Calculate the value of K in the following reaction: HCN (aq) + F- (aq) <-------> CN-...

Calculate the value of K in the following reaction:

HCN (aq) + F- (aq) <-------> CN- (aq) + HF (aq)

Ka for HCN = 6.2 x 10^-10 Ka for HF = 7.2 x 10^-4

Solutions

Expert Solution

HCN <=> H+ + CN- ( ka1 = 6.2*10^-10 )

HF <=> H+ + F- ( ka2 = 7.2*10^-4)

Reversing the reaction

H+ + F- <=> HF ( 1/ ka2 = 1388.89)

Adding the reaction

HCN + F- = CN- + HF

K(net) = k1 * 1/K2

= 6.2*10^-10 * 1388.89

= 8.611*10^-7

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queen_honey_blossom answered 1 year ago

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