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.

1 answer:

8 0

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

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