Question

A 110.0 −mL sample of a solution that is 2.7×10−3 M in AgNO3 is mixed with a 225.0 −mL sample of a solution that is 0.14 M in NaCN. A complex ion forms. After the solution reaches equilibrium, what concentration of Ag+(aq) remains?

Answer 1

we need formation constant of Ag(CN)2

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

Kf = 5.3 X 10^18

Now , Kf = [Ag(CN)2-] / [Ag+] [ CN-]^2

As Kf is very very high so the reaction will go towards formation of complex

As per stoichiometry , 1 mole of Ag+ will react with 2 moles of CN-

Moles of Ag+ = Molarity of AgNO3 X volume in litres = 2.7 X 10^-3 X 0.11 = 0.297 X 10^-3 moles

Moles of CN- added = Molarity of NaCN X volume in litres = 0.14 X 0.225 = 0.0315 CN-

For 0.297 X 10^-3 moles of Ag+ , 2 X 0.297 X 10^-3 moles of CN- = 0.594 X 10^-3 moles

so the Ag+ will be completely utilized, and there will be no Ag+ will left in the solution.