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What is the molar solubility of AgCl in 0.10 M NaCN if the colorless complex ion Ag(CN)2- forms? Ksp for AgCl is 1.8 x 10^-10 Kf for Ag(CN)2- is 1.0 x 10^21

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Answer 1

AgCl<==>Ag+ + Cl-

Ag+ + 2CN-<==>Ag(CN)2-

overall reaction is

AgCl+2CN-<==>Cl- + Ag(CN)2- K=Ksp*Kf=1.8x10^11

1.8x10^11=[Cl-][Ag(CN)2-]/[CN-]

=x/(0.10-2x)

x=0.050

Molar solubility of AgCl is 0.050 mol/L

[d] By: Guest
Date: Unknown---
Response
What is 1 + 100



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