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
