If the pK_b for CN^- is 4.7 at 25,^oC,then th | vedclass

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(C) $NaCN$ is a salt of a weak acid $(HCN)$ and a strong base $(NaOH)$.For the salt of a weak acid and strong base,the $pH$ is given by the formula:$p

Vedclass · Accepted Answer

$11.5$

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(C) $NaCN$ is a salt of a weak acid $(HCN)$ and a strong base $(NaOH)$.For the salt of a weak acid and strong base,the $pH$ is given by the formula:$pH = 7 + \frac{1}{2} pK_a + \frac{1}{2} \log C$Given $pK_b(CN^-) = 4.7$,we know $pK_a(HCN) = 14 - 4.7 = 9.3$.Concentration $C = 0.5\,M = 5 	imes 10^{-1}\,M$.Substituting the values:$pH = 7 + \frac{1}{2}(9.3) + \frac{1}{2} \log(5 	imes 10^{-1})$$pH = 7 + 4.65 + \frac{1}{2}(\log 5 + \log 10^{-1})$$pH = 11.65 + \frac{1}{2}(0.699 - 1)$$pH = 11.65 + \frac{1}{2}(-0.301)$$pH = 11.65 - 0.15 = 11.5$.

If the $pK_b$ for $CN^-$ is $4.7$ at $25\,^oC$,then the $pH$ of a $0.5\,M \,NaCN$ solution is:

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