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1) A solution that is 500 molar HCN, K.-6.2 x 10-10 is in equilibrium with solid silver (1) cyanide, Ksp - 2.2 x 10. However,
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Answer #1

HCN(aq) <----> H+(aq) + CN-(aq); Ka = 6.2*10-10

AgCN(s) <----> Ag+(aq) + CN-(aq); Ksp = 2.2*10-16

Ag+(aq) + 2CN-(aq) <----> [Ag(CN)2]-(aq); Kf = 1020.48 = 3.02*1020

Yes, there is a single equilibrium that describes all the above equilibria.

HCN(aq) + AgCN(s) <----> [Ag(CN)2]-(aq) + H+(aq); K = Ka * Ksp * Kf = 6.2*10-10 * 2.2*10-16 * 3.02*1020 = 4.12*10-5

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