The dissociation constant of HCN is 5 times 1 | vedclass

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Question

(C) The given solution is a buffer solution consisting of a weak acid $(HCN)$ and its salt with a strong base $(KCN)$.Using the Henderson-Hasselbalch

Vedclass · Accepted Answer

$8$

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(C) The given solution is a buffer solution consisting of a weak acid $(HCN)$ and its salt with a strong base $(KCN)$.Using the Henderson-Hasselbalch equation:$pH = pK_a + \log \frac{[Salt]}{[Acid]}$Given:$K_a = 5 	imes 10^{-10}$$pK_a = - \log(5 	imes 10^{-10}) = 10 - \log 5 = 10 - 0.698 = 9.302$Concentration of salt $(KCN)$ = $\frac{0.15 \, mole}{0.5 \, dm^3} = 0.3 \, M$Concentration of acid $(HCN)$ = $\frac{1.5 \, mole}{0.5 \, dm^3} = 3.0 \, M$$pH = 9.302 + \log \left( \frac{0.3}{3.0} ight)$$pH = 9.302 + \log(0.1)$$pH = 9.302 - 1 = 8.302$

The dissociation constant of $HCN$ is $5 \times 10^{-10}$. The $pH$ of the solution prepared by mixing $1.5 \, mole$ of $HCN$ and $0.15 \, mole$ of $KCN$ in water and making up the total volume to $0.5 \, dm^3$ is (in $.302$)

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