Question

A 125.0 −mL sample of a solution that is 3.0×10^−3 M in AgNO3 is mixed with a 230.0 −mL sample of a solution that is 0.12 M in NaCN.

After the solution reaches equilibrium, what concentration of Ag+(aq) remains?

Answer 1

125.0 mL sample of a solution that is 3×10−3 M in AgNO3 is
(0.125 L ) (3×10−3 mol / L) = 0.000375 moles of Ag+



230.0 mL sample of a solution that is 0.12 M in NaCN is
(0.230 L) (0.12 M in NaCN) = 0.0276 moles of CN-


when 0.000375 moles of Ag+ is mixed with 0.0276 moles of CN-
we say that essentially all of the 0.000375 moles of Ag+ is converted to [Ag(CN)2]^-1

the 0.000375 moles of [Ag(CN)2]^-1 has been diluted to a total volume of 355.0 ml:
(0.000375 moles of [Ag(CN)2]^-1) / (0.3550L) = 0.00106 Molar [Ag(CN)2]^-


by the equation:
1 Ag+ & 2 CN^- --> [Ag+(CN^-)₂]^-
twice as much CN- is consumed, when it reacts with the 0.000375 moles of Ag+:
(0.0276 moles of CN-) - (2) (0.000375 moles lost) = 0.02685 mol CN- remains

which has also been diluted to 355.0 ml
(0.02685 mol CN- remains) / (0.355L) = 0.0756 Molar CN-

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

1 X 10^21 = [ 0.00106] / [Ag+] [0.0756]²

Ag = 1.855 X 10^-22 M