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Home Q & A Board Science-Related Homework Help Chemistry (Moderator: Laser_3)
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softballmom softballmom
wrote...
12 years ago
Given the following acid dissociation constants
Ka (HF) = 7.2 x 10^-4
ka (HCN) = 4.0 x 10^-10

determine the equilibrium constant for the reaction below
HCN (aq) + F (aq) ------> CN(aq) + HF (aq)Leftwards Arrow-----
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wrote...
#1
12 years ago
HF <=> H+ + F-
Ka = [H+][F-]/ [HF]

HCN < => H+ + CN-
Ka = [H+][CN-]/ [HCN]

for the reaction
H+ + CN- <=> HCN
K =[HCN]/ [H+][CN-] = 1 / Ka = 2.5 x 10^9

K = [HCN][F-]/ [CN-][HF] = 2.5 x 10^9 x 7.2 x 10^-4 =1.8 x 10^6
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