? asked in Science & MathematicsChemistry · 6 years ago

# Ag+ reacts w/ cn- to form ag(cn)2 ^- kf=3*10^20 Conc of Ag+ in a solution of equal volumes of 2*10^-3M AgNO3 & .2M NaCN? please show how!?

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• Fern
Lv 7
6 years ago

Ag+ + 2CN- ==⇒ Ag(CN)2^-1

Assume you mix 1 liter of 2.0 x 0 10^-3 M AgNO3 with 1 liter of 0.20 M NaCN.

Moles of Ag+ = 2.0 x 10^-3 ; Moles of CN- = 0.20

Since Kf is very, very large, assume all the Ag+ reacts with CN- to form Ag(CN)2^-

Moles of CN- consumed = 0.002 moles of Ag+ x 2 moles CN- /1mole Ag+ =0.004

Moles of CN- remaining: 0.20 – 0.004 = 0.196

Moles of Ag(CN)2^-1 produced: 0.002 moles of Ag+ x 1 mole Ag(CN)2^-1 / 1 mole Ag+ = 0.002

Final Volume = 2.0 liters

[Ag(CN)2^-1] = 0.002 moles / 2 liters = 0.001

[CN-] = 0.196 / 2 liters = 0.098

Kf = [Ag(CN)2^-1] / [Ag+] [CN-]^2

3.0 x 10^20 = (0.001) / [Ag+] (0.098)^2

[Ag+] = 0.001 / (0.098)^2(3.0 x 10^20)

[Ag+] = 3.47 x 10^-22

The ratio of [Ag(CN)2^-1] / [CN-]^2 would remain the same no matter what volume of solutions were mixed.

• david
Lv 7
6 years ago

Your question is confusing. Try complete sentences. "Please show how!" What are you wanting someone to show you? Is this Ksp ... kf ??? what is kf?

AgCN is silver cyanide ... what is Ag(CN)2?? Are you talking about the formation of the complex ion [Ag(CN)2] - upon further addition of cyanide which cause the precipitate to dissolve?