20 , mL of 0.2 , M , HCN is mixed with 10 , m | vedclass

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(C) Number of moles $=$ concentration $	imes$ volume.Moles of $NaOH = (0.2 \, M) 	imes (10 \, mL) = 2 \, 	ext{millimoles}$.Moles of $HCN = (0.2 \,

Vedclass · Accepted Answer

$5$

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(C) Number of moles $=$ concentration $	imes$ volume.Moles of $NaOH = (0.2 \, M) 	imes (10 \, mL) = 2 \, 	ext{millimoles}$.Moles of $HCN = (0.2 \, M) 	imes (20 \, mL) = 4 \, 	ext{millimoles}$.The reaction is: $HCN + NaOH \longrightarrow NaCN + H_2O$.Since $NaOH$ is the limiting reagent,$2 \, 	ext{millimoles}$ of $NaOH$ will react with $2 \, 	ext{millimoles}$ of $HCN$ to produce $2 \, 	ext{millimoles}$ of $NaCN$.Remaining $HCN = 4 - 2 = 2 \, 	ext{millimoles}$.This forms an acidic buffer solution. Using the Henderson-Hasselbalch equation: $pH = pKa + \log \frac{[Salt]}{[Acid]}$.Since the volume is the same for both,we can use the ratio of moles: $pH = 5 + \log \frac{2}{2} = 5 + \log(1) = 5 + 0 = 5$.Therefore,the $pH$ of the resulting mixture is $5$.

$20 \, mL$ of $0.2 \, M \, HCN$ is mixed with $10 \, mL$ of $0.2 \, M \, NaOH$. Calculate the $pH$ of the resulting mixture. The $pKa$ value of $HCN$ is $5$.

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