If pK_b for CN^{-} at 25,^{circ}C is 3.7,what | vedclass

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(A) For a salt of a weak acid and a strong base $(NaCN)$,the $pH$ is given by the formula: $pH = 7 + \frac{1}{2} pK_a + \frac{1}{2} \log C$.First,calc

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$12$

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(A) For a salt of a weak acid and a strong base $(NaCN)$,the $pH$ is given by the formula: $pH = 7 + \frac{1}{2} pK_a + \frac{1}{2} \log C$.First,calculate $pK_a$ for $HCN$: $pK_a = 14 - pK_b = 14 - 3.7 = 10.3$.Now,substitute the values into the $pH$ formula:$pH = 7 + \frac{1}{2}(10.3) + \frac{1}{2} \log(0.5)$.$pH = 7 + 5.15 + \frac{1}{2}(-0.301)$.$pH = 12.15 - 0.15 = 12$.

If $pK_b$ for $CN^{-}$ at $25\,^{\circ}C$ is $3.7$,what is the $pH$ of a $0.5\, M$ aqueous $NaCN$ solution?

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