 \ M \ NH_4Cl$ solution. If the $pH$ of the buffer formed is $8.2$,what is the $pK_b$ of $NH_4OH$?

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