A 130.0-mL sample of a solution that is 3.0*10-3M in AgNO3 is mixed with a 225.0-mL sample of a solution that is 0.14M in NaCN.

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

130.0ml sample solution that is 3.0 * 10-3M in AgNo3 is (0.130L) (3.0*10-3mol/L) = 0.00039mols of Ag+

225.0ml sample in 0.14m Nacl

(0.225) (0.14M of NaCN) = 0.0315

0.0039 moles of [Ag (CN) 2]^-] is diluted in a total volume of 335.0ml:

(0.00039m) / (0.3550L) = 0.001099 molar[Ag (CN) 2]-

Ag + & 2 CN - 1 > [Ag + (CN-) 2] ^-1

twice as much CN - is consumed, when it reacts with 0.00039 moles of Ag+

(0.0315 moles of CN-) - (2) (0.0039 moles lost) = 0.03702

amount which has been diluted in 335.0ml

(0.03702 of CN-) / (0.355L) = 0.104 molar CN-

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

1*10^21=[0.001099]/[Ag+][0.104^2]

Ag+=[0.001099] / (1*10^21) (0.0108)

Ag+=1.18692E^26 molar