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
    5 years ago
    Favorite Answer

    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.

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  • david
    Lv 7
    5 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?

    • LOVEBBB5 years agoReport

      Sorry! It was limiting my space so I had to shorten things. I want to know the concentration of Ag+ in a solution of equal Vol of the agno3 and NaCN. The kf is 3x10^20

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