Question

A 110.0 −mL sample of a solution that is 3.0×10−3 M in AgNO3 is mixed with a 220.0 −mL sample of a solution that is 0.14 M in NaCN. After the solution reaches equilibrium, what concentration of Ag+(aq) remains?

Answer 1

110.0 mL sample of a solution that is 3.0×10−3 M in AgNO3 is
(0.110 L ) (3.0×10−3 mol / L) = 0.00033 moles of Ag+



220.0 mL sample of a solution that is 0.14 M in NaCN is
(0.220 L) (0.14 M in NaCN) = 0.0308 moles of CN-


when 0.00033 moles of Ag+ is mixed with 0.0308 moles of CN-
then 0.00033 moles of Ag+ is converted to [Ag(CN)2]^-1

the 0.00033 moles of [Ag(CN)2]^-1 has been diluted to a total volume of 330.0 ml:
(0.00033 moles of [Ag(CN)2]^-1) / (0.330 L) = 0.001 Molar [Ag(CN)2]^-


by the equation:
1 Ag+ & 2 CN-1 --> [Ag+(CN-)2]^-1
twice as much CN- is consumed, when it reacts with the 0.00033 moles of Ag+:
(0.0308 moles of CN-) - (2) (0.00033 moles lost) = 0.03014 mol CN- remains

which has also been diluted to 330.0 ml
(0.03014 mol CN- remains) / (0.330 L) = 0.0 9133Molar CN-

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

1 X 10^21 = [0.001] / [Ag+] [0.09133]^2

[Ag+] = [0.001] / (1 X 10^21 ) [0.09133]^2

Ag = 1.198 X 10^-22 Molar