What percentage of total cyanide (CN- +HCN) is present as HCN at a pH = 7?

Question

Question

Answer 1

Answer #1

The Reaction is HCN+H2O---> CN- +H3O+

Ka= 4.9X10-10

PH=7 , -log [H+] =7 [H+] =10^(-7) let [HCN]=M

Ka = 10^(-7)2/M M is molarity of HCN

M =10^(-7)2/ 4.9X10-10=2.04X10-5M=[HCN], [CN-] =10^(-7)

Percentage of HCN = 2.04X10-5*100/{10^(-7)+2.04X10-5}=99.5%

Answer 2

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