Question

A 110.0 −mL sample of a solution that is 2.8×10^−3 M in AgNO3 is mixed with a 220.0 −mL sample of a solution that is 0.11 M in NaCN. A complex ion forms. After the solution reaches equilibrium, what concentration of Ag+(aq) remains?

Answer 1

110.0 mL sample of a solution that is 2.8×10−3 M in AgNO3 is
(0.110 L ) (2.8×10−3 mol / L) = 0.275x10^-3 moles of Ag+

220.0 mL sample of a solution that is 0.11 M in NaCN is
(0.220 L) (0.11 M in NaCN) = 0.0242 moles of CN-

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when 0.000275 moles of Ag+ is mixed with 0.0242 moles of CN-
we say that essentially all of the 0.000275 moles of Ag+ is converted to [Ag(CN)2]^-1

the 0.000275 moles of [Ag(CN)2]^-1 has been diluted to a total volume of 330.0 ml:
(0.000275 moles of [Ag(CN)2]^-1) / (0.3300L) = 0.000833 Molar [Ag(CN)2]^- = 8.33x10^-4 M

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by the equation:
1 Ag+ & 2 CN-1 --> [Ag+(CN-)2]^-1
twice as much CN- is consumed, when it reacts with the 0.000275 moles of Ag+:
(0.0242 moles of CN-) - (2) (0.000275 moles lost) = 0.02365 mol CN- remains

which has also been diluted to 330.0 ml
(0.02365 mol CN- remains) / (0.330L) = 0.07166 Molar CN-

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

1 X 10^21 = [0.000833] / [Ag+] [0.07166]^2

[Ag+] = [0.000833] / (1 X 10^21 ) [0.07166]^2

[Ag+] = [0.000833] / (1 X 10^21 ) (0.005136)

Ag = 1.6218 X 10^-22 Molar


however you might be expected to round that off to Ag = 1.6 X 10^-22 Molar
(since the Kf had only one sig fig showing in its 1x10^21)