## Organic Chemistry 9th Edition

$K_{a}$ of HCN is $4.8977$$\times$$10^{-10}$ or upon more simplification it is $4.90$$\times$$10^{-10}$
$pK_{a}$ = $-log(K_{a})$ $pK_{a}$ = $9.31$ $K_{a}$ = $antilog$ $(-pK_{a})$ $OR$ $K_{a}$ = $10^{(-pK_{a})}$ Upon calculating, $antilog$ $(-9.31)$ = $4.8977$$\times$$10^{-10}$