The pK_b of CN^{Theta} is 4.7. The pH of a so | vedclass

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(B) The mixture of a weak acid $(HCN)$ and its salt with a strong base $(KCN)$ forms an acidic buffer solution.For an acidic buffer,the $pH$ is given

Vedclass · Accepted Answer

$9.3$

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(B) The mixture of a weak acid $(HCN)$ and its salt with a strong base $(KCN)$ forms an acidic buffer solution.For an acidic buffer,the $pH$ is given by the Henderson-Hasselbalch equation:$pH = pK_a + \log \left(\frac{[Salt]}{[Acid]}\right)$Given $pK_b$ of $CN^{\Theta} = 4.7$,we calculate $pK_a$ of $HCN$:$pK_a = 14 - pK_b = 14 - 4.7 = 9.3$Since the number of moles of salt $(KCN)$ and acid $(HCN)$ are equal ($2.5 \ mol$ each),the ratio $\frac{[Salt]}{[Acid]} = 1$.Therefore,$pH = 9.3 + \log(1) = 9.3 + 0 = 9.3$.

The $pK_b$ of $CN^{\Theta}$ is $4.7$. The $pH$ of a solution prepared by mixing $2.5 \ mol$ of $KCN$ and $2.5 \ mol$ of $HCN$ in water and making the total volume up to $500 \ mL$ is:

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The $pH$ of a buffer solution made by mixing $25 \ mL$ of $0.02 \ M$ $NH_4OH$ and $25 \ mL$ of $0.2 \ M$ $NH_4Cl$ at $25^{\circ}C$ is ($pK_b$ of $NH_4OH = 4.8$).

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