NH_4CN is a salt of a weak acid HCN (K_a = 6. | vedclass

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(D) For a salt of a weak acid and a weak base,the $pH$ is given by the formula: $pH = 7 + \frac{1}{2}(pK_a - pK_b)$.First,calculate $pK_a$ and $pK_b$:

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weakly basic

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(D) For a salt of a weak acid and a weak base,the $pH$ is given by the formula: $pH = 7 + \frac{1}{2}(pK_a - pK_b)$.First,calculate $pK_a$ and $pK_b$:$pK_a = -\log(6.2 	imes 10^{-10}) \approx 9.21$$pK_b = -\log(1.8 	imes 10^{-5}) \approx 4.74$Now,substitute these values into the formula:$pH = 7 + \frac{1}{2}(9.21 - 4.74)$$pH = 7 + \frac{1}{2}(4.47)$$pH = 7 + 2.235 = 9.235$Since the $pH$ is greater than $7$ but close to neutral,the solution is weakly basic.

$NH_4CN$ is a salt of a weak acid $HCN$ $(K_a = 6.2 \times 10^{-10})$ and a weak base $NH_4OH$ $(K_b = 1.8 \times 10^{-5})$. $A$ $1 \ M$ solution of $NH_4CN$ will be:

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