Organic Chemistry 9th Edition

Published by Brooks Cole
ISBN 10: 1305080483
ISBN 13: 978-1-30508-048-5

Chapter 2 - Polar Covalent Bonds; Acids and Bases - Problem: 16

Answer

$K_{a}$ of HCN is $4.8977$$\times$$10^{-10}$ or upon more simplification it is $4.90$$\times$$10^{-10}$

Work Step by Step

$pK_{a}$ = $-log(K_{a})$ $pK_{a}$ = $9.31$ $K_{a}$ = $antilog$ $(-pK_{a})$ $OR$ $K_{a}$ = $10^{(-pK_{a})}$ Upon calculating, $antilog$ $(-9.31)$ = $4.8977$$\times$$10^{-10}$
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